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http://www.ck12.org/tebook/Nanoleap-Teacher-Guide/r1/section/6.1/ | 1,481,420,723,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543614.1/warc/CC-MAIN-20161202170903-00430-ip-10-31-129-80.ec2.internal.warc.gz | 382,635,842 | 42,641 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.1: Investigating Static Forces in Nature: The Mystery of the Gecko
Difficulty Level: At Grade Created by: CK-12
Student Learning Objectives:
• Explain that a net force of zero or greater is necessary for objects to adhere to a surface (wall or ceiling)
• Identify different variables and the constants that affect adhesive forces
• Explain how the amount of adhesion changes when the conditions of the surfaces change
Note: Some questions in the Student Journal are underlined as formative assessment checkpoints for you to check students’ understanding of lesson objectives.
At a Glance for Teachers:
• Review what students know about forces
• Teacher demonstration on balanced forces
• Determine the amount of force needed for objects of varying masses to adhere to a ceiling and maintain a net force of zero
• Activity: Tape Pull—Measure the amount of force required to remove a piece of transparent tape with varying amounts of dirt
Estimated Time: 80Minutes\begin{align*}80 \;\mathrm{Minutes}\end{align*}
Vocabulary: Adhere, Adhesive, Balanced Forces, Dependent Variable, Force, Independent Variable, Mass, Net Force, Newton, Unbalanced Force, Volume
Refer to the end of this Teacher Guide for definitions.
Materials:
• PowerPoint for Lesson 6
• Student Journals for Lesson 6
• Computer with LCD or overhead projector
• Duct tape
• 50N\begin{align*}50\;\mathrm{N}\end{align*} spring scale
• Transparent tape
• Hole punch
• Ruler, protractor
Safety Note
Have students wear safety goggles in accordance with district safety policy.
Slide #
Student Journal Page #
Teacher Background Information and Pedagogy
“Teacher Script”
Slide 1
Title
Student Journal Page: 6–1
1. Review with students that a force is a push or pull. See definitions in Appendix B.
“What is the meaning of the word force in science?”
2. Demonstrate balanced forces by partially filling a small jar with water. Place an index card underneath the rim of the jar and invert the jar while holding the index card. Next, release your hand from the card while carefully holding the jar. Students should note that the card remains in contact with the rim of the jar. Have students identify the balanced forces at work in this demonstration. Answer: air pressure is the dominate force, it is equal to the weight of the water plus the force of gravity. Other forces that are present but less dominate are gravity and capillary wet adhesion (if the card got wet when it came in contact with the rim of the glass).
“Is this demonstrating a balanced or unbalanced force? Why?”
“What would happen if this was an unbalanced force?”
3. Have students look around the room and identify pairs of objects that are at rest and represent balanced forces and record these in the box on the left side of the student journal. Students should identify the forces acting on each object and that the net force is zero. Have students draw one of the examples of balanced forces and indicate the amount of each force acting on the object using arrows in Student Journal page 6–1. Have students repeat this exercise for unbalanced forces in the box on the right side of the student journal. Note to teacher: if the object sits on a table, there is the upward normal force of the table on the object. Research has shown that students often don’t recognize this as a force, they just indicate the table is in the way.]
4. Provide examples of objects in motion such as objects speeding up, slowing down, or at constant speed.
A field test teacher used a Frayer model for balanced and unbalanced forces for this lesson (refer to Lesson 1 for directions on the use of the Frayer model).
Slide 2
Student Journal Page: 6-1
5. Display Slide 2
“In this image, there are two forces at work: one that is holding the shoe onto the ceiling and another that is pulling the shoe towards the floor. In order for the shoe to remain on the ceiling, what must be true about these two forces?”
Students should state that these represent balanced forces (i.e., the net force is zero), or that the force holding a shoe is greater than the force of gravity. An example of the latter would be if the shoe contained a magnet that was attracted to a steel ceiling. This force could be greater than gravity, and then there would be an additional normal force acting in the direction of gravity to counteract the excess magnetic force.
Slide 3
Student Journal Page: 6–2
Calculations In Appendix A
“Imagine an ant, like the one on this slide, walking on the ceiling. Draw a picture representing the forces of the ant on the ceiling in your journal. Determine the force required for each ant foot (divide total force by six).”
Explain the following assumptions that are important for this problem:
“We are assuming in this problem that the total force required is equally divided among the six ant feet, and that ONLY the contact between feet and ceiling gives rise to the force.”
The weight of the ant is provided in Newtons (N), a derived unit which is the force needed to increase the speed of (or accelerate) one kilogram of mass one meter per second every second.
A field test teacher passed around objects (e.g., a one Newton weight, an eight Newton cell phone) for students to be able to relate to this unit of measure.
For this module, there is no need to calculate force with Newton’s Second Law of Motion. However, there may be a need to explain how an object’s weight can be expressed in Newtons. Explain that in the metric system forces are measured in units of Newtons (using the symbol “N”). Provide students with the definition found in Appendix B along with the following illustration. Use these along with the direct vocabulary instruction strategy as described in the preface. Weight is action of the force of gravity on an object. A standard kilogram mass would therefore have a weight of 9.8 Newtons on Earth since the acceleration due to gravity is 9.8 m/s/s.
6. Point out to students that the weight is the minimum amount of force that must be provided by the feet of the ant on the ceiling in order for there to be balanced forces and thus have the ant adhere to the ceiling. See Appendix A for the answers.
Note: During the pilot test, students thought this activity was interesting. The calculations took a bit to understand, and it was valuable to review unit conversions. Use Appendix A to assist students in solving the first problem.
Slide 4
Student Journal Page: 6–3
7. Display slide 4.
“Repeat the calculation—this time for an imaginary object that is larger in every dimension and whose mass and volume is ten times larger.”
Determine how many “ant feet” it would take for this imaginary object to remain adhered to the ceiling. Compare and discuss the difference between the two calculations in class. See Appendix A for calculations.
Teacher Demonstration:
Optional: One pilot teacher added a calculation for a two-ton elephant as well. Actual weight for an African male elephant in Newtons is 122,580 Newtons. Refer to optional notes in Slide 5.
Slide 5
Student Journal Pages: 6–3 6–4
“Let’s return our attention to the gecko.
Repeat your calculations from the imaginary animal for the Tokay Gecko, which has an average weight of 2.2 Newtons.”
8. Have students write a statement and/or draw pictures that describe the relationship between size (mass) and weight and, therefore, the adhesive forces required for an animal to remain on a ceiling.
Slide 6
9. Explain to students that they will be using the following terms in this lesson.
“Adhere describes how something sticks to something else.
Separation force is the amount of pull that is required to detach two objects.”
Slide 7
“What are the tools that we can use in the laboratory to measure the amount of force that an object exerts? What are the units used when measuring with this tool?”
Forces can be measured with a spring scale that changes when a force is applied. Forces are measured in Newtons (N).
Slide 8
Student Journal Page: 6-4
“As you have observed a gecko adhering to a wall, you may have wondered about the types of surfaces that are required to accomplish this feat.
Can the gecko adhere to any surface?
Does the surface need to be clean or can the gecko adhere to dirty surfaces too?
What if the surface is wet? Will that affect how well the gecko can adhere?
To better understand how the gecko can adhere to different surfaces, we will be exploring the forces involved in the adhesion of transparent tape on a table top.”
10. Tell students that over the next day or so, they will be able to refine this question based on how they set up their experiment.
11. Prior to showing slide 9 introduce the Tape Pull activity by having the students answer the question in their journal on page 6–4.
Slide 9
Student Journal Pages: 6–5 6–6
“You will be working with transparent tape on the tabletop and measuring the force required to remove the tape with different amounts of dirt. This force, as stated previously, is actually GREATER THAN the adhesive force.”
12. Before beginning the experiment have students work with the materials and practice the tape pull procedure as described on Student Journal page 6–5.
“Write down the independent variable (manipulated variable) and the dependent variable (responding variable).” Allow students to identify the amount of dirt as the independent variable and the force that it takes to remove or break the adhesion as the dependent variable.
Optional: Use the “sticky hands” toy (the one that initially sticks to glass then slowly falls/rolls down the glass) as a demonstration of dirt’s effect. This toy’s ability to stick decreases rapidly when it becomes dirty.
13. Hold a discussion about how to vary the “amount of dirt.” For starters, students could test fresh (never before used) tape. Then, rather than adding dirt to the tape, students could make a finger print on the tape and test its adhesion. Other ideas: drop chalk dust onto tape and blow it off, touch the tape to the floor, etc. This then becomes the operational definition for the independent variable.
14. Hold a class discussion about how to keep certain variables constant, such as the amount of surface area in which there is contact between surfaces and the angle of pull. Based on this discussion, students should write a research question and a hypothesis before completing the activity. An example of a research question is given on this slide.
Slide 10
Student Journal Pages: 6–6 6–7 6–8 6–9
“On this slide, you see how the materials are set up for the experiment. Image 6.8 shows a piece of tape on a table. The end of the tape that is pulled is reinforced with some electrical tape that has a hole punched through it. The hook end of the spring scale is then placed through the hole. Image 6.9 shows the spring scale being pulled at an angle (make sure this is the same each time). During the pull, a second student should carefully observe the force readings on the spring scale.”
15. Allow students time to complete the activity as shown in the journal. As students are completing the procedure, make sure they refine their initial question and use their findings in order to provide explanations and further questions.
Student Journal pages 6–5 through 6–8 can be completed for homework and graded for use as a formative assessment.
Classroom Management Tip: One pilot teacher assigned jobs for the experiment:
• Tape handler and assembly
• Measurer
• Equipment Manager
16. After students are done with the experiment, have them answer the questions in their journal on page 6–9 and 6–10.
Question 7:
Describe how you made your observations in today’s lesson.
a. “What tools did you use?” (spring scale)
b. “Were your observations at the visible or invisible scale?” (invisible)
c. “What is the dominant force at this scale?” (adhesive force/unknown)
Slide 11
“What do you know about the effectiveness of transparent tape underwater, and how tape gets dirty over time? (Display slide 11) This is a quote from researcher Kellar Autumn, Assistant Professor of biology at Lewis & Clark College, about the self-cleaning ability of the gecko.”
Students may state that when transparent tape is placed underwater, it will eventually lose its adhesiveness. Likewise, transparent tape does not work well on dirty surfaces.
17. It should be noted that ants leave a residue behind as they walk, whereas geckos do not.
18. Draw students’ attention to the note on the slide about the gecko adhesion working underwater.
Optional: Students could test other variables: amount of tape contact area, cleanliness of the surface, etc.
Slide 12
19. As a culminating class discussion, ask students to respond to the questions in “Making Connections.”
“Let’s review.
1. Describe one or two ideas that you learned during this lesson.
2. What factors contribute to the amount of force to remove a sticky substance?
3. How does dirt affect adhesion?
4. Do you think that a sticky substance is a possible method for the gecko adhesion?
5. How do you think the gecko sticks to the ceiling?
6. What should we explore next?”
Slide 13
20. The pilot-test teachers highly recommend using this flow chart at the end and/or beginning of each lesson. The end of each lesson contains this flow chart that provides an opportunity to show students the “big picture” and where they are in the lesson sequence. The following color code is used:
Yellow: Past Lessons
Blue: Current Lesson
Green: Next Lesson
White: Future Lesson
Appendix A: Calculations and Possible Responses to Accompany PowerPoint Slides
Slide 3 Calculations
Ant
Weight of Ant=0.00004Newtons\begin{align*}\mathrm{Weight\ of\ Ant} = 0.00004 \;\mathrm{Newtons}\end{align*} or 4×105Newtons\begin{align*}4 \times 10^{-5} \;\mathrm{Newtons}\end{align*}
Weight of Ant/6Ant Feet=Force for each foot=0.0000067Newtons\begin{align*}\mathrm{Weight\ of\ Ant}/6 \;\mathrm{Ant\ Feet} = \mathrm{Force\ for\ each\ foot} = 0.0000067 \;\mathrm{Newtons}\end{align*} per Ant Foot or 6.7×106Newtons\begin{align*}6.7 \times 10^{-6}\;\mathrm{Newtons}\end{align*} per Ant Foot
Slide 4 Calculations
Ant Mass Times 10times\begin{align*}10 \;\mathrm{times}\end{align*} what it was before
Then IF the Ant Foot can ONLY support 6.7×106N\begin{align*}6.7 \times 10^{-6}\;\mathrm{N}\end{align*}, how many ant feet would be required?
Weight of Imaginary object / Force for each Ant Foot
4×104Newtons/6.7×106Newtons per Ant Foot\begin{align*}4 \times 10^{-4} \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
59.7ant feet=60ant feet\begin{align*}59.7 \;\mathrm{ant\ feet} = 60 \;\mathrm{ant\ feet}\end{align*}
Slide 5 Calculations
Gecko
Weight of Gecko=2.2Newtons\begin{align*}\mathrm{Weight\ of\ Gecko} = 2.2 \;\mathrm{Newtons}\end{align*}
2.2Newtons/6.7×106Newtons per Ant Foot\begin{align*}2.2 \;\mathrm{Newtons}/6.7 \times 10^{-6}\;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
328,358ant feet\begin{align*}328,358 \;\mathrm{ant\ feet}\end{align*}
From Liang, Autumn, Hsieh, Zesch, Chan, Fearing, Full, Kenny5\begin{align*}^5\end{align*}:
43.4N\begin{align*}43.4 \;\mathrm{N}\end{align*} average sustained clinging force of gecko with 227.1mm2\begin{align*}227.1 \;\mathrm{mm}^2\end{align*} pad area
5\begin{align*}^5\end{align*} Autumn, K., Liang, Y. A., Hsieh, S. T., Zesch, W., Chan, W. P., Kenny, T. W., Fearing, R., & Full, R. J. (2000). Adhesive force of a single gecko foot-hair.
Nature, 405, 681-684.
Slide 5 Calculations (Optional)
200lb\begin{align*}200 \;\mathrm{lb}\end{align*} adult
Weight of adult in Newtons=888.9Newtons\begin{align*}\mathrm{Weight\ of\ adult\ in\ Newtons} = 888.9 \;\mathrm{Newtons}\end{align*}
888.9Newtons/6.7×106Newtons per Ant Foot\begin{align*}888.9 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
32,671,642ant feet\begin{align*}32,671,642 \;\mathrm{ant\ feet}\end{align*}
27,000lb\begin{align*}27,000 \;\mathrm{lb}\end{align*} African male elephant
Weight of elephant in Newtons=122,580Newtons\begin{align*}\mathrm{Weight\ of\ elephant\ in\ Newtons} = 122,580 \;\mathrm{Newtons}\end{align*}
122,580Newtons/6.7×106Newtons per Ant Foot\begin{align*}122,580 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*}
18,295,522,390ant feet\begin{align*}18,295,522,390 \;\mathrm{ant\ feet}\end{align*}
Appendix B: NanoLeap Physical Science Vocabulary
Adhere
1. To hold fast or to stick
2. To bind to
Adhesive
A substance that helps objects stick together
Balanced Forces
For each force acting on a body, there is another force on the same body equal in magnitude and opposite in direction. A body is said to be at rest if it is being acted on by balanced forces.
Dependent Variable
A factor or condition that might be affected as a result of a change in the independent variable (also called a responding variable)
Force
1. Energy exerted
2. A push or a pull that acts on an object
Independent Variable
A factor or condition that is intentionally changed by an investigator or experiment to explore its effects on other factors (also called a manipulated variable)
Mass
1. A quantity of matter
2. A measurement of the quantity
Net Force
The resultant non-zero force due to an unbalanced force
Newton
A unit of force needed to change the speed of a kilogram of mass by one meter per second for every second that the force is acting on the mass
Unbalanced
When there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction. A body is said to be in motion if acted upon by unbalanced forces.
Volume
The amount of space occupied by a three-dimensional object (Length times Width times Height for a rectangular object)
Investigating Static Forces in Nature: The Mystery of the Gecko
Lesson 6: How MUCH Force Is Needed to Make an Object Stick?
Teacher Guide
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Here | 4,547 | 18,659 | {"found_math": true, "script_math_tex": 25, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2016-50 | longest | en | 0.786544 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # 6.1: Investigating Static Forces in Nature: The Mystery of the Gecko Difficulty Level: At Grade Created by: CK-12 Student Learning Objectives: • Explain that a net force of zero or greater is necessary for objects to adhere to a surface (wall or ceiling) • Identify different variables and the constants that affect adhesive forces • Explain how the amount of adhesion changes when the conditions of the surfaces change Note: Some questions in the Student Journal are underlined as formative assessment checkpoints for you to check students’ understanding of lesson objectives. At a Glance for Teachers: • Review what students know about forces • Teacher demonstration on balanced forces • Determine the amount of force needed for objects of varying masses to adhere to a ceiling and maintain a net force of zero • Activity: Tape Pull—Measure the amount of force required to remove a piece of transparent tape with varying amounts of dirt Estimated Time: 80Minutes\begin{align*}80 \;\mathrm{Minutes}\end{align*} Vocabulary: Adhere, Adhesive, Balanced Forces, Dependent Variable, Force, Independent Variable, Mass, Net Force, Newton, Unbalanced Force, Volume Refer to the end of this Teacher Guide for definitions. Materials: • PowerPoint for Lesson 6 • Student Journals for Lesson 6 • Computer with LCD or overhead projector • Duct tape • 50N\begin{align*}50\;\mathrm{N}\end{align*} spring scale • Transparent tape • Hole punch • Ruler, protractor Safety Note Have students wear safety goggles in accordance with district safety policy. Slide # Student Journal Page # Teacher Background Information and Pedagogy “Teacher Script” Slide 1 Title Student Journal Page: 6–1 1. Review with students that a force is a push or pull. See definitions in Appendix B. “What is the meaning of the word force in science?” 2. Demonstrate balanced forces by partially filling a small jar with water. Place an index card underneath the rim of the jar and invert the jar while holding the index card. Next, release your hand from the card while carefully holding the jar. Students should note that the card remains in contact with the rim of the jar. Have students identify the balanced forces at work in this demonstration. Answer: air pressure is the dominate force, it is equal to the weight of the water plus the force of gravity. Other forces that are present but less dominate are gravity and capillary wet adhesion (if the card got wet when it came in contact with the rim of the glass). “Is this demonstrating a balanced or unbalanced force? Why?” “What would happen if this was an unbalanced force?” 3. Have students look around the room and identify pairs of objects that are at rest and represent balanced forces and record these in the box on the left side of the student journal. Students should identify the forces acting on each object and that the net force is zero. Have students draw one of the examples of balanced forces and indicate the amount of each force acting on the object using arrows in Student Journal page 6–1. Have students repeat this exercise for unbalanced forces in the box on the right side of the student journal. Note to teacher: if the object sits on a table, there is the upward normal force of the table on the object. Research has shown that students often don’t recognize this as a force, they just indicate the table is in the way.] 4. Provide examples of objects in motion such as objects speeding up, slowing down, or at constant speed. A field test teacher used a Frayer model for balanced and unbalanced forces for this lesson (refer to Lesson 1 for directions on the use of the Frayer model). Slide 2 Student Journal Page: 6-1 5. Display Slide 2 “In this image, there are two forces at work: one that is holding the shoe onto the ceiling and another that is pulling the shoe towards the floor. In order for the shoe to remain on the ceiling, what must be true about these two forces?” Students should state that these represent balanced forces (i.e., the net force is zero), or that the force holding a shoe is greater than the force of gravity. An example of the latter would be if the shoe contained a magnet that was attracted to a steel ceiling. This force could be greater than gravity, and then there would be an additional normal force acting in the direction of gravity to counteract the excess magnetic force. Slide 3 Student Journal Page: 6–2 Calculations In Appendix A “Imagine an ant, like the one on this slide, walking on the ceiling. Draw a picture representing the forces of the ant on the ceiling in your journal. Determine the force required for each ant foot (divide total force by six).” Explain the following assumptions that are important for this problem: “We are assuming in this problem that the total force required is equally divided among the six ant feet, and that ONLY the contact between feet and ceiling gives rise to the force.” The weight of the ant is provided in Newtons (N), a derived unit which is the force needed to increase the speed of (or accelerate) one kilogram of mass one meter per second every second. A field test teacher passed around objects (e.g., a one Newton weight, an eight Newton cell phone) for students to be able to relate to this unit of measure. For this module, there is no need to calculate force with Newton’s Second Law of Motion. However, there may be a need to explain how an object’s weight can be expressed in Newtons. Explain that in the metric system forces are measured in units of Newtons (using the symbol “N”). Provide students with the definition found in Appendix B along with the following illustration. Use these along with the direct vocabulary instruction strategy as described in the preface. Weight is action of the force of gravity on an object. A standard kilogram mass would therefore have a weight of 9.8 Newtons on Earth since the acceleration due to gravity is 9.8 m/s/s. 6. Point out to students that the weight is the minimum amount of force that must be provided by the feet of the ant on the ceiling in order for there to be balanced forces and thus have the ant adhere to the ceiling. See Appendix A for the answers. Note: During the pilot test, students thought this activity was interesting. The calculations took a bit to understand, and it was valuable to review unit conversions. Use Appendix A to assist students in solving the first problem. Slide 4 Student Journal Page: 6–3 7. Display slide 4. “Repeat the calculation—this time for an imaginary object that is larger in every dimension and whose mass and volume is ten times larger.” Determine how many “ant feet” it would take for this imaginary object to remain adhered to the ceiling. Compare and discuss the difference between the two calculations in class. See Appendix A for calculations. Teacher Demonstration: Optional: One pilot teacher added a calculation for a two-ton elephant as well. Actual weight for an African male elephant in Newtons is 122,580 Newtons. Refer to optional notes in Slide 5. Slide 5 Student Journal Pages: 6–3 6–4 “Let’s return our attention to the gecko. Repeat your calculations from the imaginary animal for the Tokay Gecko, which has an average weight of 2.2 Newtons.” 8. Have students write a statement and/or draw pictures that describe the relationship between size (mass) and weight and, therefore, the adhesive forces required for an animal to remain on a ceiling. Slide 6 9. Explain to students that they will be using the following terms in this lesson. “Adhere describes how something sticks to something else. Separation force is the amount of pull that is required to detach two objects.” Slide 7 “What are the tools that we can use in the laboratory to measure the amount of force that an object exerts? What are the units used when measuring with this tool?” Forces can be measured with a spring scale that changes when a force is applied. Forces are measured in Newtons (N). Slide 8 Student Journal Page: 6-4 “As you have observed a gecko adhering to a wall, you may have wondered about the types of surfaces that are required to accomplish this feat. Can the gecko adhere to any surface? Does the surface need to be clean or can the gecko adhere to dirty surfaces too? What if the surface is wet? Will that affect how well the gecko can adhere? To better understand how the gecko can adhere to different surfaces, we will be exploring the forces involved in the adhesion of transparent tape on a table top.” 10. Tell students that over the next day or so, they will be able to refine this question based on how they set up their experiment. 11. Prior to showing slide 9 introduce the Tape Pull activity by having the students answer the question in their journal on page 6–4. Slide 9 Student Journal Pages: 6–5 6–6 “You will be working with transparent tape on the tabletop and measuring the force required to remove the tape with different amounts of dirt. This force, as stated previously, is actually GREATER THAN the adhesive force.” 12. Before beginning the experiment have students work with the materials and practice the tape pull procedure as described on Student Journal page 6–5. “Write down the independent variable (manipulated variable) and the dependent variable (responding variable).” Allow students to identify the amount of dirt as the independent variable and the force that it takes to remove or break the adhesion as the dependent variable. Optional: Use the “sticky hands” toy (the one that initially sticks to glass then slowly falls/rolls down the glass) as a demonstration of dirt’s effect. This toy’s ability to stick decreases rapidly when it becomes dirty. 13. Hold a discussion about how to vary the “amount of dirt.” For starters, students could test fresh (never before used) tape. Then, rather than adding dirt to the tape, students could make a finger print on the tape and test its adhesion. Other ideas: drop chalk dust onto tape and blow it off, touch the tape to the floor, etc. This then becomes the operational definition for the independent variable. 14. Hold a class discussion about how to keep certain variables constant, such as the amount of surface area in which there is contact between surfaces and the angle of pull. Based on this discussion, students should write a research question and a hypothesis before completing the activity. An example of a research question is given on this slide. Slide 10 Student Journal Pages: 6–6 6–7 6–8 6–9 “On this slide, you see how the materials are set up for the experiment. Image 6.8 shows a piece of tape on a table. The end of the tape that is pulled is reinforced with some electrical tape that has a hole punched through it. The hook end of the spring scale is then placed through the hole. Image 6.9 shows the spring scale being pulled at an angle (make sure this is the same each time). During the pull, a second student should carefully observe the force readings on the spring scale.” 15. Allow students time to complete the activity as shown in the journal. As students are completing the procedure, make sure they refine their initial question and use their findings in order to provide explanations and further questions. Student Journal pages 6–5 through 6–8 can be completed for homework and graded for use as a formative assessment. Classroom Management Tip: One pilot teacher assigned jobs for the experiment: • Tape handler and assembly • Measurer • Equipment Manager 16. After students are done with the experiment, have them answer the questions in their journal on page 6–9 and 6–10. Question 7: Describe how you made your observations in today’s lesson. a. “What tools did you use?” (spring scale) b. “Were your observations at the visible or invisible scale?” (invisible) c. “What is the dominant force at this scale?” (adhesive force/unknown) Slide 11 “What do you know about the effectiveness of transparent tape underwater, and how tape gets dirty over time? (Display slide 11) This is a quote from researcher Kellar Autumn, Assistant Professor of biology at Lewis & Clark College, about the self-cleaning ability of the gecko.” Students may state that when transparent tape is placed underwater, it will eventually lose its adhesiveness. Likewise, transparent tape does not work well on dirty surfaces. 17. It should be noted that ants leave a residue behind as they walk, whereas geckos do not. 18. Draw students’ attention to the note on the slide about the gecko adhesion working underwater. Optional: Students could test other variables: amount of tape contact area, cleanliness of the surface, etc. Slide 12 19. As a culminating class discussion, ask students to respond to the questions in “Making Connections.” “Let’s review. 1. Describe one or two ideas that you learned during this lesson. 2. What factors contribute to the amount of force to remove a sticky substance? 3. How does dirt affect adhesion? 4. Do you think that a sticky substance is a possible method for the gecko adhesion? 5. How do you think the gecko sticks to the ceiling? 6. What should we explore next?” Slide 13 20. The pilot-test teachers highly recommend using this flow chart at the end and/or beginning of each lesson. The end of each lesson contains this flow chart that provides an opportunity to show students the “big picture” and where they are in the lesson sequence. The following color code is used: Yellow: Past Lessons Blue: Current Lesson Green: Next Lesson White: Future Lesson Appendix A: Calculations and Possible Responses to Accompany PowerPoint Slides Slide 3 Calculations Ant Weight of Ant=0.00004Newtons\begin{align*}\mathrm{Weight\ of\ Ant} = 0.00004 \;\mathrm{Newtons}\end{align*} or 4×105Newtons\begin{align*}4 \times 10^{-5} \;\mathrm{Newtons}\end{align*} Weight of Ant/6Ant Feet=Force for each foot=0.0000067Newtons\begin{align*}\mathrm{Weight\ of\ Ant}/6 \;\mathrm{Ant\ Feet} = \mathrm{Force\ for\ each\ foot} = 0.0000067 \;\mathrm{Newtons}\end{align*} per Ant Foot or 6.7×106Newtons\begin{align*}6.7 \times 10^{-6}\;\mathrm{Newtons}\end{align*} per Ant Foot Slide 4 Calculations Ant Mass Times 10times\begin{align*}10 \;\mathrm{times}\end{align*} what it was before Then IF the Ant Foot can ONLY support 6.7×106N\begin{align*}6.7 \times 10^{-6}\;\mathrm{N}\end{align*}, how many ant feet would be required? Weight of Imaginary object / Force for each Ant Foot 4×104Newtons/6.7×106Newtons per Ant Foot\begin{align*}4 \times 10^{-4} \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*} 59.7ant feet=60ant feet\begin{align*}59.7 \;\mathrm{ant\ feet} = 60 \;\mathrm{ant\ feet}\end{align*} Slide 5 Calculations Gecko Weight of Gecko=2.2Newtons\begin{align*}\mathrm{Weight\ of\ Gecko} = 2.2 \;\mathrm{Newtons}\end{align*} 2.2Newtons/6.7×106Newtons per Ant Foot\begin{align*}2.2 \;\mathrm{Newtons}/6.7 \times 10^{-6}\;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*} 328,358ant feet\begin{align*}328,358 \;\mathrm{ant\ feet}\end{align*} From Liang, Autumn, Hsieh, Zesch, Chan, Fearing, Full, Kenny5\begin{align*}^5\end{align*}: 43.4N\begin{align*}43.4 \;\mathrm{N}\end{align*} average sustained clinging force of gecko with 227.1mm2\begin{align*}227.1 \;\mathrm{mm}^2\end{align*} pad area 5\begin{align*}^5\end{align*} Autumn, K., Liang, Y. A., Hsieh, S. T., Zesch, W., Chan, W. P., Kenny, T. W., Fearing, R., & Full, R. J. (2000). Adhesive force of a single gecko foot-hair. Nature, 405, 681-684. Slide 5 Calculations (Optional) 200lb\begin{align*}200 \;\mathrm{lb}\end{align*} adult Weight of adult in Newtons=888.9Newtons\begin{align*}\mathrm{Weight\ of\ adult\ in\ Newtons} = 888.9 \;\mathrm{Newtons}\end{align*} 888.9Newtons/6.7×106Newtons per Ant Foot\begin{align*}888.9 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*} 32,671,642ant feet\begin{align*}32,671,642 \;\mathrm{ant\ feet}\end{align*} 27,000lb\begin{align*}27,000 \;\mathrm{lb}\end{align*} African male elephant Weight of elephant in Newtons=122,580Newtons\begin{align*}\mathrm{Weight\ of\ elephant\ in\ Newtons} = 122,580 \;\mathrm{Newtons}\end{align*} 122,580Newtons/6.7×106Newtons per Ant Foot\begin{align*}122,580 \;\mathrm{Newtons}/6.7 \times 10^{-6} \;\mathrm{Newtons\ per\ Ant\ Foot}\end{align*} 18,295,522,390ant feet\begin{align*}18,295,522,390 \;\mathrm{ant\ feet}\end{align*} Appendix B: NanoLeap Physical Science Vocabulary Adhere 1. To hold fast or to stick 2. To bind to Adhesive A substance that helps objects stick together Balanced Forces For each force acting on a body, there is another force on the same body equal in magnitude and opposite in direction. A body is said to be at rest if it is being | acted on by balanced forces. Dependent Variable A factor or condition that might be affected as a result of a change in the independent variable (also called a responding variable) Force 1. Energy exerted 2. A push or a pull that acts on an object Independent Variable A factor or condition that is intentionally changed by an investigator or experiment to explore its effects on other factors (also called a manipulated variable) Mass 1. A quantity of matter 2. A measurement of the quantity Net Force The resultant non-zero force due to an unbalanced force Newton A unit of force needed to change the speed of a kilogram of mass by one meter per second for every second that the force is acting on the mass Unbalanced When there is an individual force that is not being balanced by a force of equal magnitude and in the opposite direction. A body is said to be in motion if acted upon by unbalanced forces. Volume The amount of space occupied by a three-dimensional object (Length times Width times Height for a rectangular object) Investigating Static Forces in Nature: The Mystery of the Gecko Lesson 6: How MUCH Force Is Needed to Make an Object Stick? Teacher Guide © 2009 McREL ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Please to create your own Highlights / Notes Show More ### Image Attributions Show Hide Details Description Authors: Tags: Subjects: Grades: Date Created: Feb 23, 2012 Last Modified: Apr 29, 2014 Save or share your relevant files like activites, homework and worksheet. To add resources, you must be the owner of the section. Click Customize to make your own copy. Please wait... Please wait... Image Detail Sizes: Medium | Original CK.SCI.ENG.TE.1.Nanotechnology-NanoLeap.6.1 Here |
https://lifewithdata.com/2023/06/20/leetcode-maximum-product-of-three-numbers-solution-in-python/ | 1,695,938,697,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00082.warc.gz | 400,888,868 | 25,282 | # Leetcode – Maximum Product of Three Numbers Solution in Python
## Introduction
In this article, we will explore a classic problem from Leetcode known as “Maximum Product of Three Numbers”. This problem tests our understanding of array manipulation, mathematical properties, and algorithm optimization. We will discuss the problem statement, explore multiple approaches to solve it, and analyze their complexity.
## Problem Statement (Leetcode #628)
Given an integer array, find three numbers whose product is maximum and output the maximum product.
### Example:
Input: [1,2,3,4] Output: 24
### Note:
The length of the given array will be in range [3, 10^4] and all elements are in the range [-10^3, 10^3].
## Solution Approaches
### 1. Brute Force Approach
The most straightforward approach is to check the product of every possible triplet in the array.
#### Algorithm:
1. Iterate through each element, a, in the array.
2. For each a, iterate through each element, b, after a.
3. For each pair a and b, iterate through each element, c, after b.
4. For each triplet a, b, and c, calculate their product and keep track of the maximum product.
def maximum_product(nums):
max_product = float('-inf')
n = len(nums)
for i in range(n):
for j in range(i + 1, n):
for k in range(j + 1, n):
max_product = max(max_product, nums[i] * nums[j] * nums[k])
return max_product
#### Time Complexity:
• O(n^3), as there are three nested loops iterating through the array.
#### Space Complexity:
• O(1), as we are only using a constant amount of extra space.
### 2. Sorting Approach
An optimized approach involves sorting the array. Once the array is sorted, the maximum product can be the product of the last three elements (in case they are all positive or all negative) or the product of the first two elements and the last element (in case the first two elements are negative).
#### Algorithm:
1. Sort the array.
2. Calculate the product of the last three elements.
3. Calculate the product of the first two elements and the last element.
4. Return the maximum of the two products.
def maximum_product(nums):
nums.sort()
return max(nums[-1] * nums[-2] * nums[-3], nums[0] * nums[1] * nums[-1])
#### Time Complexity:
• O(n log n), as the most time-consuming operation is sorting the array.
#### Space Complexity:
• O(1), as we are using a constant amount of extra space.
### 3. Linear Scan Approach
We can optimize the solution further by realizing that we only need the three largest numbers and the two smallest numbers. We can find these five numbers in a single linear scan.
#### Algorithm:
1. Initialize variables to keep track of the three largest numbers (max1, max2, max3) and the two smallest numbers (min1, min2).
2. Iterate through each element in the array, updating the maximum and minimum values.
3. Return the maximum between the product of the three largest numbers and the product of the two smallest numbers and the largest number.
def maximum_product(nums):
max1 = max2 = max3 = float('-inf')
min1 = min2 = float('inf')
for num in nums:
if num > max1:
max3, max2, max1 = max2, max1, num
elif num > max2:
max3, max2 = max2, num
elif num > max3:
max3 = num
if num < min1:
min2, min1 = min1, num
elif num < min2:
min2 = num
return max(max1 * max2 * max3, max1 * min1 * min2)
#### Time Complexity:
• O(n), as we are iterating through the array once.
#### Space Complexity:
• O(1), as we are using a constant amount of extra space.
## Conclusion
In this article, we delved into the “Maximum Product of Three Numbers” problem on LeetCode and examined three different solutions in Python. The brute-force approach, though simple, is inefficient. By understanding the mathematical properties of the problem, we improved the performance using sorting. Finally, we achieved the most optimized solution through a single linear scan. | 925 | 3,867 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-40 | latest | en | 0.83488 | # Leetcode – Maximum Product of Three Numbers Solution in Python ## Introduction In this article, we will explore a classic problem from Leetcode known as “Maximum Product of Three Numbers”. This problem tests our understanding of array manipulation, mathematical properties, and algorithm optimization. We will discuss the problem statement, explore multiple approaches to solve it, and analyze their complexity. ## Problem Statement (Leetcode #628) Given an integer array, find three numbers whose product is maximum and output the maximum product. ### Example: Input: [1,2,3,4] Output: 24 ### Note: The length of the given array will be in range [3, 10^4] and all elements are in the range [-10^3, 10^3]. ## Solution Approaches ### 1. Brute Force Approach The most straightforward approach is to check the product of every possible triplet in the array. #### Algorithm: 1. Iterate through each element, a, in the array. 2. For each a, iterate through each element, b, after a. 3. For each pair a and b, iterate through each element, c, after b. 4. For each triplet a, b, and c, calculate their product and keep track of the maximum product. def maximum_product(nums): max_product = float('-inf') n = len(nums) for i in range(n): for j in range(i + 1, n): for k in range(j + 1, n): max_product = max(max_product, nums[i] * nums[j] * nums[k]) return max_product #### Time Complexity: • O(n^3), as there are three nested loops iterating through the array. #### Space Complexity: • O(1), as we are only using a constant amount of extra space. ### 2. Sorting Approach An optimized approach involves sorting the array. Once the array is sorted, the maximum product can be the product of the last three elements (in case they are all positive or all negative) or the product of the first two elements and the last element (in case the first two elements are negative). #### Algorithm: 1. Sort the array. 2. Calculate the product of the last three elements. 3. Calculate the product of the first two elements and the last element. 4. Return the maximum of the two products. def maximum_product(nums): nums.sort() return max(nums[-1] * nums[-2] * nums[-3], nums[0] * nums[1] * nums[-1]) #### Time Complexity: • O(n log n), as the most time-consuming operation is sorting the array. #### Space Complexity: • O(1), as we are using a constant amount of extra space. ### 3. Linear Scan Approach We can optimize the solution further by realizing that we only need the three largest numbers and the two smallest numbers. We can find these five numbers in a single linear scan. #### Algorithm: 1. Initialize variables to keep track of the three largest numbers (max1, max2, max3) and the two smallest numbers (min1, min2). 2. Iterate through each element in the array, updating the maximum and minimum values. 3. Return the maximum between the product of the three largest numbers and the product of the two smallest numbers and the largest number. def maximum_product(nums): max1 = max2 = max3 = float('-inf') min1 = min2 = float('inf') for num in nums: if num > max1: max3, max2, max1 = max2, max1, num elif num > max2: max3, max2 = max2, num elif num > max3: max3 = num if num < min1: min2, min1 = min1, num elif num < min2: min2 = num return max(max1 * max2 * max3, max1 * min1 * min2) #### Time Complexity: • O(n), as we are iterating through the array once. #### Space Complexity: • O(1), as we | are using a constant amount of extra space. ## Conclusion In this article, we delved into the “Maximum Product of Three Numbers” problem on LeetCode and examined three different solutions in Python. The brute-force approach, though simple, is inefficient. By understanding the mathematical properties of the problem, we improved the performance using sorting. Finally, we achieved the most optimized solution through a single linear scan. |
https://en.wikibooks.org/wiki/Modular_Arithmetic/What_is_a_Modulus%3F | 1,716,234,099,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058293.53/warc/CC-MAIN-20240520173148-20240520203148-00718.warc.gz | 188,231,083 | 12,127 | Modular Arithmetic/What is a Modulus?
Modular Arithmetic What is a Modulus? Modular Arithmetic →
In modular arithmetic, 38 can equal 14 — what??
You might be wondering how so! (Or you might already know how so, but we will assume that you don't.) Well, modular arithmetic works as follows:
Think about military time which ranges from 0000 to 2359. For the sake of this lesson, we will only consider the hour portion, so let us consider how hourly time works from 0 to 23. After a 24 hour period, the time restarts at 0 and builds again to 23. So, using this reasoning, the 27th hour would be equivalent to the third hour in military time, as the remainder when 27 is divided by 24 is 3 (make sure you understand why this is true as it is vital to understanding everything that follows in this book). So to get any hour into military time we just find the number's remainder when divided by 24. What we really are doing is finding that number modulo 24.
To write the earlier example mathematically, we would write:
${\displaystyle 27\equiv 3{\pmod {24}}}$
which reads as "27 is congruent to 3 modulo 24". Let us try one more example before setting you loose: what is 34 congruent to modulo 4, which can also be written as "34 (mod 4)" which is read as "what is 34 congruent to modulo 4?". We find the remainder when 34 is divided by 4; the remainder is 2. So, ${\displaystyle 34\equiv 2{\pmod {4}}}$.
Now, try some on your own:
Exercises: Definition of Modular Arithmetic Replace the '?' by the correct value. ${\displaystyle 25\equiv {\text{?}}{\pmod {7}}}$ ${\displaystyle 10^{10}\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {729}}}$
Challenge: Definition of Modular Arithmetic These two are a lot harder. Simplify: ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {11}}}$ You do not need to work out what (936 + 3) is to solve these, but they will require some calculation. | 547 | 1,984 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 7, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-22 | latest | en | 0.916799 | Modular Arithmetic/What is a Modulus? Modular Arithmetic What is a Modulus? Modular Arithmetic → In modular arithmetic, 38 can equal 14 — what?? You might be wondering how so! (Or you might already know how so, but we will assume that you don't.) Well, modular arithmetic works as follows: Think about military time which ranges from 0000 to 2359. For the sake of this lesson, we will only consider the hour portion, so let us consider how hourly time works from 0 to 23. After a 24 hour period, the time restarts at 0 and builds again to 23. So, using this reasoning, the 27th hour would be equivalent to the third hour in military time, as the remainder when 27 is divided by 24 is 3 (make sure you understand why this is true as it is vital to understanding everything that follows in this book). So to get any hour into military time we just find the number's remainder when divided by 24. What we really are doing is finding that number modulo 24. To write the earlier example mathematically, we would write: ${\displaystyle 27\equiv 3{\pmod {24}}}$ which reads as "27 is congruent to 3 modulo 24". Let us try one more example before setting you loose: what is 34 congruent to modulo 4, which can also be written as "34 (mod 4)" which is read as "what is 34 congruent to modulo 4?". We find the remainder when 34 is divided by 4; the remainder is 2. So, ${\displaystyle 34\equiv 2{\pmod {4}}}$. Now, try some on your own: Exercises: Definition of Modular Arithmetic Replace the '?' by the correct value. ${\displaystyle 25\equiv {\text{?}}{\pmod {7}}}$ ${\displaystyle 10^{10}\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {729}}}$ Challenge: Definition of Modular Arithmetic These two are | a lot harder. Simplify: ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {10}}}$ ${\displaystyle (9^{36}+3)\equiv {\text{?}}{\pmod {11}}}$ You do not need to work out what (936 + 3) is to solve these, but they will require some calculation. |
https://forum.mrmoneymustache.com/ask-a-mustachian/401k-math-test!/ | 1,709,078,705,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00026.warc.gz | 260,587,439 | 6,914 | ### Author Topic: 401k Math Test! (Read 2347 times)
#### tcbonline
• Posts: 1
##### 401k Math Test!
« on: July 25, 2016, 10:35:08 AM »
Somebody please help me do a calculation example using this forumla below. I'm a bit perplexed by it. When I asked a potential employer what they matched for 401k this is what they wrote back...
"Company matches 100% for the 1st percent and then matches 50% for 2-6% for a total match of 3.5%"
I'm getting hung up on the 2-6% part. I'm more used to seeing company matching stated as something like 50% at 3.5% ... simple, right? I don't get all the 2-6% part at all. Should I ignore that and just figure the 3.5% total?
So let's take some concrete numbers to help me understand this:
if you take a yearly salary of 100k, say I contribute 6% myself at \$6000. What would be the company match contribution for the year given the formula above?
Sorry if I'm totally brain dead, let me know if you guys can figure out what this means better than me.
Thanks,
tcb
#### jorjor
• Bristles
• Posts: 351
##### Re: 401k Math Test!
« Reply #1 on: July 25, 2016, 10:44:36 AM »
If someone contributes 1%, the company matches that 1%. If someone contributes 2%, the company matches the first percent and half of the second percent, for a total of 1.5%. And so on, up to 6%. For someone who contributes 6%, the company would do a full match of the first percent and half of the rest up to 6%, for a total of 3.5% contribution by the company.
#### RWD
• Walrus Stache
• Posts: 6455
• Location: Arizona
##### Re: 401k Math Test!
« Reply #2 on: July 25, 2016, 10:44:51 AM »
So let's take some concrete numbers to help me understand this:
if you take a yearly salary of 100k, say I contribute 6% myself at \$6000. What would be the company match contribution for the year given the formula above?
In this scenario your company would contribute \$3500. (\$1000 to match your first 1% and \$2500 to half match your next 5%). This lines up with the 3.5% number too.
#### desk_jockey
• CM*MW 2023 Attendees
• Bristles
• Posts: 326
##### Re: 401k Math Test!
« Reply #3 on: July 25, 2016, 10:55:59 AM »
So let's take some concrete numbers to help me understand this:
if you take a yearly salary of 100k, say I contribute 6% myself at \$6000. What would be the company match contribution for the year given the formula above?
In this scenario your company would contribute \$3500. (\$1000 to match your first 1% and \$2500 to half match your next 5%). This lines up with the 3.5% number too.
And if you contribute the 18% so as to put the maximum of your allowed portion in 2016 (i.e. \$18,000) the company will still contribute \$3500.
#### Elle 8
• Stubble
• Posts: 167
##### Re: 401k Math Test!
« Reply #4 on: July 25, 2016, 10:59:49 AM »
The first percent is fully matched (1%). Then the next 2-6% are matched at 50%. So, for example, if you were to contribute 6%, you'll get 1% for your first 1% plus 2.5% which is 50% your next 5%. So 3.5% total. Here's a little chart showing your contribution and what you would get for the first 1%, then the next 2-6% and your total match.
Cont First % Next 2-6% Total Match
1% 1 0 1.0%
2% 1 .5 1.5%
3% 1 1.0 2.0%
4% 1 1.5 2.5%
5% 1 2.0 3.0%
6%+ 1 2.5 3.5% | 1,063 | 3,432 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2024-10 | latest | en | 0.891562 | ### Author Topic: 401k Math Test! (Read 2347 times) #### tcbonline • Posts: 1 ##### 401k Math Test! « on: July 25, 2016, 10:35:08 AM » Somebody please help me do a calculation example using this forumla below. I'm a bit perplexed by it. When I asked a potential employer what they matched for 401k this is what they wrote back... "Company matches 100% for the 1st percent and then matches 50% for 2-6% for a total match of 3.5%" I'm getting hung up on the 2-6% part. I'm more used to seeing company matching stated as something like 50% at 3.5% ... simple, right? I don't get all the 2-6% part at all. Should I ignore that and just figure the 3.5% total? So let's take some concrete numbers to help me understand this: if you take a yearly salary of 100k, say I contribute 6% myself at \$6000. What would be the company match contribution for the year given the formula above? Sorry if I'm totally brain dead, let me know if you guys can figure out what this means better than me. Thanks, tcb #### jorjor • Bristles • Posts: 351 ##### Re: 401k Math Test! « Reply #1 on: July 25, 2016, 10:44:36 AM » If someone contributes 1%, the company matches that 1%. If someone contributes 2%, the company matches the first percent and half of the second percent, for a total of 1.5%. And so on, up to 6%. For someone who contributes 6%, the company would do a full match of the first percent and half of the rest up to 6%, for a total of 3.5% contribution by the company. #### RWD • Walrus Stache • Posts: 6455 • Location: Arizona ##### Re: 401k Math Test! « Reply #2 on: July 25, 2016, 10:44:51 AM » So let's take some concrete numbers to help me understand this: if you take a yearly salary of 100k, say I contribute 6% myself at \$6000. What would be the company match contribution for the year given the formula above? In this scenario your company would contribute \$3500. (\$1000 to match your first 1% and \$2500 to half match your next 5%). This lines up with the 3.5% number too. #### desk_jockey • CM*MW 2023 Attendees • Bristles • Posts: 326 ##### Re: 401k Math Test! « Reply #3 on: July 25, 2016, 10:55:59 AM » So let's take some concrete numbers to help me understand this: if you take a yearly salary of 100k, say I contribute 6% myself at \$6000. What would be the company match contribution for the year given the formula above? In this scenario your company would contribute \$3500. (\$1000 to match your first 1% and \$2500 to half match your next 5%). This lines up with the 3.5% number too. And if you contribute the 18% so as to put the maximum of your allowed portion in 2016 (i.e. \$18,000) the company will still contribute \$3500. #### Elle 8 • Stubble • Posts: 167 ##### Re: 401k Math Test! « Reply #4 on: July 25, 2016, 10:59:49 AM » The first percent is fully matched (1%). Then the next 2-6% are matched at 50%. So, for example, if you were to contribute 6%, you'll get 1% for your first 1% plus 2.5% which is | 50% your next 5%. So 3.5% total. Here's a little chart showing your contribution and what you would get for the first 1%, then the next 2-6% and your total match. Cont First % Next 2-6% Total Match 1% 1 0 1.0% 2% 1 .5 1.5% 3% 1 1.0 2.0% 4% 1 1.5 2.5% 5% 1 2.0 3.0% 6%+ 1 2.5 3.5% |
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# Completing the square - GCSE Maths (MathsWatch) watch
1. Currently on MathsWatch doing some completing the square questions and it is asking me to explain how I know x^2 - 6x + 12 is always positive.
I've got absolutely no clue on how to answer this, and if you've used MathsWatch before you'll know how difficult it is to actually get marked right when doing explanation questions.
Any help will be appreciated!
Question attached as an image.
2. Image of question
Attached Images
3. anything squared is always positive or equal to zero, and if you add 3 to it it must be above zero, so positive.
4. (Original post by Angel_Chen)
anything squared is always positive or equal to zero, and if you add 3 to it it must be above zero, so positive.
Thank you!! Makes sense and gave me the marks
5. (Original post by Milli22)
Thank you!! Makes sense and gave me the marks
Np
6. I would explain using the minimum. we that completing the square for f (x) directly shows us the maximum or minimum value of f (x) and the value of x for which this minimum occurs. for our graphs we already know that we have a minimum point since a>0 and from completing the square we know the minimum point of f (x) is 3 hence all other values of f (x) will be 3 or greater than 3 which makes all values positive. so I would write "the minimum value of 3 so f (x) cannot go below 3 hence all values are positive"
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 645 | 2,739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-13 | latest | en | 0.913044 | Hey there! Sign in to join this conversationNew here? Join for free x Turn on thread page Beta You are Here: Home >< Maths # Completing the square - GCSE Maths (MathsWatch) watch 1. Currently on MathsWatch doing some completing the square questions and it is asking me to explain how I know x^2 - 6x + 12 is always positive. I've got absolutely no clue on how to answer this, and if you've used MathsWatch before you'll know how difficult it is to actually get marked right when doing explanation questions. Any help will be appreciated! Question attached as an image. 2. Image of question Attached Images 3. anything squared is always positive or equal to zero, and if you add 3 to it it must be above zero, so positive. 4. (Original post by Angel_Chen) anything squared is always positive or equal to zero, and if you add 3 to it it must be above zero, so positive. Thank you!! Makes sense and gave me the marks 5. (Original post by Milli22) Thank you!! Makes sense and gave me the marks Np 6. I would explain using the minimum. we that completing the square for f (x) directly shows us the maximum or minimum value of f (x) and the value of x for which this minimum occurs. for our graphs we already know that we have a minimum point since a>0 and from completing the square we know the minimum point of f (x) is 3 hence all other values of f (x) will be 3 or greater than 3 which makes all values positive. so I would write "the minimum value of 3 so f (x) cannot go below 3 hence all values are positive" Posted on the TSR App. Download from Apple or Google Play Reply Submit reply Turn on thread page Beta TSR Support Team We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out. This forum is supported by: Updated: February 11, 2018 Today on TSR ### Four things top students are doing Over the Easter break Poll Useful resources ## Make your revision easier ### Maths Forum posting guidelines Not sure where to post? Read the updated guidelines here ### How to use LaTex Writing equations the easy way ### Study habits of A* students Top tips from students who have already aced their exams Can you help? Study help unanswered threads ## Groups associated with this forum: View associated groups The Student Room, Get Revising and Marked by Teachers are trading names of The | Student Room Group Ltd. Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE Write a reply... Reply Hide Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. |
https://www.jiskha.com/display.cgi?id=1347476627 | 1,501,268,810,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550977093.96/warc/CC-MAIN-20170728183650-20170728203650-00452.warc.gz | 779,635,132 | 3,983 | # Physics
posted by .
If your car accelerates from rest at a steady rate of 4 m/s2, how soon will it reach 84 km/h
• Physics -
V = 81km/h = 81000m/3600s = 22.5 m/s.
V = Vo + at = 22.5 m/s.
0 + 4*t = 22.5
t = 5.625 s.
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Post a New Question | 646 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2017-30 | longest | en | 0.899063 | # Physics posted by . If your car accelerates from rest at a steady rate of 4 m/s2, how soon will it reach 84 km/h • Physics - V = 81km/h = 81000m/3600s = 22.5 m/s. V = Vo + at = 22.5 m/s. 0 + 4*t = 22.5 t = 5.625 s. ## Similar Questions 1. ### physics A car and a motorcycle start from rest at the same time on a straight section of the road. However, the motorcycle starts 29.6 m behind the car. The car accelerates at a uniform rate of 3.9 m/s^2 while the motorcycle accelerates at … 2. ### Physics A car starts from rest and accelerates at a constant rate of 1.1 m/s2. If the radius of the car wheels is 29 cm, and the final angular speed of the car wheel is 48 rad/s, how far does the car travel 3. ### Math A refrigerated truck leaves a rest stop traveling at a steady rate of 56 mi/h. A car leaves the same rest stop 1/4 h later followinf rhe truck at a steady rate of 64 mi/h. How long after the truck leaves the rest stop wil the car overtake … 4. ### physics If your car accelerates from rest at a steady rate of 4 m/s2, how soon will it reach 81 km/h (50.3 mph or 22.5 m/s)? 5. ### physics a speeder is going at a constant speed of 20m/s when a police car accelerates from rest at a rate of 4.5,/ssquared to catch them. How long did it take for the officer to reach the speeding car? 6. ### physics(Quick!) if a car accelerates from rest at a constant 5.5 m/s^2, how long will it take for the car to reach a velocity of 28 m/s 7. ### physics A car accelerates from rest at a steady rate of 1 ms-2. Calculate: (a) the time taken to reach 15 ms-1 (b) the distance travelled during this time (c) the velocity of the car when it was 100 m from the start point 8. ### physics A car accelerates from rest at a steady rate of 1 ms-2. Calculate: (a) the time taken to reach 15 ms-1 (b) the distance travelled during this time (c) the velocity of the car when it was 100 m from the start point 9. ### Physics If your car accelerates from rest at a steady rate | of 4 m/s2, how soon will it reach 84 km/h 10. ### Science A car of mass 1200kg accelerates from rest at a rate of 2m/s^-2. Calculate how long it should take to reach a velocity of 50m/s. More Similar Questions Post a New Question |
http://www.jiskha.com/display.cgi?id=1319416343 | 1,495,935,221,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463609404.11/warc/CC-MAIN-20170528004908-20170528024908-00123.warc.gz | 675,741,156 | 4,021 | # Physics
posted by on .
A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65kg and its length is 0.55m. What is the rotational energy of the blade at its operating angular speed of 3450 rpm? If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
• Physics - ,
Use the rotational energy equation, E = 0.5Iω², where I is the moment of inertia for a long thin rod with an axis through it's midpoint (I = (1/12)mL²).
E = 0.5Iω²
E = 0.5 [1/12)mL²] ω²
E = (1/24) mL² ω²
Find ω by multiplying the rpm by 2π and then dividing by 60 to put units into rad/sec (or multiply ω by π/30)
3450 * 2π = 21 676.9... rad/min
(21 676.9... rad/min)/60 = 361.28... rad/sec
Insert into energy eqn and solve
E = (1/24) mL² (361.28...rad/s)²
E = (1/24)(0.65kg)(0.55m²)(361.28...rad/s)²
E = ~1069 J
### Answer This Question
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Post a New Question | 334 | 1,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-22 | latest | en | 0.903085 | # Physics posted by on . A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65kg and its length is 0.55m. What is the rotational energy of the blade at its operating angular speed of 3450 rpm? If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise? • Physics - , Use the rotational energy equation, E = 0.5Iω², where I is the moment of inertia for a long thin rod with an axis through it's midpoint (I = (1/12)mL²). E = 0.5Iω² E = 0.5 [1/12)mL²] ω² E = (1/24) mL² ω² Find ω by multiplying the rpm by 2π and then dividing by 60 to put units into rad/sec (or multiply ω by π/30) 3450 * 2π = 21 676.9... rad/min (21 676.9... rad/min)/60 = 361.28... rad/sec Insert into energy eqn and solve E = (1/24) mL² (361.28...rad/s)² E = (1/24)(0.65kg)(0.55m²)(361.28...rad/s)² E = ~1069 | J ### Answer This Question First Name: School Subject: Answer: ### Related Questions More Related Questions Post a New Question |
https://es.mathworks.com/matlabcentral/answers/883904-how-to-calculate-the-standard-error-estimation-when-using-fit-from-curve-fitting-toolbox | 1,632,705,118,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058222.43/warc/CC-MAIN-20210926235727-20210927025727-00311.warc.gz | 295,104,473 | 24,929 | # How to calculate the standard error estimation when using fit from curve fitting toolbox?
10 views (last 30 days)
Albert Bing on 22 Jul 2021
Answered: Star Strider on 22 Jul 2021
Is is possible to calculate the standard error estimation when using fit from curve fitting toolbox as in polyfit?
Suppose I have 2 vector (x, y). Using polyfit and polyval gives the standard error estimation for all predictions.
How to calculate delta in fit? I need the prediction interval like examples below.
I assume the delta in polyval is not a scalar but varies with x. (Purhaps it is not?)
Example from the documention,
x = 1:100;
y = -0.3*x + 2*randn(1,100);
[p,S] = polyfit(x,y,1);
[y_fit,delta] = polyval(p,x,S);
plot(x,y,'bo')
hold on
plot(x,y_fit,'r-')
plot(x,y_fit+2*delta,'m--',x,y_fit-2*delta,'m--')
title('Linear Fit of Data with 95% Prediction Interval')
legend('Data','Linear Fit','95% Prediction Interval')
Star Strider on 22 Jul 2021
Yes. Use the predint function.
x = linspace(0, 100, 100);
y = -0.3*x + 2*randn(1,100);
[f,gof,out] = fit(x(:), y(:), 'poly1')
f =
Linear model Poly1: f(x) = p1*x + p2 Coefficients (with 95% confidence bounds): p1 = -0.2946 (-0.3081, -0.2811) p2 = -0.5298 (-1.311, 0.251)
gof = struct with fields:
sse: 384.9559 rsquare: 0.9504 dfe: 98 adjrsquare: 0.9499 rmse: 1.9819
out = struct with fields:
numobs: 100 numparam: 2 residuals: [100×1 double] Jacobian: [100×2 double] exitflag: 1 algorithm: 'QR factorization and solve' iterations: 1
ci = predint(f, x);
figure
plot(f, x, y)
hold on
plot(x, ci, '--')
hold off
grid
hl = legend;
hl.String{3} = 'Lower 95% CI';
hl.String{4} = 'Upper 95% CI';
.
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Translated by | 573 | 1,773 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-39 | latest | en | 0.727539 | # How to calculate the standard error estimation when using fit from curve fitting toolbox? 10 views (last 30 days) Albert Bing on 22 Jul 2021 Answered: Star Strider on 22 Jul 2021 Is is possible to calculate the standard error estimation when using fit from curve fitting toolbox as in polyfit? Suppose I have 2 vector (x, y). Using polyfit and polyval gives the standard error estimation for all predictions. How to calculate delta in fit? I need the prediction interval like examples below. I assume the delta in polyval is not a scalar but varies with x. (Purhaps it is not?) Example from the documention, x = 1:100; y = -0.3*x + 2*randn(1,100); [p,S] = polyfit(x,y,1); [y_fit,delta] = polyval(p,x,S); plot(x,y,'bo') hold on plot(x,y_fit,'r-') plot(x,y_fit+2*delta,'m--',x,y_fit-2*delta,'m--') title('Linear Fit of Data with 95% Prediction Interval') legend('Data','Linear Fit','95% Prediction Interval') Star Strider on 22 Jul 2021 Yes. Use the predint function. x = linspace(0, 100, 100); y = -0.3*x + 2*randn(1,100); [f,gof,out] = fit(x(:), y(:), 'poly1') f = Linear model Poly1: f(x) = p1*x + p2 Coefficients (with 95% confidence bounds): p1 = -0.2946 (-0.3081, -0.2811) p2 = -0.5298 (-1.311, 0.251) gof = struct with fields: sse: 384.9559 rsquare: 0.9504 dfe: 98 adjrsquare: 0.9499 rmse: 1.9819 out = struct with fields: numobs: 100 numparam: 2 residuals: [100×1 double] Jacobian: [100×2 double] exitflag: 1 algorithm: 'QR factorization and solve' iterations: 1 ci = predint(f, x); figure plot(f, x, y) hold on plot(x, ci, '--') hold off grid hl = legend; hl.String{3} = 'Lower 95% CI'; | hl.String{4} = 'Upper 95% CI'; . ### Community Treasure Hunt Find the treasures in MATLAB Central and discover how the community can help you! Start Hunting! Translated by |
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Nova's DS prep guide. Jumps to defining an equation: Interme
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Nova's DS prep guide. Jumps to defining an equation: Interme [#permalink]
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08 May 2011, 02:34
I laid my hands on Nova's DS prep Course. and started seeing some places where a mathematical equation was given straight away without the intermediate steps. Does any one know how I can fill in the blanks in the following content of that guide?
The author
Every time x increases (decreases) by a particular value, y decreases (increases) by two timesthat value (this is the system description). What is the value of x when y is 3?(1) When y is 1, x = 3.(2) When y is 2, x = 1.• Statements (1) and (2) above are constraints that limit the system. For example, the shape of thehidden system in the problem (question setup) is x + 2y = c. So, the system blah blah...
Speaking intuitively too, the relation between x and y appears to be positive monotonous and hence the line will have a positive slope. This is the first proof of contradiction for the equation above.
To be slightly more mathematical, if (x1,y1) and (x2,y2) are two points that follow the above rule,then (y2-y1) = 2(x2-x1) and indirectly this means that the equation for the line is y = 2x + k for some real number k.
How does the author arrive at the equation (which I highlighted in italic bold above? The main problem is that my equation has a slope that is inverse to that of the author's claim. But as the author does not show the intermediate steps, this part of the book defeats me.
Any help?
Regards
Rahul
_________________
Regards
Rahul
Nova's DS prep guide. Jumps to defining an equation: Interme [#permalink] 08 May 2011, 02:34
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6 Quadratic equations in DS problems 6 19 Nov 2011, 07:05
equations 8 27 Mar 2011, 08:14
Display posts from previous: Sort by | 778 | 2,871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2016-40 | latest | en | 0.902283 | Find all School-related info fast with the new School-Specific MBA Forum It is currently 26 Sep 2016, 04:27 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History Events & Promotions Events & Promotions in June Open Detailed Calendar Nova's DS prep guide. Jumps to defining an equation: Interme Author Message TAGS: Hide Tags Manager Status: 700 (q47,v40); AWA 6.0 Joined: 16 Mar 2011 Posts: 82 GMAT 1: 700 Q47 V40 Followers: 1 Kudos [?]: 92 [0], given: 3 Nova's DS prep guide. Jumps to defining an equation: Interme [#permalink] Show Tags 08 May 2011, 02:34 I laid my hands on Nova's DS prep Course. and started seeing some places where a mathematical equation was given straight away without the intermediate steps. Does any one know how I can fill in the blanks in the following content of that guide? The author Every time x increases (decreases) by a particular value, y decreases (increases) by two timesthat value (this is the system description). What is the value of x when y is 3?(1) When y is 1, x = 3.(2) When y is 2, x = 1.• Statements (1) and (2) above are constraints that limit the system. For example, the shape of thehidden system in the problem (question setup) is x + 2y = c. So, the system blah blah... Speaking intuitively too, the relation between x and y appears to be positive monotonous and hence the line will have a positive slope. This is the first proof of contradiction for the equation above. To be slightly more mathematical, if (x1,y1) and (x2,y2) are two points that follow the above rule,then (y2-y1) = 2(x2-x1) and indirectly this means that the equation for the line is y = 2x + k for some real number k. How does the author arrive at the equation (which I highlighted in italic bold above? The main problem is that my equation has a slope that is inverse to that of the author's claim. But as the author does not show the intermediate steps, this part of the book defeats me. Any help? Regards Rahul _________________ Regards Rahul Nova's DS prep guide. Jumps to defining an equation: Interme [#permalink] 08 May 2011, 02:34 Similar topics Replies Last post Similar Topics: 4 All GMAT Prep DS questions 8 29 | Oct 2014, 07:46 Official Guide 12ed DS Explanations 1 16 Feb 2014, 16:29 103 Consolodited QUANT GUIDES of Forum Most Helpful in Preps 25 17 Apr 2013, 12:57 6 Quadratic equations in DS problems 6 19 Nov 2011, 07:05 equations 8 27 Mar 2011, 08:14 Display posts from previous: Sort by |
https://neacollege.com/confidence-intervals-essay/ | 1,721,789,828,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00641.warc.gz | 363,542,046 | 21,515 | # Confidence Intervals Essay
While hypothesized, high cholesterol levels in children can cause their children staying affected with hyperlipidemia.
Remember: This is just a sample from a fellow student. Your time is important. Let us write you an essay from scratch
Research is carried out to approximate the indicate cholesterol in children between the ages of 2 – six years of age. It also attempted to set up a correlation for the effect family history has on the onset of the disease. From info collected since shown in Table one of the spreadsheet attached, a sample size of 9 (n=9) participants enrolled in the study. Total cholesterol amounts measured in children between ages two – 6 years was summarized at you, 765.
The sample indicate (X) and standard deviation (S) calculated as (1765/90) =196. 1 and rectangular root summation (X-X) rectangular / n-1 =29. zero respectively. Now to generate a 95% assurance interval pertaining to the true imply total hypercholesteria levels in children from data gathered, we used the z value to get 95% as (z= 1 ) 96). From sample stats the confidence interval for 95% calculated from the solution is (196. 1 +/- 1 . ninety six X 29/3) we now have 196.
1 +/-19. 0. At this point, by adding and subtracting the margin of error, we certainly have (215. you, 177. 1) respectively. A spot estimate pertaining to the true indicate cholesterol levels in the populace is 196.
1 and this we are 95% confident that the true imply is among 215. you and 177. 1 . The margin or perhaps is significant because of the little sample size. A pilot study with 10 members was executed to assess how systolic blood pressure changes overtime if still left. Clinical trial compared fresh medication created to lower to this of a placebo.
The test mean (X) for the in bloodstream pressures more than a four weeks period is computed as (9/10) =0. 09 or 90%. Also, the normal deviation pertaining to the difference over the same a month period computed as [(X) 2/square root n-1] = 183/20. 33= 4. your five, and.
Z . value =1. 96. The confident span (CI) intended for 95% difference in blood pressures by sample info collected as seen in stand 2 on the spreadsheet attached is calculated as [(0. 9 +/- 1 ) 96 (4.
5)/3. 162 = 0. 9 +/- 2 . 78. By adding and subtracting the margin of error, we have (3. 68, -1. 88) respectively.
The point estimation for the actual difference in blood pressures over a 4 weeks period inside the population is definitely 0. being unfaithful and we are 95% confident that the accurate mean intended for the difference in blood pressures over a month is between 3. sixty-eight and -1. 88. Now a main trial is carried out with 200 patients signed up and at random assigned to either one from the two groups: experimental medicine group or placebo group. At the end with the six weeks treatment period the information displayed equally groups while assigned in table 3 on the schedule attached.
Both the comparison teams physically segregated where 75 patients had been assigned to receive the trial and error drug and another 100 patients to obtain the placebo. We initial compute the descriptive statistics on each in the of the two samples my spouse and i. e. sample size, imply, and normal deviation denoted as n1, X1, s1 for test 1 and n2, X2, s2 to get sample two respectively. The purpose estimate intended for the difference in the population are the differences in the sample means indicated as (X1-X2) and square root of 1/n1 + 1/n2 as the pooled estimate of the prevalent standard deviation (Sp). Sp = sq . root (n1-1) s1 square + (n2-1) square / n1 + n2 -2; thus Sp = seventeen.
07. Considering the fact that Sp (17. 07), test deviation (-11. 2) and z worth at 95% (1. 96). we now figure out the 95% confident interval (CI) pertaining to the difference in mean systolic blood pressure among groups is usually (-6.
52, -15. 8). The CI is construed as follows: Were 95% confident that that the difference inside the mean systolic blood demands between pool one and group two can be between -6. 52 and -15.
8 respectively.
## Related essay
Category: Writing
Words: 631
Views: 1659
f42bd64bfbd359e7 | 1,028 | 4,174 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-30 | latest | en | 0.927844 | # Confidence Intervals Essay While hypothesized, high cholesterol levels in children can cause their children staying affected with hyperlipidemia. Remember: This is just a sample from a fellow student. Your time is important. Let us write you an essay from scratch Research is carried out to approximate the indicate cholesterol in children between the ages of 2 – six years of age. It also attempted to set up a correlation for the effect family history has on the onset of the disease. From info collected since shown in Table one of the spreadsheet attached, a sample size of 9 (n=9) participants enrolled in the study. Total cholesterol amounts measured in children between ages two – 6 years was summarized at you, 765. The sample indicate (X) and standard deviation (S) calculated as (1765/90) =196. 1 and rectangular root summation (X-X) rectangular / n-1 =29. zero respectively. Now to generate a 95% assurance interval pertaining to the true imply total hypercholesteria levels in children from data gathered, we used the z value to get 95% as (z= 1 ) 96). From sample stats the confidence interval for 95% calculated from the solution is (196. 1 +/- 1 . ninety six X 29/3) we now have 196. 1 +/-19. 0. At this point, by adding and subtracting the margin of error, we certainly have (215. you, 177. 1) respectively. A spot estimate pertaining to the true indicate cholesterol levels in the populace is 196. 1 and this we are 95% confident that the true imply is among 215. you and 177. 1 . The margin or perhaps is significant because of the little sample size. A pilot study with 10 members was executed to assess how systolic blood pressure changes overtime if still left. Clinical trial compared fresh medication created to lower to this of a placebo. The test mean (X) for the in bloodstream pressures more than a four weeks period is computed as (9/10) =0. 09 or 90%. Also, the normal deviation pertaining to the difference over the same a month period computed as [(X) 2/square root n-1] = 183/20. 33= 4. your five, and. Z . value =1. 96. The confident span (CI) intended for 95% difference in blood pressures by sample info collected as seen in stand 2 on the spreadsheet attached is calculated as [(0. 9 +/- 1 ) 96 (4. 5)/3. 162 = 0. 9 +/- 2 . 78. By adding and subtracting the margin of error, we have (3. 68, -1. 88) respectively. The point estimation for the actual difference in blood pressures over a 4 weeks period inside the population is definitely 0. being unfaithful and we are 95% confident that the accurate mean intended for the difference in blood pressures over a month is between 3. sixty-eight and -1. 88. Now a main trial is carried out with 200 patients signed up and at random assigned to either one from the two groups: experimental medicine group or placebo group. At the end with the six weeks treatment period the information displayed equally groups while assigned in table 3 on the schedule attached. Both the comparison teams physically segregated where 75 patients had been assigned to receive the trial and error drug and another 100 patients to obtain the placebo. We initial compute the descriptive statistics on each in the of the two samples my spouse and i. e. sample size, imply, and normal deviation denoted as n1, X1, s1 for test 1 and n2, X2, s2 to get sample two respectively. The purpose estimate intended for the difference in the population are the differences in the sample means indicated as (X1-X2) and square root of 1/n1 + 1/n2 as the pooled estimate of the prevalent standard deviation (Sp). Sp = sq . root (n1-1) s1 square + (n2-1) square / n1 + n2 -2; thus Sp = seventeen. 07. Considering the fact that Sp (17. 07), test deviation (-11. 2) and z worth | at 95% (1. 96). we now figure out the 95% confident interval (CI) pertaining to the difference in mean systolic blood pressure among groups is usually (-6. 52, -15. 8). The CI is construed as follows: Were 95% confident that that the difference inside the mean systolic blood demands between pool one and group two can be between -6. 52 and -15. 8 respectively. ## Related essay Category: Writing Words: 631 Views: 1659 f42bd64bfbd359e7 |
https://www.physicsforums.com/threads/bead-on-a-spinning-hoop.575903/ | 1,527,303,758,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867277.64/warc/CC-MAIN-20180526014543-20180526034543-00532.warc.gz | 814,769,868 | 19,213 | # Homework Help: Bead on a Spinning Hoop
1. Feb 9, 2012
### zaper
A bead of mass m can slide on a frictionless circular hoop in the vertical plane with radius R. When the hoop spins around the vertical axis with a rate of ω, the bead moves up the hoop by an angle θ which depends on the angular velocity of the hoop. (Pic attached)
a) Find θ in terms of ω, R, m and g.
b) If there is friction between the bead and hoop given by μ, within what range can ω change without changing θ?
For part a what I have so far is that arad=R*ω2. The centripetal force (F) then is m*arad and the normal force is F/sinθ.
The weight needs to cancel out the normal force then so the weight in the direction of the normal force is mg*cosθ = F/sinθ
Am I anywhere close to being right?
File size:
35.9 KB
Views:
581
2. Feb 9, 2012
### tiny-tim
hi zaper!
no
keep it simple
do F = ma vertically to find N
then do F= ma horizontally to find ω
(and remember that the centripetal acceleration is horizontal, so use the correct r )
3. Feb 9, 2012
### zaper
Hey there tiny tim!
Sorry, but I'm really struggling to keep up with this problem so forgive me for anything dumb that I do here.
So you said to find N as F=ma vertically so that means that N=mg*cosθ
By you're second part are you saying m*arad=m*ω2*r (r being R*sinθ)
4. Feb 9, 2012
### tiny-tim
hey there zaper!
that's not in the vertical direction, is it?
no, that's ma = ma, you need F = ma
5. Feb 9, 2012
### zaper
Ok so then N=mg*sinθ I assume. Could you please explain why this is and what exactly you want for the second equation. I'm not really following you here.
6. Feb 9, 2012
### tiny-tim
no
start again …
Fx = max
Fy = may
what are ax and ay ?
7. Feb 9, 2012
### zaper
I'm really really sorry. I guess I'm way overthinking this or something. Is ax the radial acceleration?
8. Feb 9, 2012
### tiny-tim
a is the acceleration
ax and ay are its x and y components
(i'm assuming you're using y for up and x for radial)
9. Feb 9, 2012
### zaper
Ok so ax is the radial acceleration and the total acceleration is pointing up and to the left. The triangle formed by the accelerations has top angle θ and ay=ax/tanθ. Is this correct? Also does ay actually describe something here or is it just a component of the total a?
10. Feb 9, 2012
### tiny-tim
what??
why?
11. Feb 9, 2012
### zaper
Wait, I mean to the right
12. Feb 10, 2012
### tiny-tim
do you mean up and to the right, or only to the right?
13. Feb 10, 2012
### abhinav111
hi zapper i have an easily understandable solution, if u have not come to the answer i may help.
14. Feb 10, 2012
### zaper
Up and to the right tiny tim and what's your idea abhinav?
15. Feb 10, 2012
### tiny-tim
(along the normal?)
nooo
i think you're confusing physics and maths
force is physics
acceleration is maths (geometry)
if you know how the object is moving, then the acceleration follows automatically as a matter of maths
it doesn't matter what forces are producing that motion, the acceleration depends only on the motion, not on the forces
in this case, in equilibrium, the bead is moving uniformly in a horizontal circle …
so what is its acceleration?
16. Feb 10, 2012
### zaper
Since there's no angular acceleration then the only acceleration is radial right?
17. Feb 10, 2012
### tiny-tim
yes, but you're looking at the wrong radius …
the circle to look at is the circle of the bead's motion, which is horizontal, isn't it?
18. Feb 10, 2012
### zaper
Yes. That last post was supposed to be in terms of the horizontal circle. Sorry that I forgot to explain that
19. Feb 10, 2012
### tiny-tim
ok, so now do Ftotal = ma in both the horizontal and vertical directions
20. Feb 10, 2012
### zaper
Ok I think we're getting back to my point of confusion from before. We've established that the only acceleration in the horizontal circle is the radial acceleration so I'm not quite seeing where this vertical force can come from unless you mean the weight
21. Feb 10, 2012
### tiny-tim
yes, the weight is vertical
but the bead has no vertical acceleration
so what's the other vertical force (that makes F=ma)?
22. Feb 10, 2012
### zaper
The normal force? That's about all that's left
23. Feb 10, 2012
### tiny-tim
yes, N
N and mg are the only two forces on the bead
so their components in the vertical direction have to equal the component of ma in the vertical direction,
and their components in the horizontal direction have to equal the component of ma in the horizontal direction
24. Feb 10, 2012
### zaper
Ok so N is the force pulling to the center of the hoop. Would it's x component have the force m*arad? Also the weight obviously has no x component so the only x force is from N.
25. Feb 10, 2012
### tiny-tim
that's right!
so the x and y equations for F = ma are … ? | 1,340 | 4,860 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2018-22 | latest | en | 0.91831 | # Homework Help: Bead on a Spinning Hoop 1. Feb 9, 2012 ### zaper A bead of mass m can slide on a frictionless circular hoop in the vertical plane with radius R. When the hoop spins around the vertical axis with a rate of ω, the bead moves up the hoop by an angle θ which depends on the angular velocity of the hoop. (Pic attached) a) Find θ in terms of ω, R, m and g. b) If there is friction between the bead and hoop given by μ, within what range can ω change without changing θ? For part a what I have so far is that arad=R*ω2. The centripetal force (F) then is m*arad and the normal force is F/sinθ. The weight needs to cancel out the normal force then so the weight in the direction of the normal force is mg*cosθ = F/sinθ Am I anywhere close to being right? File size: 35.9 KB Views: 581 2. Feb 9, 2012 ### tiny-tim hi zaper! no keep it simple do F = ma vertically to find N then do F= ma horizontally to find ω (and remember that the centripetal acceleration is horizontal, so use the correct r ) 3. Feb 9, 2012 ### zaper Hey there tiny tim! Sorry, but I'm really struggling to keep up with this problem so forgive me for anything dumb that I do here. So you said to find N as F=ma vertically so that means that N=mg*cosθ By you're second part are you saying m*arad=m*ω2*r (r being R*sinθ) 4. Feb 9, 2012 ### tiny-tim hey there zaper! that's not in the vertical direction, is it? no, that's ma = ma, you need F = ma 5. Feb 9, 2012 ### zaper Ok so then N=mg*sinθ I assume. Could you please explain why this is and what exactly you want for the second equation. I'm not really following you here. 6. Feb 9, 2012 ### tiny-tim no start again … Fx = max Fy = may what are ax and ay ? 7. Feb 9, 2012 ### zaper I'm really really sorry. I guess I'm way overthinking this or something. Is ax the radial acceleration? 8. Feb 9, 2012 ### tiny-tim a is the acceleration ax and ay are its x and y components (i'm assuming you're using y for up and x for radial) 9. Feb 9, 2012 ### zaper Ok so ax is the radial acceleration and the total acceleration is pointing up and to the left. The triangle formed by the accelerations has top angle θ and ay=ax/tanθ. Is this correct? Also does ay actually describe something here or is it just a component of the total a? 10. Feb 9, 2012 ### tiny-tim what?? why? 11. Feb 9, 2012 ### zaper Wait, I mean to the right 12. Feb 10, 2012 ### tiny-tim do you mean up and to the right, or only to the right? 13. Feb 10, 2012 ### abhinav111 hi zapper i have an easily understandable solution, if u have not come to the answer i may help. 14. Feb 10, 2012 ### zaper Up and to the right tiny tim and what's your idea abhinav? 15. Feb 10, 2012 ### tiny-tim (along the normal?) nooo i think you're confusing physics and maths force is physics acceleration is maths (geometry) if you know how the object is moving, then the acceleration follows automatically as a matter of maths it doesn't matter what forces are producing that motion, the acceleration depends only on the motion, not on the forces in this case, in equilibrium, the bead is moving uniformly in a horizontal circle … so what is its acceleration? 16. Feb 10, 2012 ### zaper Since there's no angular acceleration then the only acceleration is radial right? 17. Feb 10, 2012 ### tiny-tim yes, but you're looking at the wrong radius … the circle to look at is the circle of the bead's motion, which is horizontal, isn't it? 18. Feb 10, 2012 ### zaper Yes. That last post was supposed to be in terms of the horizontal circle. Sorry that I forgot to explain that 19. Feb 10, 2012 ### tiny-tim ok, so now do Ftotal = ma in both the horizontal and vertical directions 20. Feb 10, 2012 ### zaper Ok I think we're getting back to my point of confusion from before. We've established that the only acceleration in the horizontal circle is the radial acceleration so I'm not quite seeing where this vertical force can come from unless you mean the weight 21. Feb 10, 2012 ### tiny-tim yes, the weight is vertical but the bead has no vertical acceleration so what's the other vertical force (that makes F=ma)? 22. Feb 10, 2012 ### zaper The normal force? That's about all that's left 23. Feb 10, 2012 ### tiny-tim yes, N N and mg are the only two forces on the bead so their components in the vertical direction | have to equal the component of ma in the vertical direction, and their components in the horizontal direction have to equal the component of ma in the horizontal direction 24. Feb 10, 2012 ### zaper Ok so N is the force pulling to the center of the hoop. Would it's x component have the force m*arad? Also the weight obviously has no x component so the only x force is from N. 25. Feb 10, 2012 ### tiny-tim that's right! so the x and y equations for F = ma are … ? |
http://mathhelpforum.com/trigonometry/165949-some-complex-trigo-equations-print.html | 1,503,457,586,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117519.82/warc/CC-MAIN-20170823020201-20170823040201-00078.warc.gz | 271,829,135 | 3,210 | # Some complex trigo equations
• Dec 11th 2010, 09:00 AM
liukawa
Some complex trigo equations
Hi Guys.
am stuck with a few trigo questions here ...
1) Find value of sin10 + sin50 - sin70 without using calculator
2) Prove that sin(A+2B) + sin(A-2B) = 4sinAsinBcosB
3) If A =36 , show that sin3A = Sin 2A and find cos 36
Thanks !
• Dec 11th 2010, 10:27 AM
worc3247
1) Note that 10=30-20, 50=30+20 and 70=90-20. Then think sin(A+B)=sinAcosB+sinBcosA.
2) Again this is just trig formula sin(A+B)=sinAcosB+sinBcosA and the one for cos as well.
3) 3A=108 2A=72. Note that 90+18=108 and 90-18=72 and then look at a graph of y=sin(x).
Let me know how you get on.
• Dec 15th 2010, 05:24 AM
liukawa
But i kinda dont understand qn 3
how does looking at the sin curve help me to deduce the value of cos36.
• Dec 15th 2010, 07:53 PM
SammyS
Quote:
Originally Posted by liukawa
But i kinda don't understand qn 3
how does looking at the sin curve help me to deduce the value of cos36.
$36=2\times 18$
• Dec 17th 2010, 12:39 PM
worc3247
$sin2A=sin3A$
$2sinAcosA=sin(2A+A)$
$2sinAcosA=sin2AcosA+sinAcos2A$
$2sinAcosA=2sinA\cos^2 A+sinA\cos^2 A -\sin^3 A$
$sinA\neq0$
$\uptherefore 2cosA=3\cos^2 A - \sin^2 A$
You should now be able to solve this by using trig identities and then applying the quadratic formula. | 508 | 1,306 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-34 | longest | en | 0.767979 | # Some complex trigo equations • Dec 11th 2010, 09:00 AM liukawa Some complex trigo equations Hi Guys. am stuck with a few trigo questions here ... 1) Find value of sin10 + sin50 - sin70 without using calculator 2) Prove that sin(A+2B) + sin(A-2B) = 4sinAsinBcosB 3) If A =36 , show that sin3A = Sin 2A and find cos 36 Thanks ! • Dec 11th 2010, 10:27 AM worc3247 1) Note that 10=30-20, 50=30+20 and 70=90-20. Then think sin(A+B)=sinAcosB+sinBcosA. 2) Again this is just trig formula sin(A+B)=sinAcosB+sinBcosA and the one for cos as well. 3) 3A=108 2A=72. Note that 90+18=108 and 90-18=72 and then look at a graph of y=sin(x). Let me know how you get on. • Dec 15th 2010, 05:24 AM liukawa But i kinda dont understand qn 3 how does looking at the sin curve help me to deduce the value of cos36. • Dec 15th 2010, 07:53 PM SammyS Quote: Originally Posted by liukawa But i kinda don't understand qn 3 how does looking at the sin curve help me to deduce the value of cos36. $36=2\times 18$ • Dec 17th 2010, 12:39 PM worc3247 $sin2A=sin3A$ $2sinAcosA=sin(2A+A)$ $2sinAcosA=sin2AcosA+sinAcos2A$ $2sinAcosA=2sinA\cos^2 A+sinA\cos^2 A -\sin^3 A$ $sinA\neq0$ $\uptherefore | 2cosA=3\cos^2 A - \sin^2 A$ You should now be able to solve this by using trig identities and then applying the quadratic formula. |
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Q:
# The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence.
Q:
The sum of the present ages of a father and his son is 126 years. 8 years ago their respective ages were in the ratio of 7 : 4. After 7 years what will be the ratio of ages of father and son?
A) 23 : 19 B) 17 : 11 C) 27 : 20 D) 34 : 23
Explanation:
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4 514
Q:
Jeremy is 26 years younger than his father. Eight years hence his father's age will be two years less than twice his age. What is Jeremy's present age?
A) 20 years B) 18 years C) 24 years D) 22 years
Explanation:
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4 636
Q:
The average age of Sita and Gita is 30 years. If Rita replaces Sita, then the average age will become 28 years and if Rita replaces Gita, then the average age will become 32 years. What are the respective ages (in years) of Sita, Gita and Rita?
A) 40, 44, 36 B) 44, 40, 36 C) 34, 26, 30 D) 30, 26, 34
Explanation:
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2 568
Q:
The ratio of present ages of Sheetal and Divya is 6 : 5. After 8 years, their ages will be in the ratio of 22 : 19. What is the present age (in years) of Divya?
A) 22 B) 38 C) 34 D) 30
Explanation:
Filed Under: Problems on Ages
Exam Prep: Bank Exams
1 452
Q:
Tom's father is thrice as old as Tom. 10 years ago Tom's father's age was 7 times his age. What is Tom's current age?
A) 17 years B) 16 years C) 15 years D) 14 years
Explanation:
Filed Under: Problems on Ages
Exam Prep: Bank Exams
1 633
Q:
Average age of 4 daughters of a family is 12 years. The average age of daughters and their parents is 26 years. If the mother is 4 years older than the father, then what is the age (in years) of the father?
A) 56 B) 52 C) 48 D) 44
Explanation:
Filed Under: Problems on Ages
Exam Prep: Bank Exams
0 494
Q:
The present ages of Kavitha, Rajitha and Haritha are in the ratio of 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
A) 16,36,28 B) 16,28,36 C) 20,35,45 D) 12,21,27
Explanation:
Filed Under: Problems on Ages
Exam Prep: Bank Exams
9 774
Q:
Satheesh is two years older than Goutham who is twice as old as Sai. If the total of the ages of Satheesh, Goutham and Sai is 27, then how old is Goutham?
A) 12 B) 10 C) 11 D) 13 | 783 | 2,425 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-18 | latest | en | 0.945429 | 69 Q: # The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence. Q: The sum of the present ages of a father and his son is 126 years. 8 years ago their respective ages were in the ratio of 7 : 4. After 7 years what will be the ratio of ages of father and son? A) 23 : 19 B) 17 : 11 C) 27 : 20 D) 34 : 23 Explanation: Filed Under: Problems on Ages Exam Prep: Bank Exams 4 514 Q: Jeremy is 26 years younger than his father. Eight years hence his father's age will be two years less than twice his age. What is Jeremy's present age? A) 20 years B) 18 years C) 24 years D) 22 years Explanation: Filed Under: Problems on Ages Exam Prep: Bank Exams 4 636 Q: The average age of Sita and Gita is 30 years. If Rita replaces Sita, then the average age will become 28 years and if Rita replaces Gita, then the average age will become 32 years. What are the respective ages (in years) of Sita, Gita and Rita? A) 40, 44, 36 B) 44, 40, 36 C) 34, 26, 30 D) 30, 26, 34 Explanation: Filed Under: Problems on Ages Exam Prep: Bank Exams 2 568 Q: The ratio of present ages of Sheetal and Divya is 6 : 5. After 8 years, their ages will be in the ratio of 22 : 19. What is the present age (in years) of Divya? A) 22 B) 38 C) 34 D) 30 Explanation: Filed Under: Problems on Ages Exam Prep: Bank Exams 1 452 Q: Tom's father is thrice as old as Tom. 10 years ago Tom's father's age was 7 times his age. What is Tom's current age? A) 17 years B) 16 years C) 15 years D) 14 years Explanation: Filed Under: Problems on Ages Exam Prep: Bank Exams 1 633 Q: Average age of 4 daughters of a family is 12 years. The average age of daughters and their parents is 26 years. If the mother is 4 years older than the father, then what is the age (in years) of the father? A) 56 B) 52 C) 48 D) 44 Explanation: Filed Under: Problems on Ages Exam Prep: Bank Exams 0 494 Q: The present ages of Kavitha, Rajitha and Haritha are in the ratio of 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years). A) 16,36,28 B) 16,28,36 C) 20,35,45 D) 12,21,27 Explanation: Filed | Under: Problems on Ages Exam Prep: Bank Exams 9 774 Q: Satheesh is two years older than Goutham who is twice as old as Sai. If the total of the ages of Satheesh, Goutham and Sai is 27, then how old is Goutham? A) 12 B) 10 C) 11 D) 13 |
https://zh.m.wikipedia.org/wiki/%E5%8B%92%E5%A3%A4%E5%BE%97%E5%A4%9A%E9%A0%85%E5%BC%8F | 1,718,568,424,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861670.48/warc/CC-MAIN-20240616172129-20240616202129-00727.warc.gz | 970,474,268 | 32,623 | # 勒让德多项式
（重定向自勒壤得多項式
${\displaystyle (1-x^{2}){\frac {\mathrm {d} ^{2}P(x)}{\mathrm {d} x^{2}}}-2x{\frac {\mathrm {d} P(x)}{\mathrm {d} x}}+n(n+1)P(x)=0.}$
${\displaystyle {\mathrm {d} \over \mathrm {d} x}\left[(1-x^{2}){\mathrm {d} \over \mathrm {d} x}P(x)\right]+n(n+1)P(x)=0.}$
## 正交性
${\displaystyle \int _{-1}^{1}P_{m}(x)P_{n}(x)\,\mathrm {d} x={2 \over {2n+1}}\delta _{mn}}$
${\displaystyle {\mathrm {d} \over \mathrm {d} x}\left[(1-x^{2}){\mathrm {d} \over \mathrm {d} x}P(x)\right]=-\lambda P(x),}$
## 部分实例
n ${\displaystyle P_{n}(x)\,}$ 0 ${\displaystyle 1\,}$ 1 ${\displaystyle x\,}$ 2 ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}(3x^{2}-1)\,}$ 3 ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}(5x^{3}-3x)\,}$ 4 ${\displaystyle {\begin{matrix}{\frac {1}{8}}\end{matrix}}(35x^{4}-30x^{2}+3)\,}$ 5 ${\displaystyle {\begin{matrix}{\frac {1}{8}}\end{matrix}}(63x^{5}-70x^{3}+15x)\,}$ 6 ${\displaystyle {\begin{matrix}{\frac {1}{16}}\end{matrix}}(231x^{6}-315x^{4}+105x^{2}-5)\,}$ 7 ${\displaystyle {\begin{matrix}{\frac {1}{16}}\end{matrix}}(429x^{7}-693x^{5}+315x^{3}-35x)\,}$ 8 ${\displaystyle {\begin{matrix}{\frac {1}{128}}\end{matrix}}(6435x^{8}-12012x^{6}+6930x^{4}-1260x^{2}+35)\,}$ 9 ${\displaystyle {\begin{matrix}{\frac {1}{128}}\end{matrix}}(12155x^{9}-25740x^{7}+18018x^{5}-4620x^{3}+315x)\,}$ 10 ${\displaystyle {\begin{matrix}{\frac {1}{256}}\end{matrix}}(46189x^{10}-109395x^{8}+90090x^{6}-30030x^{4}+3465x^{2}-63)\,}$
## 在物理学中的应用
${\displaystyle {\frac {1}{\left|\mathbf {x} -\mathbf {x} ^{\prime }\right|}}={\frac {1}{\sqrt {r^{2}+r^{\prime 2}-2rr'\cos \gamma }}}=\sum _{\ell =0}^{\infty }{\frac {r^{\prime \ell }}{r^{\ell +1}}}P_{\ell }(\cos \gamma )}$
${\displaystyle \Phi (r,\theta )=\sum _{\ell =0}^{\infty }\left[A_{\ell }r^{\ell }+B_{\ell }r^{-(\ell +1)}\right]P_{\ell }(\cos \theta ).}$
## 其他性质
${\displaystyle P_{k}(-x)=(-1)^{k}P_{k}(x).\,}$
### 递推关系
${\displaystyle (n+1)P_{n+1}=(2n+1)xP_{n}-nP_{n-1}\,}$
${\displaystyle {x^{2}-1 \over n}{\mathrm {d} \over \mathrm {d} x}P_{n}=xP_{n}-P_{n-1}.}$
${\displaystyle (2n+1)P_{n}={\mathrm {d} \over \mathrm {d} x}\left[P_{n+1}-P_{n-1}\right].}$
#include <iostream>
using namespace std;
int main()
{
float n,x;
float polyaendl;
return 0;
}
float polya(float n, float x)
{
if (n == 0) return 1.0;
eurn x;
else return ((2.0 * n - 1.0) * x * polya(n - 1.0, x) - (n - 1.0) * polya(n - 2.0, x)) / n;
}
## 移位勒让德多项式
${\displaystyle \int _{0}^{1}{\tilde {P_{m}}}(x){\tilde {P_{n}}}(x)\,\mathrm {d} x={1 \over {2n+1}}\delta _{mn}.}$
${\displaystyle {\tilde {P_{n}}}(x)=(-1)^{n}\sum _{k=0}^{n}{n \choose k}{n+k \choose k}(-x)^{k}.}$
${\displaystyle {\tilde {P_{n}}}(x)=(n!)^{-1}{\mathrm {d} ^{n} \over \mathrm {d} x^{n}}\left[(x^{2}-x)^{n}\right].\,}$
n ${\displaystyle {\tilde {P_{n}}}(x)}$ 0 1 1 ${\displaystyle 2x-1}$ 2 ${\displaystyle 6x^{2}-6x+1}$ 3 ${\displaystyle 20x^{3}-30x^{2}+12x-1}$
## 极限关系
${\displaystyle \lim _{q\to 1}P_{n}(x|q)=P_{n}(x)}$
${\displaystyle \lim _{q\to 1}p_{n}(x|q)=P_{n}(1-2x)}$
## 参考文献
1. ^ 严镇军编，《数学物理方程》，第二版，中国科学技术大学出版社，合肥，2002，ISBN 7-312-00799-6，第140页 | 1,493 | 3,107 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 63, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-26 | latest | en | 0.253811 | # 勒让德多项式 （重定向自勒壤得多項式 ${\displaystyle (1-x^{2}){\frac {\mathrm {d} ^{2}P(x)}{\mathrm {d} x^{2}}}-2x{\frac {\mathrm {d} P(x)}{\mathrm {d} x}}+n(n+1)P(x)=0.}$ ${\displaystyle {\mathrm {d} \over \mathrm {d} x}\left[(1-x^{2}){\mathrm {d} \over \mathrm {d} x}P(x)\right]+n(n+1)P(x)=0.}$ ## 正交性 ${\displaystyle \int _{-1}^{1}P_{m}(x)P_{n}(x)\,\mathrm {d} x={2 \over {2n+1}}\delta _{mn}}$ ${\displaystyle {\mathrm {d} \over \mathrm {d} x}\left[(1-x^{2}){\mathrm {d} \over \mathrm {d} x}P(x)\right]=-\lambda P(x),}$ ## 部分实例 n ${\displaystyle P_{n}(x)\,}$ 0 ${\displaystyle 1\,}$ 1 ${\displaystyle x\,}$ 2 ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}(3x^{2}-1)\,}$ 3 ${\displaystyle {\begin{matrix}{\frac {1}{2}}\end{matrix}}(5x^{3}-3x)\,}$ 4 ${\displaystyle {\begin{matrix}{\frac {1}{8}}\end{matrix}}(35x^{4}-30x^{2}+3)\,}$ 5 ${\displaystyle {\begin{matrix}{\frac {1}{8}}\end{matrix}}(63x^{5}-70x^{3}+15x)\,}$ 6 ${\displaystyle {\begin{matrix}{\frac {1}{16}}\end{matrix}}(231x^{6}-315x^{4}+105x^{2}-5)\,}$ 7 ${\displaystyle {\begin{matrix}{\frac {1}{16}}\end{matrix}}(429x^{7}-693x^{5}+315x^{3}-35x)\,}$ 8 ${\displaystyle {\begin{matrix}{\frac {1}{128}}\end{matrix}}(6435x^{8}-12012x^{6}+6930x^{4}-1260x^{2}+35)\,}$ 9 ${\displaystyle {\begin{matrix}{\frac {1}{128}}\end{matrix}}(12155x^{9}-25740x^{7}+18018x^{5}-4620x^{3}+315x)\,}$ 10 ${\displaystyle {\begin{matrix}{\frac {1}{256}}\end{matrix}}(46189x^{10}-109395x^{8}+90090x^{6}-30030x^{4}+3465x^{2}-63)\,}$ ## 在物理学中的应用 ${\displaystyle {\frac {1}{\left|\mathbf {x} -\mathbf {x} ^{\prime }\right|}}={\frac {1}{\sqrt {r^{2}+r^{\prime 2}-2rr'\cos \gamma }}}=\sum _{\ell =0}^{\infty }{\frac {r^{\prime \ell }}{r^{\ell +1}}}P_{\ell }(\cos \gamma )}$ ${\displaystyle \Phi (r,\theta )=\sum _{\ell =0}^{\infty }\left[A_{\ell }r^{\ell }+B_{\ell }r^{-(\ell +1)}\right]P_{\ell }(\cos \theta ).}$ ## 其他性质 ${\displaystyle P_{k}(-x)=(-1)^{k}P_{k}(x).\,}$ ### 递推关系 ${\displaystyle (n+1)P_{n+1}=(2n+1)xP_{n}-nP_{n-1}\,}$ ${\displaystyle {x^{2}-1 \over n}{\mathrm {d} \over \mathrm {d} x}P_{n}=xP_{n}-P_{n-1}.}$ ${\displaystyle (2n+1)P_{n}={\mathrm {d} \over \mathrm {d} x}\left[P_{n+1}-P_{n-1}\right].}$ #include <iostream> using namespace std; int main() { float n,x; float polyaendl; return 0; } float polya(float n, float x) { if (n == 0) return 1.0; eurn x; else return ((2.0 * n - 1.0) * x * polya(n - 1.0, x) - (n - 1.0) * polya(n - 2.0, x)) / n; } ## 移位勒让德多项式 ${\displaystyle \int _{0}^{1}{\tilde {P_{m}}}(x){\tilde {P_{n}}}(x)\,\mathrm {d} x={1 \over {2n+1}}\delta _{mn}.}$ ${\displaystyle {\tilde {P_{n}}}(x)=(-1)^{n}\sum _{k=0}^{n}{n \choose k}{n+k \choose k}(-x)^{k}.}$ ${\displaystyle {\tilde {P_{n}}}(x)=(n!)^{-1}{\mathrm {d} ^{n} \over \mathrm {d} x^{n}}\left[(x^{2}-x)^{n}\right].\,}$ n ${\displaystyle {\tilde | {P_{n}}}(x)}$ 0 1 1 ${\displaystyle 2x-1}$ 2 ${\displaystyle 6x^{2}-6x+1}$ 3 ${\displaystyle 20x^{3}-30x^{2}+12x-1}$ ## 极限关系 ${\displaystyle \lim _{q\to 1}P_{n}(x|q)=P_{n}(x)}$ ${\displaystyle \lim _{q\to 1}p_{n}(x|q)=P_{n}(1-2x)}$ ## 参考文献 1. ^ 严镇军编，《数学物理方程》，第二版，中国科学技术大学出版社，合肥，2002，ISBN 7-312-00799-6，第140页 |
https://medium.com/towards-data-science/hair-today-gone-tomorrow-cfcaa7c31f3b | 1,506,375,852,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818693363.77/warc/CC-MAIN-20170925201601-20170925221601-00422.warc.gz | 684,087,742 | 25,239 | # Hair Today, Gone Tomorrow
It’s been quite a long time since I last cut my hair. It’s now past my shoulders, where it has remained in length for the last several years.
Even though it’s still growing, it isn’t getting any longer because it also falls out. It has reached a dynamic equilibrium, where the rate of hair growing in equals the rate falling out.
To better understand this process, I looked up a few facts about hair growth. Hair grows in at a rate of ~0.5 inches/month. There are about 100,000 hairs on a human head, of which about 100 fall out every day. This comes out to 3% of hairs falling out per month. There are other nuances to hair, but I am simplifying the model to assume that growth is constant and that all strands have the same chance of falling out.
I simulated this dynamic in R and plotted a graph to see how hair length changes over time. It starts without hair (imagine a buzz cut) and converges to a steady state length of about 16 inches. The y-axis is reversed to make the plot itself look like hanging strands of hair.
Suppose T is a random variable that represents time until a strand of hair falls out, or the age of a hair. In statistics, the hazard rate is the instantaneous rate at which some event, like failure or death, happens given that it has not yet happened.
By applying the definition of conditional probability,
and taking the limit as dt → 0, the hazard rate is the ratio of the probability density function (pdf) to 1- the cumulative density function (cdf). A hazard rate uniquely characterizes a distribution, and vice versa. A constant hazard rate is a unique property of the exponential distribution.
Here, the hazard rate is .03/month, so T ~ Exp(rate = .03/month). Hair ages are then exponentially distributed with a mean age of 1/.03 = 33 months. Since length grows linearly with age, hair length ~ Exp(rate = .06/inch) and has mean length 16.7 inches. This agrees with the mean hair length converged to in the simulation. The Q-Q plot compares hair lengths at the last time step of the simulation with randomly generated exponentials, and confirms they have the same distribution. | 478 | 2,147 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-39 | latest | en | 0.944688 | # Hair Today, Gone Tomorrow It’s been quite a long time since I last cut my hair. It’s now past my shoulders, where it has remained in length for the last several years. Even though it’s still growing, it isn’t getting any longer because it also falls out. It has reached a dynamic equilibrium, where the rate of hair growing in equals the rate falling out. To better understand this process, I looked up a few facts about hair growth. Hair grows in at a rate of ~0.5 inches/month. There are about 100,000 hairs on a human head, of which about 100 fall out every day. This comes out to 3% of hairs falling out per month. There are other nuances to hair, but I am simplifying the model to assume that growth is constant and that all strands have the same chance of falling out. I simulated this dynamic in R and plotted a graph to see how hair length changes over time. It starts without hair (imagine a buzz cut) and converges to a steady state length of about 16 inches. The y-axis is reversed to make the plot itself look like hanging strands of hair. Suppose T is a random variable that represents time until a strand of hair falls out, or the age of a hair. In statistics, the hazard rate is the instantaneous rate at which some event, like failure or death, happens given that it has not yet happened. By applying the definition of conditional probability, and taking the limit as dt → 0, the hazard rate is the ratio of the probability density function (pdf) to 1- the cumulative density function (cdf). A hazard rate uniquely characterizes a distribution, and vice versa. A constant hazard rate is a unique property of the exponential distribution. Here, the hazard rate is .03/month, so T ~ Exp(rate = .03/month). Hair ages are then exponentially distributed with a mean age of 1/.03 = 33 months. Since length grows linearly with age, hair length ~ Exp(rate = .06/inch) and has mean length | 16.7 inches. This agrees with the mean hair length converged to in the simulation. The Q-Q plot compares hair lengths at the last time step of the simulation with randomly generated exponentials, and confirms they have the same distribution. |
https://www.nagwa.com/en/worksheets/676186769546/ | 1,550,627,199,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494125.62/warc/CC-MAIN-20190220003821-20190220025821-00175.warc.gz | 934,883,888 | 32,715 | # Worksheet: Two-Step Linear Inequalities
Q1:
What can you say about ?
• A It is an inequality.
• B It is neither an equation nor an inequality.
• C It is an equation.
Q2:
What can you say about ?
• A It is neither an equation nor an inequality.
• B It is an equation.
• C It is an inequality.
Q3:
What can you say about ?
• A It is an equation.
• B It is an inequality.
• C It is neither an equation nor an inequality.
Q4:
Given that , solve the inequality .
• A
• B
• C
• D
• E
Q5:
Find the solution set of given that .
• A
• B
• C
• D
Q6:
Find the solution set of the inequality in . Give your answer in interval notation.
• A
• B
• C
• D
Q7:
Given that the solution set of the inequality is , find the values of and .
• A ,
• B ,
• C ,
• D ,
• E ,
Q8:
Find the solution set of given that .
• A
• B
• C
• D
Q9:
Solve the inequality in .
• A
• B
• C
• D
Q10:
Find the solution set of the inequality in . Give your answer in interval notation.
• A
• B
• C
• D
• E
Q11:
Determine the solution set of , where .
• A
• B
• C
• D
Q12:
Find all values of that satisfy . Write your answer as an interval.
• A
• B
• C
• D
• E
Q13:
In the figure, the perimeter of the rectangle is less than that of the triangle.
Write an inequality that can be used to find the range of values that can take.
• A
• B
• C
• D
• E
• A
• B
• C
• D
• E
Q14:
Find the solution set of the inequality in . Give your answer in interval notation.
• A
• B
• C
• D
Q15:
Solve .
• A
• B
• C
• D
Q16:
Solve .
• A
• B
• C
• D
Q17:
Mrs. Olivia tells her math class, βFive more than four times a number is more than 12.β Let represent the number, and write an inequality that represents her statement.
• A
• B
• C
• D
• E | 553 | 1,738 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2019-09 | longest | en | 0.845414 | # Worksheet: Two-Step Linear Inequalities Q1: What can you say about ? • A It is an inequality. • B It is neither an equation nor an inequality. • C It is an equation. Q2: What can you say about ? • A It is neither an equation nor an inequality. • B It is an equation. • C It is an inequality. Q3: What can you say about ? • A It is an equation. • B It is an inequality. • C It is neither an equation nor an inequality. Q4: Given that , solve the inequality . • A • B • C • D • E Q5: Find the solution set of given that . • A • B • C • D Q6: Find the solution set of the inequality in . Give your answer in interval notation. • A • B • C • D Q7: Given that the solution set of the inequality is , find the values of and . • A , • B , • C , • D , • E , Q8: Find the solution set of given that . • A • B • C • D Q9: Solve the inequality in . • A • B • C • D Q10: Find the solution set of the inequality in . Give your answer in interval notation. • A • B • C • D • E Q11: Determine the solution set of , where . • A • B • C • D Q12: Find all values of that satisfy . Write your answer as an interval. • A • B • C • D • E Q13: In the figure, the perimeter of the rectangle is less than that of the triangle. Write an inequality that can be used to find the range of values that can take. • A • B • C • D • E • A • B • C • D • E Q14: Find the solution set of the inequality in . Give your answer in interval notation. • A • B • C • D Q15: Solve . • A • B • C • D Q16: Solve . • A • B • C | • D Q17: Mrs. Olivia tells her math class, βFive more than four times a number is more than 12.β Let represent the number, and write an inequality that represents her statement. • A • B • C • D • E |
https://edustrings.com/mathematics/1312542.html | 1,701,551,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100452.79/warc/CC-MAIN-20231202203800-20231202233800-00622.warc.gz | 278,751,709 | 7,160 | 1 October, 03:25
# How many positive integers less than 2018 are divisible by at least 3, 11, or 61?
+2
1. 1 October, 04:56
0
To find all the positive integers less than 2018 that are divisible by 3, 11, and 61, you will use what you know about factors.
3, 11, and 61 are all answers. So are 33, 183, 671, and 2013.
If you put these in factors, the product will be divisible by them!
3 x 11 = 33
3 x 61 = 183
11 x 61 = 671
3 x 11 x 61 = 2013
Take each number and square it, cube it, etc ...
9, 27, 81, 243, 729
121, 1331
9 x 11 = 99
27 x 11 = 297
81 x 11 = 891
121 x 9 = 1089
121 x 3 = 363
61 x 9 = 549
61 x 27 = 1647
Everything in bold is a correct answer. | 261 | 676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-50 | latest | en | 0.906552 | 1 October, 03:25 # How many positive integers less than 2018 are divisible by at least 3, 11, or 61? +2 1. 1 October, 04:56 0 To find all the positive integers less than 2018 that are divisible by 3, 11, and 61, you will use what you know about factors. 3, 11, and 61 are all answers. So are 33, 183, 671, and 2013. If you put these in factors, the product will be divisible by them! 3 x 11 = 33 3 x 61 = 183 11 x 61 = 671 3 x 11 x 61 = 2013 Take each number and square it, cube it, etc ... 9, 27, 81, 243, 729 121, 1331 9 x 11 = 99 27 x 11 = 297 81 x 11 = 891 121 x 9 = 1089 121 x 3 = 363 61 | x 9 = 549 61 x 27 = 1647 Everything in bold is a correct answer. |
https://www.physicsforums.com/threads/calculate-h.51566/ | 1,542,554,052,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00499.warc.gz | 959,677,040 | 13,462 | # Calculate [H+]?
1. Nov 5, 2004
### parwana
Calculate [H+]?????
Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.4e-10
2. Nov 5, 2004
### Sirus
Consider that for a reaction
$$HA_{(aq)}\rightleftharpoons~H^{+}_{(aq)}~+~A^{-}_{(aq)}$$
we can use
$$K_{a}=\frac{[H^{+}_{(aq)}][A^{-}_{(aq)}]}{[HA_{(aq)}]}$$
How can you find $K_{a}$ when given $K_{b}$?
3. Nov 5, 2004
### chem_tr
Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only $\displaystyle x$ of it is ionized to give some $\displaystyle H^+$. We know the equilibrium constant of this reaction, i.e., $$\displaystyle \frac {10^{-14}}{K_b}$$.
$$\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}$$
Now it is better for you to consider the magnitude of $$\displaystyle \frac {10^{-14}}{K_b}$$; for the sake of simplification, you may omit the $\displaystyle x$ in $\displaystyle 0.220-x$, as we may omit values generally less than 5%. Of course, you may also want to solve the quadratic equation to find the exact answer, but believe me this is not very necessary.
What you'll do next is to find the $\displaystyle -\log x$.
4. Nov 6, 2004
### ShawnD
Aniline is a base, not an acid. Hydrogen will stick to the lone pair on the nitrogen so the reaction is like this:
$$C_6H_5NH_2 + H_2O \rightarrow C_6H_5NH_3^+ + OH^-$$
5. Nov 6, 2004
### Sirus
In that case, for a reaction
$$B_{(aq)}~+~H_{2}O_{(l)}\rightleftharpoons~BH^{+}_{(aq)}~+~OH^{-}_{(aq)}$$
we can use
$$K_{b}=\frac{[BH^{+}_{(aq)}][OH^{-}_{(aq)}]}{[B_{(aq)}]}$$
6. Nov 6, 2004
### chem_tr
Dear ShawnD, you are right about aqueous reactions, but don't forget that there are very powerful bases for using in non-aqueous phases like sodium hydride, lithium diisopropylamide, etc, which can take a proton to give anilinide anion.
In most cases, your reaction is sufficient, and same things may be said for Sirus' last post. | 714 | 2,038 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2018-47 | latest | en | 0.733647 | # Calculate [H+]? 1. Nov 5, 2004 ### parwana Calculate [H+]????? Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.4e-10 2. Nov 5, 2004 ### Sirus Consider that for a reaction $$HA_{(aq)}\rightleftharpoons~H^{+}_{(aq)}~+~A^{-}_{(aq)}$$ we can use $$K_{a}=\frac{[H^{+}_{(aq)}][A^{-}_{(aq)}]}{[HA_{(aq)}]}$$ How can you find $K_{a}$ when given $K_{b}$? 3. Nov 5, 2004 ### chem_tr Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only $\displaystyle x$ of it is ionized to give some $\displaystyle H^+$. We know the equilibrium constant of this reaction, i.e., $$\displaystyle \frac {10^{-14}}{K_b}$$. $$\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}$$ Now it is better for you to consider the magnitude of $$\displaystyle \frac {10^{-14}}{K_b}$$; for the sake of simplification, you may omit the $\displaystyle x$ in $\displaystyle 0.220-x$, as we may omit values generally less than 5%. Of course, you may also want to solve the quadratic equation to find the exact answer, but believe me this is not very necessary. What you'll do next is to find the $\displaystyle -\log x$. 4. Nov 6, 2004 ### ShawnD Aniline is a base, not an acid. Hydrogen will stick to the lone pair on the nitrogen so the reaction is like this: $$C_6H_5NH_2 + H_2O \rightarrow C_6H_5NH_3^+ + OH^-$$ 5. Nov 6, 2004 ### Sirus In that case, for a reaction $$B_{(aq)}~+~H_{2}O_{(l)}\rightleftharpoons~BH^{+}_{(aq)}~+~OH^{-}_{(aq)}$$ we can use $$K_{b}=\frac{[BH^{+}_{(aq)}][OH^{-}_{(aq)}]}{[B_{(aq)}]}$$ 6. Nov 6, 2004 ### chem_tr Dear ShawnD, you are right about aqueous reactions, but don't forget that there are very powerful bases for using in non-aqueous phases like sodium | hydride, lithium diisopropylamide, etc, which can take a proton to give anilinide anion. In most cases, your reaction is sufficient, and same things may be said for Sirus' last post. |
https://discussmain.com/q/531054 | 1,603,371,041,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107879537.28/warc/CC-MAIN-20201022111909-20201022141909-00717.warc.gz | 289,593,168 | 2,565 | # Length of a rectangle is 3 times more party of a square, and width - is 5 cm less than the party of a square. Find the party of a square if its area is 50 cm² less than the area of a rectangle. Help to solve please. In advance, bolshooooy thank you. Very urgently it is necessary.
301
The party of quarter - the X Dl. it is direct. - 3X Scheer. it is direct. - (H-5) of Skv = X² Spr =3X×(X-5) x²+50=3x(x-5) x²-3x²+15x²+50=0 - 2x²+15x+50=0 D=225+400=625 x ₁= (-15+25)/-4=-10/4 x ₂= (-15-25)/-4=10 Party of a square =10sm.
102 | 204 | 527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-45 | latest | en | 0.886984 | # Length of a rectangle is 3 times more party of a square, and width - is 5 cm less than the party of a square. Find the party of a square if its area is 50 cm² less than the area of a rectangle. Help to solve please. In advance, bolshooooy thank you. Very urgently it is necessary. 301 The party of quarter - the X Dl. it is direct. - 3X Scheer. it is direct. - (H-5) of Skv = X² Spr =3X×(X-5) x²+50=3x(x-5) x²-3x²+15x²+50=0 - 2x²+15x+50=0 D=225+400=625 x | ₁= (-15+25)/-4=-10/4 x ₂= (-15-25)/-4=10 Party of a square =10sm. 102 |
https://www.physicsforums.com/threads/electric-field-in-a-square.734314/ | 1,531,955,434,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.62/warc/CC-MAIN-20180718213135-20180718233135-00065.warc.gz | 944,479,716 | 16,018 | # Homework Help: Electric field in a square
1. Jan 23, 2014
### puzzledup
1. The problem statement, all variables and given/known data
A square has a charge at each corner. The square is 2cm x 2cm. In the upper left corner there is 1μc, in the upper right corner there is 2μc, the lower right corner there is 3μc, and the lower left corner there is 4μc.
A. What is the magnitude and direction of the electric field at the 4μc charge?
B. wha Tia the magnitude and direction of the force on the 4μc charge?
2. Relevant equations
E=kq/r^2. And F=EQ
3. The attempt at a solution
1. (9*10^9)(1*10^-6)/(.02)^2 = 2.25*10^7 N/C
2. Same process with 2*10^-6 to get. 4.5*10^7 N/ C
3. Same with 3*10^-6 to get 6.75*10^7 N/C
E net = 2.25+4.5+6.75=13.5*10^7 N/C.
But not sure how to get direction. Since they are all positive, would it be 45° SE?
For force, F=(13.5*10^7)(1.6*10^-19)=21.6*10^-12N.
How to get direction? Pythagorean to be used but how?
Any help is appreciated.
2. Jan 23, 2014
### Curious3141
The question hasn't explicitly stated that all charges are positive. This has to be assume to make it doable.
Right.
Wrong. What's the distance between the upper right and lower left corners?
Right.
It's not simple addition. Electric field strength is a vector. Do vector addition.
Why SE? Shouldn't it be SW?
Isn't the charge at the lower left corner 4uC?
Draw a force diagram. How would you add vectors?
3. Jan 23, 2014
### puzzledup
So for 2. I should use Pythagorean - Sqrt(.02^2+.02^2)?
Ok for vector addition, tip to tail for E.
I believe SW would be right.
But what do you mean about the charge in lower left? Am I not supposed to use E times Q the constant at 1.6*10^-19? Or am I wrong about Q?
4. Jan 23, 2014
### Curious3141
SW is the general direction. It's not exactly SW because the magnitudes of the field strength in a Southward direction (due to the upper left corner) and that in a Westward direction (due to the lower right corner) are not identical. So you need to use trig to get the correct orientation of the resultant vector.
It's easiest to work out the equal component vectors (Southward and Westward) of the field strength due to the top right hand corner charge first. Then add up the Southern and Westward components of the three field strengths individually. Finally, use Pythagorean theorem to find the magnitude of the resultant, and use trig to find the direction.
Your (correct) units for field strength are N/C. Your charge is 4uC. How would you get the force (in Newton)?
5. Jan 23, 2014
### puzzledup
Q the constant is in C, so would leave me with N.
This helps me a lot, Thanks so much! | 773 | 2,656 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-30 | latest | en | 0.889173 | # Homework Help: Electric field in a square 1. Jan 23, 2014 ### puzzledup 1. The problem statement, all variables and given/known data A square has a charge at each corner. The square is 2cm x 2cm. In the upper left corner there is 1μc, in the upper right corner there is 2μc, the lower right corner there is 3μc, and the lower left corner there is 4μc. A. What is the magnitude and direction of the electric field at the 4μc charge? B. wha Tia the magnitude and direction of the force on the 4μc charge? 2. Relevant equations E=kq/r^2. And F=EQ 3. The attempt at a solution 1. (9*10^9)(1*10^-6)/(.02)^2 = 2.25*10^7 N/C 2. Same process with 2*10^-6 to get. 4.5*10^7 N/ C 3. Same with 3*10^-6 to get 6.75*10^7 N/C E net = 2.25+4.5+6.75=13.5*10^7 N/C. But not sure how to get direction. Since they are all positive, would it be 45° SE? For force, F=(13.5*10^7)(1.6*10^-19)=21.6*10^-12N. How to get direction? Pythagorean to be used but how? Any help is appreciated. 2. Jan 23, 2014 ### Curious3141 The question hasn't explicitly stated that all charges are positive. This has to be assume to make it doable. Right. Wrong. What's the distance between the upper right and lower left corners? Right. It's not simple addition. Electric field strength is a vector. Do vector addition. Why SE? Shouldn't it be SW? Isn't the charge at the lower left corner 4uC? Draw a force diagram. How would you add vectors? 3. Jan 23, 2014 ### puzzledup So for 2. I should use Pythagorean - Sqrt(.02^2+.02^2)? Ok for vector addition, tip to tail for E. I believe SW would be right. But what do you mean about the charge in lower left? Am I not supposed to use E times Q the constant at 1.6*10^-19? Or am I wrong about Q? 4. Jan 23, 2014 ### Curious3141 SW is the general direction. It's not exactly SW because the magnitudes of the field strength in a Southward direction (due to the upper left corner) and that in a Westward direction (due to the lower right corner) are not identical. So you need to use trig to get the correct orientation of the resultant vector. It's easiest to work out the equal component vectors (Southward and Westward) of the field strength due to the top right hand corner charge first. Then add up the Southern and Westward components of the three field strengths individually. Finally, use Pythagorean theorem to find the magnitude of the resultant, and use trig to find | the direction. Your (correct) units for field strength are N/C. Your charge is 4uC. How would you get the force (in Newton)? 5. Jan 23, 2014 ### puzzledup Q the constant is in C, so would leave me with N. This helps me a lot, Thanks so much! |
https://www.gradesaver.com/textbooks/math/algebra/algebra-a-combined-approach-4th-edition/chapter-6-review-page-473/19 | 1,537,422,917,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267156416.22/warc/CC-MAIN-20180920041337-20180920061337-00352.warc.gz | 689,903,239 | 13,746 | ## Algebra: A Combined Approach (4th Edition)
$-4(x^{2}-3x-8)$
$32+12x-4x^{2}=$ $-4x^{2}+12x+32=$ $-4(x^{2}-3x-8)$. No further factorization possible. | 69 | 151 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-39 | longest | en | 0.38264 | ## Algebra: A Combined Approach (4th Edition) $-4(x^{2}-3x-8)$ $32+12x-4x^{2}=$ $-4x^{2}+12x+32=$ $-4(x^{2}-3x-8)$. No further | factorization possible. |
https://homeschoolmath.blogspot.com/2009/01/little-trick-for-square-roots-mental.html | 1,623,862,777,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487625967.33/warc/CC-MAIN-20210616155529-20210616185529-00510.warc.gz | 291,501,230 | 24,421 | ### A little trick for square roots (mental math)
Someone sent me this little mental math trick for square roots. I liked it, didn't know it before, so here goes:
`I read your suggestion for calculating square root without a calculator. I teach Math for Elementary Teachers and developmental math courses (algebra) to adults. I feel that the focus should be on understanding the number rather than an exercise in following a memorized algorithm. I suggest you have the student determine the pair of perfect squares the number falls between. For example, if finding the sqrt of 645, it falls between the sqrt of 625 which equals 25 and the sqrt of 676 which equals 26. So the sqrt of 645 has to be between 25 and 26. Where does it fall between? There are 50 numbers between 676 and 625. 645 is 20 numbers beyond 625, so 20/50 = 0.4So the sqrt of 645 is very close to 25.4This method provides the student with a process that improves their understanding of numbers without expecting them to memorize an algorithm, and it provides an answer to the nearest tenth.Andrea S. Levy, Ed.D.`
She is referring to my article about the square root algorithm. However, I've never meant that kids would need to learn that long algorithm in school work. In the article, I'm actually advocating the method of finding the approximate square root by "guess and check".
vishal said…
hey maria i did not understand the trick properly can u help me out???
plz i am a student in india in std 10 in ICSE board
- vishal
Anonymous said…
Hi Maria,
It is a good effort from your part to find a strategy to get square root of numbers which are not square numbers. But it does take much of the needed time to go for other problems and the use of calculator rather would be more practical for such numbers.
Mr Sand (TIS)
Maria Miller said…
Of course calculator is what one would use, most of the time, for square roots. But the teacher could show this square root trick to students in order to help them learn better the actual concept of square root itself and to develop number sense.
Anonymous said…
I am taking the ASVAB tomorrow. Thank you for helping me understand square roots more clearly. I absolutely need to know math, especially since I am joining the Navy.
Jason Barnhard said…
This is a fantastic concept. I have been looking for a way to find square roots that is a little easier for the younger generation. the long division method is just too complicated with all its doubling and moving numbers. you definitely helped me out a lot.
THANKS!
Anonymous said…
i like it but, I like knowing te EXACT answer
Anonymous said…
625 minus 676 is 51 rather than the stated 50. 20/51 is. 39 and dead on for the sqrt(645), which everyone typed by now. | 618 | 2,728 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-25 | latest | en | 0.947153 | ### A little trick for square roots (mental math) Someone sent me this little mental math trick for square roots. I liked it, didn't know it before, so here goes: `I read your suggestion for calculating square root without a calculator. I teach Math for Elementary Teachers and developmental math courses (algebra) to adults. I feel that the focus should be on understanding the number rather than an exercise in following a memorized algorithm. I suggest you have the student determine the pair of perfect squares the number falls between. For example, if finding the sqrt of 645, it falls between the sqrt of 625 which equals 25 and the sqrt of 676 which equals 26. So the sqrt of 645 has to be between 25 and 26. Where does it fall between? There are 50 numbers between 676 and 625. 645 is 20 numbers beyond 625, so 20/50 = 0.4So the sqrt of 645 is very close to 25.4This method provides the student with a process that improves their understanding of numbers without expecting them to memorize an algorithm, and it provides an answer to the nearest tenth.Andrea S. Levy, Ed.D.` She is referring to my article about the square root algorithm. However, I've never meant that kids would need to learn that long algorithm in school work. In the article, I'm actually advocating the method of finding the approximate square root by "guess and check". vishal said… hey maria i did not understand the trick properly can u help me out??? plz i am a student in india in std 10 in ICSE board - vishal Anonymous said… Hi Maria, It is a good effort from your part to find a strategy to get square root of numbers which are not square numbers. But it does take much of the needed time to go for other problems and the use of calculator rather would be more practical for such numbers. Mr Sand (TIS) Maria Miller said… Of course calculator is what one would use, most of the time, for square roots. But the teacher could show this square root trick to students in order to help them learn better the actual concept of square root itself and to develop number sense. Anonymous said… I am taking the ASVAB tomorrow. Thank you for helping me understand square roots more clearly. I absolutely need to know math, especially since I am joining the Navy. Jason Barnhard said… This is a fantastic concept. I have been looking for a way to find square roots that is a little easier for the younger generation. the long division method is just too complicated with all its | doubling and moving numbers. you definitely helped me out a lot. THANKS! Anonymous said… i like it but, I like knowing te EXACT answer Anonymous said… 625 minus 676 is 51 rather than the stated 50. 20/51 is. 39 and dead on for the sqrt(645), which everyone typed by now. |
https://de.scribd.com/doc/246425735/Tutorial1-Dynamics | 1,561,591,256,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000575.75/warc/CC-MAIN-20190626214837-20190627000837-00420.warc.gz | 405,962,027 | 54,625 | You are on page 1of 4
# KHWOPA ENGINEERING COLLEGE
## Tutorial-1 (Rectilinear motion) (CIVIL: I/II)
1. The acceleration of a particle is defined by the relation a =
, where a and t are expressed
in m/s2 and seconds, respectively. Knowing that x = 0 and v = 0 at t = 0, determine the velocity
and position of the particle when t = 0.5 s.
2. The acceleration of point A is defined by the relation a = 5.4 sin kt, where a and t are expressed
in m/s2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0 and v = 1.8 m/s when t = 0,
determine the velocity and position of point A when t = 0.5 s.
3. The acceleration of a particle is defined by the relation a = 9 . The particle starts at t = 0 with
v = 0 and x = 5 m. Determine (a) the time when the velocity is again zero, (b) the position and
velocity when t = 4 s, (c) the total distance traveled by the particle from t = 0 to t = 4 s.
4. The acceleration of a particle is defined by the relation a =
. (a) Knowing that v = 10 m/s
when t = 0 and that v = 10 m/s when t = 2 s, determine the constant k. (b) Write the equations of
motion, knowing also that x = 0 when t = 2 s.
5. The acceleration of a particle is defined by the relation a = k (1
), where k is a constant.
Knowing that the velocity of the particle is v = +9m/s when x = 3m and that the particle comes
to rest at the origin, determine (a) the value of k, (b) the velocity of the particle when x = 2 m.
6. The acceleration of slider A is defined by the relation a = 2k
, where a and v are
expressed in m/s2 and m/s, respectively, and k is a constant. The system starts at time t = 0 with x
= 1.5 m and v = 0. Knowing that x = 1.2 m when t = 0.2 s, determine the value of k.
7. A group of students launches a model rocket in the vertical direction. Based on tracking data, they
determine that the altitude of the rocket was 27.5 m at the end of the powered portion of the flight
and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that
the rocket fell freely to the ground after reaching its maximum altitude and assuming that g =
9.81m/s2, determine (a) the speed v1 of the rocket at the end of powered flight, (b) the maximum
altitude reached by the rocket.
1
Compiled by: Dipendra Gautam, [email protected] , 9851139815
## KHWOPA ENGINEERING COLLEGE
Tutorial-1 (Rectilinear motion) (CIVIL: I/II)
8. A projectile is fired with an initial velocity of 800 ft/s at a target located 2000 ft above gun A and
at a horizontal distance of 12000 ft. neglecting air resistance, determine the value of the firing
angle . [PU, 2011, Back]
9. Write down the 3 types of acceleration, and velocity. Justify the statement, Graphical solutions
simplify the solutions of the problems in dynamics.
10. Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it
is passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s.
Knowing that bus B passes point O 20 s after automobile A started from there, determine when
and where the vehicles passed each other.
11. A projectile is fired from the edge of a 150-m cliff with an initial velocity of 180 m/s at an angle
of 30 with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun
to the point where the projectile strikes the ground, (b) the greatest elevation above the ground
reached by the projectile.
12. Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and
speeds shown. Knowing that automobile A has a constant acceleration of 0.6 m/s2 and that B has
a constant deceleration of 0.4 m/s2, determine (a) when and where A will overtake B, (b) the
speed of each automobile at that time.
13. Automobile A is traveling east at the constant speed of 36 km/h. As automobile A crosses the
intersection shown, automobile B starts from rest 35 m north of the intersection and moves south
with a constant acceleration of 1.2 m/s2. Determine the position, velocity, and acceleration of B
relative to A 5 s after A crosses the intersection.
14. Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound
lane at a constant speed of 96 km/h. At t = 0, A starts and accelerates at a constant rate aA, while
at t = 5 s, B begins to slow down with a constant deceleration of magnitude aA / 6. Knowing that
when the cars pass each other x = 90m and vA = vB, determine (a) the acceleration aA, (b) when
the vehicles pass each other, (c) the distance between the vehicles at t = 0.
15. Block A moves down with a constant velocity of 1 m/s. Determine (a) the velocity of block C, (b)
the velocity of collar B relative to block A, (c) the relative velocity of portion D of the cable with
respect to block A.
2
Compiled by: Dipendra Gautam, [email protected] , 9851139815
## KHWOPA ENGINEERING COLLEGE
Tutorial-1 (Rectilinear motion) (CIVIL: I/II)
16. Collars A and B start from rest and move with the following accelerations: aA = 2.5t in./s2
upward and aB = 15 in./s2 downward. Determine (a) the time at which the velocity of block C is
again zero, (b) the distance through which block C will have moved at that time.
17. A parachutist is in free fall at a rate of 180 ft/s when he opens his parachute at an altitude of 1900
ft. Following a rapid and constant deceleration, he then descends at a constant rate of 44 ft/s from
1800 ft to 100 ft, where he maneuvers the parachute into the wind to further slow his descent.
Knowing that the parachutist lands with a negligible downward velocity, determine (a) the time
required for the parachutist to land after opening his parachute, (b) the initial deceleration.
18. Cars A and B are traveling respectively at the constant speeds of (vA )0 = 22 mi/h and ( )0 vB = 13
mi/h on an ice-covered road. To avoid overtaking car B, the driver of car A applies his brakes so
that his car decelerates at a constant rate of 0.14 ft/s2. Determine the distance d between the cars
at which the driver of car A must apply his brakes to just avoid colliding with car B.
(Hint: Plot v-t curve and refer the solved numerical @ class).
19. A golfer aims his shot to clear the top of a tree by a distance h at the peak of the trajectory and to
miss the pond on the opposite side. Knowing that the magnitude of v0 is 30 m/s, determine the
range of values of h which must be avoided.
3
Compiled by: Dipendra Gautam, [email protected] , 9851139815
## KHWOPA ENGINEERING COLLEGE
Tutorial-1 (Rectilinear motion) (CIVIL: I/II)
20. A handball player throws a ball from A with a horizontal velocity v0. Knowing that d = 15 ft,
determine (a) the value of v0 for which the ball will strike the corner C, (b) the range of values of
v0 for which the ball will strike the corner region BCD.
21. In slow pitch softball the underhand pitch must reach a maximum height of between 1.8 m and
3.7 m above the ground. A pitch is made with an initial velocity v0 of magnitude 13 m/s at an
angle of 33 with the horizontal. Determine (a) if the pitch meets the maximum height
requirement, (b) the height of the ball as it reaches the batter.
(One is never obliged to provide figures for the queries but one has to sketch out the illustrations
thoroughly for simplification.)
4
Compiled by: Dipendra Gautam, [email protected] , 9851139815 | 1,995 | 7,325 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2019-26 | latest | en | 0.919301 | You are on page 1of 4 # KHWOPA ENGINEERING COLLEGE ## Tutorial-1 (Rectilinear motion) (CIVIL: I/II) 1. The acceleration of a particle is defined by the relation a = , where a and t are expressed in m/s2 and seconds, respectively. Knowing that x = 0 and v = 0 at t = 0, determine the velocity and position of the particle when t = 0.5 s. 2. The acceleration of point A is defined by the relation a = 5.4 sin kt, where a and t are expressed in m/s2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0 and v = 1.8 m/s when t = 0, determine the velocity and position of point A when t = 0.5 s. 3. The acceleration of a particle is defined by the relation a = 9 . The particle starts at t = 0 with v = 0 and x = 5 m. Determine (a) the time when the velocity is again zero, (b) the position and velocity when t = 4 s, (c) the total distance traveled by the particle from t = 0 to t = 4 s. 4. The acceleration of a particle is defined by the relation a = . (a) Knowing that v = 10 m/s when t = 0 and that v = 10 m/s when t = 2 s, determine the constant k. (b) Write the equations of motion, knowing also that x = 0 when t = 2 s. 5. The acceleration of a particle is defined by the relation a = k (1 ), where k is a constant. Knowing that the velocity of the particle is v = +9m/s when x = 3m and that the particle comes to rest at the origin, determine (a) the value of k, (b) the velocity of the particle when x = 2 m. 6. The acceleration of slider A is defined by the relation a = 2k , where a and v are expressed in m/s2 and m/s, respectively, and k is a constant. The system starts at time t = 0 with x = 1.5 m and v = 0. Knowing that x = 1.2 m when t = 0.2 s, determine the value of k. 7. A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 27.5 m at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g = 9.81m/s2, determine (a) the speed v1 of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket. 1 Compiled by: Dipendra Gautam, [email protected] , 9851139815 ## KHWOPA ENGINEERING COLLEGE Tutorial-1 (Rectilinear motion) (CIVIL: I/II) 8. A projectile is fired with an initial velocity of 800 ft/s at a target located 2000 ft above gun A and at a horizontal distance of 12000 ft. neglecting air resistance, determine the value of the firing angle . [PU, 2011, Back] 9. Write down the 3 types of acceleration, and velocity. Justify the statement, Graphical solutions simplify the solutions of the problems in dynamics. 10. Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other. 11. A projectile is fired from the edge of a 150-m cliff with an initial velocity of 180 m/s at an angle of 30 with the horizontal. Neglecting air resistance, find (a) the horizontal distance from the gun to the point where the projectile strikes the ground, (b) the greatest elevation above the ground reached by the projectile. 12. Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 0.6 m/s2 and that B has a constant deceleration of 0.4 m/s2, determine (a) when and where A will overtake B, (b) the speed of each automobile at that time. 13. Automobile A is traveling east at the constant speed of 36 km/h. As automobile A crosses the intersection shown, automobile B starts from rest 35 m north of the intersection and moves south with a constant acceleration of 1.2 m/s2. Determine the position, velocity, and acceleration of B relative to A 5 s after A crosses the intersection. 14. Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 96 km/h. At t = 0, A starts and accelerates at a constant rate aA, while at t = 5 s, B begins to slow down with a constant deceleration of magnitude aA / 6. Knowing that when the cars pass each other x = 90m and vA = vB, determine (a) the acceleration aA, (b) when the vehicles pass each other, (c) the distance between the vehicles at t = 0. 15. Block A moves down with a constant velocity of 1 m/s. Determine (a) the velocity of block C, (b) the velocity of collar B relative to block A, (c) the relative velocity of portion D of the cable with respect to block A. 2 Compiled by: Dipendra Gautam, [email protected] , 9851139815 ## KHWOPA ENGINEERING COLLEGE Tutorial-1 (Rectilinear motion) (CIVIL: I/II) 16. Collars A and B start from rest and move with the following accelerations: aA = 2.5t in./s2 upward and aB = 15 in./s2 downward. Determine (a) the time at which the velocity of block C is again zero, (b) the distance through which block C will have moved at that time. 17. A parachutist is in free fall at a rate of 180 ft/s when he opens his parachute at an altitude of 1900 ft. Following a rapid and constant deceleration, he then descends at a constant rate of 44 ft/s from 1800 ft to 100 ft, where he maneuvers the parachute into the wind to further slow his descent. Knowing that the parachutist lands with a negligible downward velocity, determine (a) the time required for the parachutist to land after opening his parachute, (b) the initial deceleration. 18. Cars A and B are traveling respectively at the constant speeds of (vA )0 = 22 mi/h and ( )0 vB = 13 mi/h on an ice-covered road. To avoid overtaking car B, the driver of car A applies his brakes so that his car decelerates at a constant rate of 0.14 ft/s2. Determine the distance d between the cars at which the driver of car A must apply his brakes to just avoid colliding with car B. (Hint: Plot v-t curve and refer the solved numerical @ class). 19. A golfer aims his shot to clear the top of a tree by a distance h at the peak of the trajectory and to miss the pond on the opposite side. Knowing that the magnitude of v0 is 30 m/s, determine the range of values of h which must be avoided. 3 Compiled by: Dipendra Gautam, [email protected] , 9851139815 ## KHWOPA ENGINEERING COLLEGE Tutorial-1 (Rectilinear motion) (CIVIL: I/II) 20. A handball player throws | a ball from A with a horizontal velocity v0. Knowing that d = 15 ft, determine (a) the value of v0 for which the ball will strike the corner C, (b) the range of values of v0 for which the ball will strike the corner region BCD. 21. In slow pitch softball the underhand pitch must reach a maximum height of between 1.8 m and 3.7 m above the ground. A pitch is made with an initial velocity v0 of magnitude 13 m/s at an angle of 33 with the horizontal. Determine (a) if the pitch meets the maximum height requirement, (b) the height of the ball as it reaches the batter. (One is never obliged to provide figures for the queries but one has to sketch out the illustrations thoroughly for simplification.) 4 Compiled by: Dipendra Gautam, [email protected] , 9851139815 |
http://web2.0calc.com/questions/how-to-add-radicals | 1,508,593,277,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824819.92/warc/CC-MAIN-20171021133807-20171021153807-00170.warc.gz | 367,265,509 | 6,275 | +0
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see if this video helps
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#4
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it kinda did help thanks
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I don't even know what that means
Guest May 3, 2017
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If a and b have a common non-square-number factor**, and the square root of the 2 cannot be simplified into a whole number, then $$\sqrt a$$ and $$\sqrt b$$ can be added together!!
Your question is $$\sqrt{5} + \sqrt 8$$, because 5 and 8 does not have a common non-square-number factor**, they cannot be added together. The answer is:
$$\quad\sqrt 5 + \sqrt 8 \\ =\sqrt 5 + \sqrt{2^2\cdot 2}\\ =\sqrt 5 + \sqrt{2^2} \cdot \sqrt2\\ =\sqrt 5 + 2\sqrt 2$$
Let's say $$\sqrt 8 + \sqrt {32}$$, 8 and 32 have a common non-square-number factor** 2, so they can be added together. The answer is:
$$\quad \sqrt8 + \sqrt{32}\\ =\sqrt{2^2\cdot 2}+\sqrt{2^4\cdot 2}\\ =\sqrt{2^2}\cdot \sqrt2 + \sqrt{2^4}\cdot \sqrt2\\ =2\cdot \sqrt2 + 2^2 \cdot \sqrt2\\ =2\sqrt2 + 4\sqrt2\\ =6\sqrt2$$
**: common non-square-number factor here means common factor that are not square numbers(i.e. 1,4,9,16,25,36,49,.....)
MaxWong May 3, 2017
### 7 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details | 497 | 1,374 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2017-43 | longest | en | 0.780312 | +0 0 152 6 +50 Sort: #1 +50 0 #2 0 see if this video helps Guest May 3, 2017 #4 +50 0 it kinda did help thanks #3 0 I don't even know what that means Guest May 3, 2017 #5 +50 0 #6 +6810 0 If a and b have a common non-square-number factor**, and the square root of the 2 cannot be simplified into a whole number, then $$\sqrt a$$ and $$\sqrt b$$ can be added together!! Your question is $$\sqrt{5} + \sqrt 8$$, because 5 and 8 does not have a common non-square-number factor**, they cannot be added together. The answer is: $$\quad\sqrt 5 + \sqrt 8 \\ =\sqrt 5 + \sqrt{2^2\cdot 2}\\ =\sqrt 5 + \sqrt{2^2} \cdot \sqrt2\\ =\sqrt 5 + 2\sqrt 2$$ Let's say $$\sqrt 8 + \sqrt {32}$$, 8 and 32 have a common non-square-number factor** 2, so they can be added together. The answer is: $$\quad \sqrt8 + \sqrt{32}\\ =\sqrt{2^2\cdot 2}+\sqrt{2^4\cdot 2}\\ =\sqrt{2^2}\cdot \sqrt2 + \sqrt{2^4}\cdot \sqrt2\\ =2\cdot \sqrt2 + 2^2 \cdot \sqrt2\\ =2\sqrt2 + 4\sqrt2\\ =6\sqrt2$$ **: common non-square-number factor here means common factor that are not square numbers(i.e. 1,4,9,16,25,36,49,.....) MaxWong May 3, 2017 ### 7 Online Users We use cookies to personalise content and ads, to provide social media features and | to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details |
https://www.studyadda.com/notes/11th-class/mental-ability/geometry/notes-geometry/18794 | 1,670,098,665,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710936.10/warc/CC-MAIN-20221203175958-20221203205958-00262.warc.gz | 1,066,407,409 | 23,379 | # 11th Class Mental Ability Geometry Notes - Geometry
Notes - Geometry
Category : 11th Class
# Geometry
Learning Objectives
• Geometry
Geometry
Geometry is the visual study of shapes, sizes, patterns, and positions. It occurred in all cultures, through at least one of these five strands of human activities: The following formulas and relationships are important in solving geometry problems.
Angle Relationships
1. The base angles of an isosceles triangle are equal
2. The sum of the measures of the interior angles of any n-sided polygon is 180(n - 2) degrees.
3. The sum of the measures of the exterior angles of any n-sided polygon is 360°.
4. If two parallel lines are cut by a transversal, the alternate interior angles are equal, and the corresponding angles are equal.
Angle Measurement Theorems
1. A central angle of a circle is measured by its intercepted arc.
2. An inscribed angle in a circle is measured by one-half of its intercepted arc,
3. An angle formed by two chords intersecting within a circle is measured by one-half the sum of the opposite intercepted arcs.
4. An angle formed by a tangent and a chord is measured by one-half its intercepted arc.
5. An angle formed by two secants, or by two tangents, or by a tangent and a secant, is measured by one-half the difference of the intercepted arcs.
Proportion Relationships
1. A line parallel to one side of triangle divides the other two sides proportionally.
2. In two similar triangles, corresponding sides, medians, altitudes, and angle bisectors are proportional.
3. If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other.
4. If a tangent and a secant are drawn to a circle from an outside point, the tangent is the mean proportional between the secant and the external segment.
5. In similar polygons the perimeters have the same ratio as any pair of corresponding sides.
Triangle Relationships
1. If an altitude is drawn to the hypotenuse of a right triangle, it is the mean proportional between the segments of the hypotenuse, and either leg is the mean proportional between the hypotenuse and the segment adjacent to that leg.
2. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. (Remember the Pythagorean triples: 3, 4, 5; 5, 12, 13.)
3. In a 30°- 60° right triangle, the leg opposite the 30° angle is one-half the hypotenuse, and the leg opposite the 60° angle is one-half the hypotenuse times $\sqrt{3}\,.$
4. In a right isosceles triangle the hypotenuse is equal to either leg times $\sqrt{2}\,.$
5. In an equilateral triangle of sides, the altitude equals $\frac{s}{2}\sqrt{3}\,.$
If the sides of a triangle are produced then the sum of the exterior angles i.e.
$\angle DAB+\angle EBC+\angle FCA$is equal to:
(a) 180° (b) 270°
(c) 360° (d) 240°
(e) None of these
Ans. (c)
Explanation: Sum of exterior angles = 360°
2. In a $\Delta \mathbf{ABC},\text{ }\angle \mathbf{BAC>90{}^\circ },$then $\angle \mathbf{ABC}$and $\angle \mathbf{ACB}$must be:
(a) acute (b) obtuse
(c) one acute and one obtuse (d) can't be determined
(e) None of these
Ans. (a)
Explanation: $\angle ABC+\angle ACB<90{}^\circ$
1. If the angles of a triangle are in the ratio 1:4:7, then the value of the largest is:
(a) 135° (b) 84°
(c) 105° (d) 110°
(e) None of these
Ans. (c)
Explanation: $x+4x+7x=180{}^\circ$
$\Rightarrow$ $x=15{}^\circ$
$\therefore$ $7x=105{}^\circ$
In the adjoining figure of $\Delta \mathbf{ABC}=\mathbf{12}{{\mathbf{0}}^{\mathbf{{}^\circ }}}$ and $\mathbf{AB=c,}\,\mathbf{BC=a,}\,\,\mathbf{AC=b}$then:
(a) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}+ba$ (b) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-ba$
(c) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ba$ (d) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}+2ba$
(e) None of these
2. What is the ratio of side and height of and equilateral triangle?
(a) $2:1$ (b) $1:1$
(c) $2:\sqrt{\text{3}}$ (d) $\sqrt{3}:2$
(e) None of these
In the$\Delta \mathbf{ABC}$, BD bisects$\angle \mathbf{B}$, and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, then find the value of x and y:
(a) 6, 12 (b) 10, 12
(c) 16, 8 (d) 8, 15
(e) None of these
In the given diagram AB||CD, then which one of the following is true?
(a) $\frac{AB}{AC}=\frac{AO}{OC}$ (b) $\frac{AB}{CD}=\frac{BO}{OD}$
(c) $\Delta AOB-\Delta COD$ (d) Ail of these
(e) None of these
In the figure BC||AD. Find the value of x:
(a) 9, 10 (b) 7, 8
(c) 10, 12 (d) 8, 9
(e) None of these
1. Find the maximum area that can be enclosed in a triangle of perimetre 24 cm:
(a) $32c{{m}^{2}}$
(b) $16\text{ }\sqrt{3}\text{ }c{{m}^{2}}$
(c) $16\text{ }\sqrt{\text{2}}\,\,c{{m}^{2}}$
(d) $27\text{ }c{{m}^{2}}$
(e) None of these
1. If one of the interior angles of a regular polygon is equal to 5/6 times of one of the interior angles of a regular pentagon, then the number of sides of the polygon is:
(a) 3 (b) 4
(c) 6 (d) 8
(e) None of these
1. If each interior angle of a regular polygon is 3 times its exterior angle, the number of sides of the polygon is:
(a) 4 (b) 5
(c) 6 (d) 8
(e) None of these
In the adjoining figure, O is the centre of circle and diametre AC = 26 cm. If chord AB = 10 cm, then the distance between chord AB and centre O of the circle is:
(a) 24cm (b) 16cm
(c) 12cm (d) 11cm
(e) None of these
1. A polygon has 54 diagonals. The number of sides in the polygon is :
(a) 7 (b) 9
(c) 12 (d) 11
(e) None of these
#### Other Topics
##### Notes - Geometry
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You will be redirected in 3 sec | 1,886 | 6,689 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.92511 | # 11th Class Mental Ability Geometry Notes - Geometry Notes - Geometry Category : 11th Class # Geometry Learning Objectives • Geometry Geometry Geometry is the visual study of shapes, sizes, patterns, and positions. It occurred in all cultures, through at least one of these five strands of human activities: The following formulas and relationships are important in solving geometry problems. Angle Relationships 1. The base angles of an isosceles triangle are equal 2. The sum of the measures of the interior angles of any n-sided polygon is 180(n - 2) degrees. 3. The sum of the measures of the exterior angles of any n-sided polygon is 360°. 4. If two parallel lines are cut by a transversal, the alternate interior angles are equal, and the corresponding angles are equal. Angle Measurement Theorems 1. A central angle of a circle is measured by its intercepted arc. 2. An inscribed angle in a circle is measured by one-half of its intercepted arc, 3. An angle formed by two chords intersecting within a circle is measured by one-half the sum of the opposite intercepted arcs. 4. An angle formed by a tangent and a chord is measured by one-half its intercepted arc. 5. An angle formed by two secants, or by two tangents, or by a tangent and a secant, is measured by one-half the difference of the intercepted arcs. Proportion Relationships 1. A line parallel to one side of triangle divides the other two sides proportionally. 2. In two similar triangles, corresponding sides, medians, altitudes, and angle bisectors are proportional. 3. If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other. 4. If a tangent and a secant are drawn to a circle from an outside point, the tangent is the mean proportional between the secant and the external segment. 5. In similar polygons the perimeters have the same ratio as any pair of corresponding sides. Triangle Relationships 1. If an altitude is drawn to the hypotenuse of a right triangle, it is the mean proportional between the segments of the hypotenuse, and either leg is the mean proportional between the hypotenuse and the segment adjacent to that leg. 2. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. (Remember the Pythagorean triples: 3, 4, 5; 5, 12, 13.) 3. In a 30°- 60° right triangle, the leg opposite the 30° angle is one-half the hypotenuse, and the leg opposite the 60° angle is one-half the hypotenuse times $\sqrt{3}\,.$ 4. In a right isosceles triangle the hypotenuse is equal to either leg times $\sqrt{2}\,.$ 5. In an equilateral triangle of sides, the altitude equals $\frac{s}{2}\sqrt{3}\,.$ If the sides of a triangle are produced then the sum of the exterior angles i.e. $\angle DAB+\angle EBC+\angle FCA$is equal to: (a) 180° (b) 270° (c) 360° (d) 240° (e) None of these Ans. (c) Explanation: Sum of exterior angles = 360° 2. In a $\Delta \mathbf{ABC},\text{ }\angle \mathbf{BAC>90{}^\circ },$then $\angle \mathbf{ABC}$and $\angle \mathbf{ACB}$must be: (a) acute (b) obtuse (c) one acute and one obtuse (d) can't be determined (e) None of these Ans. (a) Explanation: $\angle ABC+\angle ACB<90{}^\circ$ 1. If the angles of a triangle are in the ratio 1:4:7, then the value of the largest is: (a) 135° (b) 84° (c) 105° (d) 110° (e) None of these Ans. (c) Explanation: $x+4x+7x=180{}^\circ$ $\Rightarrow$ $x=15{}^\circ$ $\therefore$ $7x=105{}^\circ$ In the adjoining figure of $\Delta \mathbf{ABC}=\mathbf{12}{{\mathbf{0}}^{\mathbf{{}^\circ }}}$ and $\mathbf{AB=c,}\,\mathbf{BC=a,}\,\,\mathbf{AC=b}$then: (a) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}+ba$ (b) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-ba$ (c) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ba$ (d) ${{c}^{2}}={{a}^{2}}+{{b}^{2}}+2ba$ (e) None of these 2. What is the ratio of side and height of and equilateral triangle? (a) $2:1$ (b) $1:1$ (c) $2:\sqrt{\text{3}}$ (d) $\sqrt{3}:2$ (e) None of these In the$\Delta \mathbf{ABC}$, BD bisects$\angle \mathbf{B}$, and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, then find the value of x and y: (a) 6, 12 (b) 10, 12 (c) 16, 8 (d) 8, 15 (e) None of these In the given diagram AB||CD, then which one of the following is true? (a) $\frac{AB}{AC}=\frac{AO}{OC}$ (b) $\frac{AB}{CD}=\frac{BO}{OD}$ (c) $\Delta AOB-\Delta COD$ (d) Ail of these (e) None of these In the figure BC||AD. Find the value of x: (a) 9, 10 (b) 7, 8 (c) 10, 12 (d) 8, 9 (e) None of these 1. Find the maximum area that can be enclosed in a triangle of perimetre 24 cm: (a) $32c{{m}^{2}}$ (b) $16\text{ }\sqrt{3}\text{ }c{{m}^{2}}$ (c) $16\text{ }\sqrt{\text{2}}\,\,c{{m}^{2}}$ (d) $27\text{ }c{{m}^{2}}$ (e) None of these 1. If one of the interior angles of a regular polygon is equal to 5/6 times of one of the interior angles of a regular pentagon, then the number of sides of the polygon is: (a) 3 (b) 4 (c) 6 (d) 8 (e) None of these 1. If each interior angle of a regular polygon is 3 times its exterior angle, the number of sides of the polygon is: (a) 4 (b) 5 (c) 6 (d) 8 (e) None of these In | the adjoining figure, O is the centre of circle and diametre AC = 26 cm. If chord AB = 10 cm, then the distance between chord AB and centre O of the circle is: (a) 24cm (b) 16cm (c) 12cm (d) 11cm (e) None of these 1. A polygon has 54 diagonals. The number of sides in the polygon is : (a) 7 (b) 9 (c) 12 (d) 11 (e) None of these #### Other Topics ##### Notes - Geometry You need to login to perform this action. You will be redirected in 3 sec |
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## What is the meaning of consistency in statistics?
In statistics, consistency of procedures, such as computing confidence intervals or conducting hypothesis tests, is a desired property of their behaviour as the number of items in the data set to which they are applied increases indefinitely.
## What are measures of consistency?
In statistics and research, internal consistency is typically a measure based on the correlations between different items on the same test (or the same subscale on a larger test). It measures whether several items that propose to measure the same general construct produce similar scores.
What is the meaning consistency concept?
Meaning of consistency concept in English According to the consistency concept, once a business has decided on a particular method for treating an accounting item, it will treat all similar items in the same way in the future.
What is consistent in probability?
An estimator of a given parameter is said to be consistent if it converges in probability to the true value of the parameter as the sample size tends to infinity.
### How do you find consistency in statistics?
A simple test of consistency is that all frequencies should be positive. If any frequency is negative, it means that there is inconsistency in the sample data. If the data is consistent, all the ultimate class frequencies will be positive.
### Why is consistency important in statistics?
Consistency is important mainly with observational data where there is no possibility of repetition. Here, at least we want to know that if the sample is large the single estimate we will obtain will be really close to the true value with high probability, and it is consistency that guarantees that.
How do you measure consistency in statistics?
Typical measures of data consistency include statistics such as the range (i.e., the largest value minus the smallest value among a distribution of data), the variance (i.e., the sum of the squared deviations of each value in a distribution from the mean value in a distribution divided by the number of values in a …
What is consistency in accounting example?
Example of the consistency principle: Its accounting policies for depreciation are using a straight-line basis. In 2014 and 2015, it uses a straight line. But, the company subsequently wants to change its accounting policies from a straight line to a declining balance.
## Why is consistency important in accounting?
Consistency concept is important because of the need for comparability, that is, it enables investors and other users of financial statements to easily and correctly compare the financial statements of a company.
## How is data consistency calculated?
Consistency degree is a measure to quantify the degree of consistency between two data. To normalize the consistency degree, we define the value of consistency degree C which is between 0 and 1, that is, C ∈ [0,1]. The higher the consistency degree is, the more consistent the two data are.
Why do we need consistency in statistics?
4 Answers. If the estimator is not consistent, it won’t converge to the true value in probability. In other words, there is always a probability that your estimator and true value will have a difference, no matter how many data points you have.
How do you calculate consistency in statistics?
Consistency of an estimator is described by the asymptotic bias (the difference between the true parameter or estimate and the probability limit of the estimator). If this bias is zero then the estimator is consistent.
### How to measure consistency?
– For Time Series Data , Stationary Analysis can be done . If the data is non-stationary then it is likely to have some degree of inconsistency . – For multivariate data . outlier , erratic , missing value analysis can be done to find the Consistency . – Consistency cannot be computed with defining the “conf
### What is consistent in statistics?
A consistent estimate has insignificant errors (variations) as sample sizes grow larger. More specifically, the probability that those errors will vary by more than a given amount approaches zero as the sample size increases.
Why is data consistency so important?
Why is Data Consistency Important? Data consistency could be the difference between great business success or great failure. Data is the foundation for successful business decisions, and inconsistent data can lead to misinformed business decisions. It is crucial for enterprises to ensure data consistency, especially when aggregating data from
What is the meaning of ‘sufficient statistics’?
What is a Sufficient Statistic? A sufficient statistic summarizes all of the information in a sample about a chosen parameter. For example, the sample mean, x̄, estimates the population mean, μ. x̄ is a sufficient statistic if it retains all of the information about the population mean that was contained in the original data points. | 942 | 5,000 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-33 | latest | en | 0.951001 | # What is the meaning of consistency in statistics? ## What is the meaning of consistency in statistics? In statistics, consistency of procedures, such as computing confidence intervals or conducting hypothesis tests, is a desired property of their behaviour as the number of items in the data set to which they are applied increases indefinitely. ## What are measures of consistency? In statistics and research, internal consistency is typically a measure based on the correlations between different items on the same test (or the same subscale on a larger test). It measures whether several items that propose to measure the same general construct produce similar scores. What is the meaning consistency concept? Meaning of consistency concept in English According to the consistency concept, once a business has decided on a particular method for treating an accounting item, it will treat all similar items in the same way in the future. What is consistent in probability? An estimator of a given parameter is said to be consistent if it converges in probability to the true value of the parameter as the sample size tends to infinity. ### How do you find consistency in statistics? A simple test of consistency is that all frequencies should be positive. If any frequency is negative, it means that there is inconsistency in the sample data. If the data is consistent, all the ultimate class frequencies will be positive. ### Why is consistency important in statistics? Consistency is important mainly with observational data where there is no possibility of repetition. Here, at least we want to know that if the sample is large the single estimate we will obtain will be really close to the true value with high probability, and it is consistency that guarantees that. How do you measure consistency in statistics? Typical measures of data consistency include statistics such as the range (i.e., the largest value minus the smallest value among a distribution of data), the variance (i.e., the sum of the squared deviations of each value in a distribution from the mean value in a distribution divided by the number of values in a … What is consistency in accounting example? Example of the consistency principle: Its accounting policies for depreciation are using a straight-line basis. In 2014 and 2015, it uses a straight line. But, the company subsequently wants to change its accounting policies from a straight line to a declining balance. ## Why is consistency important in accounting? Consistency concept is important because of the need for comparability, that is, it enables investors and other users of financial statements to easily and correctly compare the financial statements of a company. ## How is data consistency calculated? Consistency degree is a measure to quantify the degree of consistency between two data. To normalize the consistency degree, we define the value of consistency degree C which is between 0 and 1, that is, C ∈ [0,1]. The higher the consistency degree is, the more consistent the two data are. Why do we need consistency in statistics? 4 Answers. If the estimator is not consistent, it won’t converge to the true value in probability. In other words, there is always a probability that your estimator and true value will have a difference, no matter how many data points you have. How do you calculate consistency in statistics? Consistency of an estimator is described by the asymptotic bias (the difference between the true parameter or estimate and the probability limit of the estimator). If this bias is zero then the estimator is consistent. ### How to measure consistency? – For Time Series Data , Stationary Analysis can be done . If the data is non-stationary then it is likely to have some degree of inconsistency . – For multivariate data . outlier , erratic , missing value analysis can be done to find the Consistency . – Consistency cannot be computed with defining the “conf ### What is consistent in statistics? A consistent estimate has insignificant errors (variations) as sample sizes grow larger. More specifically, the probability that those errors will vary by more than a given amount approaches zero as the sample size increases. Why is data consistency so important? Why is Data Consistency Important? Data consistency could be the difference between great business success or great failure. Data is the foundation for successful business decisions, and inconsistent data can | lead to misinformed business decisions. It is crucial for enterprises to ensure data consistency, especially when aggregating data from What is the meaning of ‘sufficient statistics’? What is a Sufficient Statistic? A sufficient statistic summarizes all of the information in a sample about a chosen parameter. For example, the sample mean, x̄, estimates the population mean, μ. x̄ is a sufficient statistic if it retains all of the information about the population mean that was contained in the original data points. |
http://www.miniphysics.com/2012/05/coordinate-transformation-under-rotation.html | 1,386,773,690,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164037762/warc/CC-MAIN-20131204133357-00053-ip-10-33-133-15.ec2.internal.warc.gz | 425,340,238 | 14,886 | # Coordinate Transformation Under Rotation
Rotation of object relative to FIXED axis:
Basic equations you can get by looking at the diagram above:
$x_{1} = r \cos {\alpha}$ $x_{2} = r \cos {(\theta + \alpha)}$
$y_{1} = r \sin {\alpha}$ $y_{2} = r \sin {(\theta + \alpha)}$
Using the equations above:
\begin{align} x_{2} &= r \cos {(\theta + \alpha)} \\ &= r\cos \theta \cos \alpha – r\sin \theta \sin \alpha \\ &= (r\cos \alpha) \cos \theta – (r \sin \alpha) \sin \theta \\ &= x_{1} \cos \theta – y_{1} \sin \theta \end{align}
\begin{align} y_{2} &= r \sin {(\theta + \alpha)} \\ &= r\sin \theta \cos \alpha + r\cos \theta \sin \alpha \\ &= (r\cos \alpha) \cos \theta + (r \sin \alpha) \cos \theta \\ &= x_{1} \sin \theta + y_{1} \cos \theta \end{align}
Hence, For an anti-clockwise rotation,
$\begin{pmatrix} x_{2} \\ y_{2} \end{pmatrix} = \begin{pmatrix} \cos \theta & – \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x_{1} \\ y_{1} \end{pmatrix}$
$\begin{pmatrix} \cos \theta & – \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ is called the rotation matrix. Its determinant is 1.
To find the clockwise rotation matrix, you can do the calculations again. OR you can just transpose the above matrix.
Hence, the clockwise rotation matrix is: $\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$
Rotation of coordinate axes:
Basic equations:
$x’ = r \cos \alpha$ $x = r \cos (\theta + \alpha)$
$y’ = r \sin \alpha$ $y = r \sin (\theta + \alpha)$
Using the basic equations:
Equation 1: \begin{align}x &= r\cos (\theta + \alpha) \\ &= r\cos \theta \cos \alpha – r \sin \theta \sin \alpha \\ &= x’ \cos \theta – y’ \sin \theta \end{align}
Equation 2: \begin{align}y &= r\sin (\theta + \alpha) \\ &= r\sin \theta \cos \alpha + r \cos \theta \sin \alpha \\ &= x’ \sin \theta + y’ \cos \theta \end{align}
From equation 2,
$x’ = \frac{y – y’ \cos \theta}{\sin \theta}$ $y’ = \frac{y – x’ \sin \theta}{\cos \theta}$
Substitute the above 2 equations into equation 1 and you will get:
$y’ = -x \sin \theta + y\cos \theta$ $x’ = x \cos \theta + y \sin \theta$
Hence,
$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \cos \theta & – \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x’ \\ y’ \end{pmatrix}$
$\begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$
Back To Mathematics For An Undergraduate Physics Course | 904 | 2,647 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2013-48 | longest | en | 0.375076 | # Coordinate Transformation Under Rotation Rotation of object relative to FIXED axis: Basic equations you can get by looking at the diagram above: $x_{1} = r \cos {\alpha}$ $x_{2} = r \cos {(\theta + \alpha)}$ $y_{1} = r \sin {\alpha}$ $y_{2} = r \sin {(\theta + \alpha)}$ Using the equations above: \begin{align} x_{2} &= r \cos {(\theta + \alpha)} \\ &= r\cos \theta \cos \alpha – r\sin \theta \sin \alpha \\ &= (r\cos \alpha) \cos \theta – (r \sin \alpha) \sin \theta \\ &= x_{1} \cos \theta – y_{1} \sin \theta \end{align} \begin{align} y_{2} &= r \sin {(\theta + \alpha)} \\ &= r\sin \theta \cos \alpha + r\cos \theta \sin \alpha \\ &= (r\cos \alpha) \cos \theta + (r \sin \alpha) \cos \theta \\ &= x_{1} \sin \theta + y_{1} \cos \theta \end{align} Hence, For an anti-clockwise rotation, $\begin{pmatrix} x_{2} \\ y_{2} \end{pmatrix} = \begin{pmatrix} \cos \theta & – \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x_{1} \\ y_{1} \end{pmatrix}$ $\begin{pmatrix} \cos \theta & – \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$ is called the rotation matrix. Its determinant is 1. To find the clockwise rotation matrix, you can do the calculations again. OR you can just transpose the above matrix. Hence, the clockwise rotation matrix is: $\begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}$ Rotation of coordinate axes: Basic equations: $x’ = r \cos \alpha$ $x = r \cos (\theta + \alpha)$ $y’ = r \sin \alpha$ $y = r \sin (\theta + \alpha)$ Using the basic equations: Equation 1: \begin{align}x &= r\cos (\theta + \alpha) \\ &= r\cos \theta \cos \alpha – r \sin \theta \sin \alpha \\ &= x’ \cos \theta – y’ \sin \theta \end{align} Equation 2: \begin{align}y &= r\sin (\theta + \alpha) \\ &= r\sin \theta \cos \alpha + r \cos \theta \sin \alpha \\ &= x’ \sin \theta + y’ \cos \theta \end{align} From equation 2, $x’ = \frac{y – y’ \cos \theta}{\sin \theta}$ $y’ = \frac{y – x’ \sin \theta}{\cos \theta}$ Substitute the above 2 equations into equation 1 and you will get: $y’ = -x \sin \theta + y\cos \theta$ $x’ = x \cos \theta + y \sin \theta$ Hence, $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \cos \theta & – \sin | \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x’ \\ y’ \end{pmatrix}$ $\begin{pmatrix} x’ \\ y’ \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}$ Back To Mathematics For An Undergraduate Physics Course |
https://dizz.com/stable-neutral-unstable-equilibrium-pdf/ | 1,719,081,732,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862410.56/warc/CC-MAIN-20240622175245-20240622205245-00210.warc.gz | 184,934,924 | 84,240 | # Condition of Equilibrium and Stability of Floating Bodies- Stable, Neutral, Unstable Equilibrium [PDF]
Contents
## Introduction
In this article, we will delve into the fascinating world of floating bodies and explore the concepts of equilibrium and stability. When a body is placed on a fluid surface, it experiences various forces that determine whether it will float, sink or stay in equilibrium. The stability of a floating body depends on its center of gravity and the position of the center of buoyancy. Understanding the conditions of stable, neutral, and unstable equilibrium is critical for engineers and designers who work with floating structures, such as ships, boats, and oil rigs. By the end of this article, you will have a deeper understanding of the principles that govern the stability of floating bodies and the factors that affect them.
## Archimedes principle
The Archimedes principle explains about the force acting on a floating body in a liquid. It states that “The body immersed or floating in a liquid are acted upon by a vertical upward liquid force equal to the weight of the liquid displaced”. This vertical upward force is called buoyancy or buoyant force. The point, through which this force acts, is known as a center of buoyancy.
A body floating or immersed in a liquid will lose its weight equal to the buoyant force of the liquid. The body has less weight in a liquid than outside. Archimedes principle is applicable to bodies floating or immersed in a fluid. This has been used by man for about 2200 years, for the problem of general floatation and naval architectural design.
## Stability of Floating Bodies
When a body floats, it is subjected to two parallel forces which are as follows:
1. The downward force of gravity acting on each of the particles.
2. The upward buoyant force of the liquid acting on various elements of the submerged surface.
If the body is to float in equilibrium in an upright position the resultant of these two forces must be collinear, equal and opposite.
Hence center of gravity of the floating body and center of buoyancy must lie in the same vertical line. B is the center of buoyancy which is the center of gravity of the area ACO, and G is the center of gravity of the body. If the ship in the (figure above) heels through an angle θ (fig b), due to tilting moments caused by wind or wave action or due to movement of loads across the deck, portion A’C’O’ will now stand immersed in water.
The center of buoyancy will shift from B to B’. The buoyant force will act through B’. The center of gravity G will of course not change and W will continue to act through it.
#### What is Metacenter?
A vertical line through the new center of buoyancy intersects the inclined axis of the ship at M which is known as Metacenter. The term metacenter is defined as the point at which a vertical line through the center of buoyancy intersects the vertical center line of the ship section, after a small angle of heal.
#### What is Metacentric Height?
The distance between the center of gravity and metacenter is called metacentric height. The metacentric height is a measure of the static stability of the ship. For the small angle of inclination, the position of M does not change materially and the metacentric height is approximately constant. Hence the ship may be considered as rotating about M. In other words, the ship may be considered as behaving like a pendulum suspended at M, the point G, corresponding to bob.
## What do you mean by equilibrium?
Equilibrium of floating bodies refers to the state of balance or stability that is achieved when a body is submerged in a fluid and is neither sinking nor rising. In other words, it is the state in which the weight of the object is equal to the buoyant force acting on it. The buoyant force is the force exerted by the fluid on the object, which is equal to the weight of the fluid displaced by the object. This phenomenon is known as Archimedes’ principle.
When a body is placed in a fluid, it experiences an upward force, which is known as the buoyant force. This force is equal to the weight of the fluid displaced by the object. If the weight of the object is less than the buoyant force, it will float on the surface of the fluid. However, if the weight of the object is greater than the buoyant force, it will sink.
In the case of floating bodies, the weight of the object is equal to the buoyant force. This means that the object is in a state of balance or equilibrium. If the object is pushed down, the buoyant force will increase, and if the object is pushed up, the buoyant force will decrease. As a result, the object will return to its original position of equilibrium.
The equilibrium of floating bodies has many practical applications, such as in the design of boats and ships. In order to ensure that a boat or ship is stable and does not tip over, the weight of the vessel must be distributed evenly and the center of gravity must be located below the center of buoyancy. This ensures that the buoyant force acting on the vessel is greater than its weight, and it remains in a state of equilibrium.
## The condition of Equilibrium of Floating Bodies:
By the condition of equilibrium of floating bodies, we mean the possible state of stability or instability of floating bodies under all odds. Following are the three conditions of equilibrium of floating bodies:
• Stable Equilibrium
• Neutral Equilibrium
• Unstable Equilibrium
Let us discuss each condition in detail.
### Stable Equilibrium:
In the above figure (b) we found that when the ship was subjected to turning moments, the center of buoyancy changed from B to B’. Further, From the same diagram, we also notice that W and F are two equal and opposite parallel forces acting at a distance X apart. Naturally, this causes anticlockwise couples WX, tending to restore the ship to its original position.
In this case, the ship is said to be in stable equilibrium. Hence it may be stated that a floating body is said to be in a state of stable equilibrium which, when subjected to turning moments, leads to regaining its original position and that M lies above G or BM>BG.
### Neutral Equilibrium:
If the ship in figure (b) is tilted or rolled over such that the new center of buoyancy B’ lies on the line of action of W, the buoyant force F and weight W would be collinear, equal and opposite and would not exert any restoring moment. In that case, the ship would neither tend to regain its original position nor would tend to heel over further.
Hence it may be stated that “a floating body is said to be in a state of neutral equilibrium which, when subjected to turning moment, neither tends to regain its original position, nor tend to heel over further but instead keeps on in the tilted position, and that M coincides with G or BM=BG”.
### Unstable Equilibrium:
In this case, the new center of buoyancy B’ lies in between B and the line of action of W. The vertical upward buoyant force F passing through B’ will intersect the inclines central line at the point M below G. The couple thus formed by W and F will be clockwise; further helping the turning moment of tilt over the ship further.
## FAQ’s
### What is the condition of equilibrium for a floating body?
It is the condition for a floating body when the body is neither sinking nor rising and is stable in its position on the fluid surface. This state occurs when the weight of the floating body is equal to the buoyant force acting on it.
### What are the factors that affect the stability of a floating body?
The stability of a floating body depends on several factors, including the position of the center of gravity, the position of the center of buoyancy, the shape and size of the body, and the density of the fluid.
### What is metacentric height, and why is it important for determining stability?
Metacentric height is the distance between the metacenter (the point at which a floating body’s center of gravity will be when it tilts) and the center of buoyancy (the center of mass of the displaced fluid). It is an important parameter for determining the stability of a floating body, as a higher metacentric height indicates greater stability.
### What is the difference between stable, neutral, and unstable for a floating body?
In stable equilibrium, the body will return to its original position when it is slightly tilted, whereas in neutral equilibrium, the body will remain in the new position after being tilted. Unstable equilibrium occurs when the body will continue to tilt until it capsizes, and it cannot return to its original position.
## Conclusion
In conclusion, the condition of equilibrium and stability of floating bodies is a crucial concept in physics. It is essential to understand the forces acting on a body and their effects on its stability to prevent accidents and ensure safety in various activities involving floating bodies. By understanding the frequently asked questions about this topic, we can develop a better understanding of the principles involved and apply them to practical situations. | 1,895 | 9,134 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.918895 | # Condition of Equilibrium and Stability of Floating Bodies- Stable, Neutral, Unstable Equilibrium [PDF] Contents ## Introduction In this article, we will delve into the fascinating world of floating bodies and explore the concepts of equilibrium and stability. When a body is placed on a fluid surface, it experiences various forces that determine whether it will float, sink or stay in equilibrium. The stability of a floating body depends on its center of gravity and the position of the center of buoyancy. Understanding the conditions of stable, neutral, and unstable equilibrium is critical for engineers and designers who work with floating structures, such as ships, boats, and oil rigs. By the end of this article, you will have a deeper understanding of the principles that govern the stability of floating bodies and the factors that affect them. ## Archimedes principle The Archimedes principle explains about the force acting on a floating body in a liquid. It states that “The body immersed or floating in a liquid are acted upon by a vertical upward liquid force equal to the weight of the liquid displaced”. This vertical upward force is called buoyancy or buoyant force. The point, through which this force acts, is known as a center of buoyancy. A body floating or immersed in a liquid will lose its weight equal to the buoyant force of the liquid. The body has less weight in a liquid than outside. Archimedes principle is applicable to bodies floating or immersed in a fluid. This has been used by man for about 2200 years, for the problem of general floatation and naval architectural design. ## Stability of Floating Bodies When a body floats, it is subjected to two parallel forces which are as follows: 1. The downward force of gravity acting on each of the particles. 2. The upward buoyant force of the liquid acting on various elements of the submerged surface. If the body is to float in equilibrium in an upright position the resultant of these two forces must be collinear, equal and opposite. Hence center of gravity of the floating body and center of buoyancy must lie in the same vertical line. B is the center of buoyancy which is the center of gravity of the area ACO, and G is the center of gravity of the body. If the ship in the (figure above) heels through an angle θ (fig b), due to tilting moments caused by wind or wave action or due to movement of loads across the deck, portion A’C’O’ will now stand immersed in water. The center of buoyancy will shift from B to B’. The buoyant force will act through B’. The center of gravity G will of course not change and W will continue to act through it. #### What is Metacenter? A vertical line through the new center of buoyancy intersects the inclined axis of the ship at M which is known as Metacenter. The term metacenter is defined as the point at which a vertical line through the center of buoyancy intersects the vertical center line of the ship section, after a small angle of heal. #### What is Metacentric Height? The distance between the center of gravity and metacenter is called metacentric height. The metacentric height is a measure of the static stability of the ship. For the small angle of inclination, the position of M does not change materially and the metacentric height is approximately constant. Hence the ship may be considered as rotating about M. In other words, the ship may be considered as behaving like a pendulum suspended at M, the point G, corresponding to bob. ## What do you mean by equilibrium? Equilibrium of floating bodies refers to the state of balance or stability that is achieved when a body is submerged in a fluid and is neither sinking nor rising. In other words, it is the state in which the weight of the object is equal to the buoyant force acting on it. The buoyant force is the force exerted by the fluid on the object, which is equal to the weight of the fluid displaced by the object. This phenomenon is known as Archimedes’ principle. When a body is placed in a fluid, it experiences an upward force, which is known as the buoyant force. This force is equal to the weight of the fluid displaced by the object. If the weight of the object is less than the buoyant force, it will float on the surface of the fluid. However, if the weight of the object is greater than the buoyant force, it will sink. In the case of floating bodies, the weight of the object is equal to the buoyant force. This means that the object is in a state of balance or equilibrium. If the object is pushed down, the buoyant force will increase, and if the object is pushed up, the buoyant force will decrease. As a result, the object will return to its original position of equilibrium. The equilibrium of floating bodies has many practical applications, such as in the design of boats and ships. In order to ensure that a boat or ship is stable and does not tip over, the weight of the vessel must be distributed evenly and the center of gravity must be located below the center of buoyancy. This ensures that the buoyant force acting on the vessel is greater than its weight, and it remains in a state of equilibrium. ## The condition of Equilibrium of Floating Bodies: By the condition of equilibrium of floating bodies, we mean the possible state of stability or instability of floating bodies under all odds. Following are the three conditions of equilibrium of floating bodies: • Stable Equilibrium • Neutral Equilibrium • Unstable Equilibrium Let us discuss each condition in detail. ### Stable Equilibrium: In the above figure (b) we found that when the ship was subjected to turning moments, the center of buoyancy changed from B to B’. Further, From the same diagram, we also notice that W and F are two equal and opposite parallel forces acting at a distance X apart. Naturally, this causes anticlockwise couples WX, tending to restore the ship to its original position. In this case, the ship is said to be in stable equilibrium. Hence it may be stated that a floating body is said to be in a state of stable equilibrium which, when subjected to turning moments, leads to regaining its original position and that M lies above G or BM>BG. ### Neutral Equilibrium: If the ship in figure (b) is tilted or rolled over such that the new center of buoyancy B’ lies on the line of action of W, the buoyant force F and weight W would be collinear, equal and opposite and would not exert any restoring moment. In that case, the ship would neither tend to regain its original position nor would tend to heel over further. Hence it may be stated that “a floating body is said to be in a state of neutral equilibrium which, when subjected to turning moment, neither tends to regain its original position, nor tend to heel over further but instead keeps on in the tilted position, and that M coincides with G or BM=BG”. ### Unstable Equilibrium: In this case, the new center of buoyancy B’ lies in between B and the line of action of W. The vertical upward buoyant force F passing through B’ will intersect the inclines central line at the point M below G. The couple thus formed by W and F will be clockwise; further helping the turning moment of tilt over the ship further. ## FAQ’s ### What is the condition of equilibrium for a floating body? It is the condition for a floating body when the body is neither sinking nor rising and is stable in its position on the fluid surface. This state occurs when the weight of the floating body is equal to the buoyant force acting on it. ### What are the factors that affect the stability of a floating body? The stability of a floating body depends on several factors, including the position of the center of gravity, the position of the center of buoyancy, the shape and size of the body, and the density of the fluid. ### What is metacentric height, and why is it important for determining stability? Metacentric height is the distance between the metacenter (the point at which a floating body’s center of gravity will be when it tilts) and the center of buoyancy (the center of mass of the displaced fluid). | It is an important parameter for determining the stability of a floating body, as a higher metacentric height indicates greater stability. ### What is the difference between stable, neutral, and unstable for a floating body? In stable equilibrium, the body will return to its original position when it is slightly tilted, whereas in neutral equilibrium, the body will remain in the new position after being tilted. Unstable equilibrium occurs when the body will continue to tilt until it capsizes, and it cannot return to its original position. ## Conclusion In conclusion, the condition of equilibrium and stability of floating bodies is a crucial concept in physics. It is essential to understand the forces acting on a body and their effects on its stability to prevent accidents and ensure safety in various activities involving floating bodies. By understanding the frequently asked questions about this topic, we can develop a better understanding of the principles involved and apply them to practical situations. |
https://collegephysicsanswers.com/openstax-solutions/600-kg-skier-initial-speed-120-ms-coasts-250-m-high-rise-shown-figure-739-find | 1,586,300,292,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371806302.78/warc/CC-MAIN-20200407214925-20200408005425-00359.warc.gz | 419,400,874 | 14,992 | Question
A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in Figure 7.39. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)
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Final Answer
$9.5 \textrm{ m/s}$
Solution Video
# OpenStax College Physics Solution, Chapter 7, Problem 24 (Problems & Exercises) (6:50)
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This is College Physics Answers with Shaun Dychko. This skier has some initial velocity of 12.0 meters per second and then they encounter a 35 degree slope. They go up the slope—a total height of 2.5 meters—and then they are skiing horizontally at the top and the question is what speed will they have on this top level assuming that there's a certain amount of friction due to a coefficient of friction of 0.0800 along this slope here. So we are going to figure out how much work is done by the non-conservative force, namely friction, along this slope and then that work is going to be the energy that is taken away from kinetic and potential energy that the skier had initially and then when they are at the top here, they will be left with just some potential and kinetic energy but the total will not be the total initial kinetic and potential energy because some of it will be taken away by the work done by the non-conservative force. There's a plus sign here which suggests that maybe this thing is adding energy but in fact, it's going to be negative because the friction and the displacement are in opposite directions and so this work will be negative. Okay! So the distance that they travel along the slope we can figure that out from this triangle here: we know that <i>sin</i> of this angle <i>Θ</i> is the opposite divided by the hypotenuse so that's <i>h</i> over <i>d</i> and then we can solve for <i>d</i> by multiplying both sides by <i>d</i> over <i>sin Θ</i>. We cancel the <i>sin Θ</i>'s there and cancel the <i>d</i>'s over here and we have <i>d</i> equals <i>h</i> over <i>sin Θ</i>. Okay! This is important because we are gonna substitute in for <i>d</i> here in this formula for the work done by the non-conservative force which is going to be the negative and I put negative there just because, you know, work is force times displacement times <i>cosine</i> of the angle between them and <i>cos</i> of 180 degrees is negative 1 or you could think of it as the displacement and the force are in opposite directions and that's why there's a negative there. Alright. The other factor here is this friction force and it's going to be their coefficient of friction multiplied by the normal force applied by the slope and the normal force is going to equal the component of gravity that is perpendicular to the slope. So gravity is straight down but there's this component here which is the force of gravity times <i>cos</i> of <i>Θ</i> which will be the perpendicular component of gravity and these two have to be equal because the skier is not accelerating perpendicular to the slope. Okay! So we have <i>μ kF N</i> and then substituting in the perpendicular component of gravity in place of normal force and we have <i>μ k</i> is times <i>mgcos Θ</i> is the friction force. So we can substitute for both friction force— <i>μ kmgcos Θ</i>—and substitute for the distance along the slope— which is <i>h</i> over <i>sin Θ</i>— and then we have this formula for the work done by friction. It's gonna be negative times coefficient of friction times <i>mgh</i> divided by <i>tangent Θ</i>— because <i>cos</i> over <i>sin</i>... there's an identity which says that <i>cos Θ</i> divided by <i>sin Θ</i> is the reciprocal of <i>tan Θ</i>. Okay! So all of this gets plugged into the work done by the non-conservative force here in our conservation of energy formula and then we can start plugging in for the other terms as well. So the initial kinetic energy is one-half mass times initial velocity squared plus the initial potential energy and we'll say that that is zero because we'll define this to be our reference level where <i>h</i> equals 0 so <i>mgh</i> would be <i>mg</i> times 0 and then adding to that... that should be a minus or maybe a plus for the formula and then a minus for the substitution but let's just resolve it into a single operation which is minus. Okay! And that equals one-half <i>mv f squared</i> plus <i>mgh</i>— this is the potential energy at the top of the slope and the kinetic energy at the top of the slope and this <i>v f</i> is what we are ultimately trying to find. So we'll make things look a little bit simpler by multiplying everything by 2 over <i>m</i>; it's kind of messy having fractions so we'll get rid of the fractions by multiplying by 2 and then we'll also get rid of the <i>m</i> which is a factor in all of the terms. So we are left with, after you switch the sides around, <i>v f squared</i> plus 2<i>gh</i>—the <i>m</i> is canceled and this 2 is canceled but the 2 appears here because it has to get distributed among both terms into the brackets— and that equals <i>v initial squared</i> minus 2<i>μ kgh</i> over <i>tan Θ</i>. I made a mistake here with the plus sign but nevertheless I corrected it anyway and it's correctly written as a minus here. Okay! So we are gonna subtract 2<i>gh</i> from both sides to solve for <i>v f squared</i> and we have this line and I factored out this common factor 2<i>gh</i> from both of these terms and it becomes minus 2<i>gh</i> times 1 plus <i>μ k</i> over <i>tan Θ</i> and then take the square root of both sides and that's all the algebra that we need to do. So <i>v f</i> is the square root of <i>v initial squared</i> minus 2<i>gh</i> times 1 plus <i>μ k</i> over <i>tan Θ</i>. So that's the square root of 12.0 meters per second squared minus 2 times 9.80 meters per second squared times 2.5 meters up the slope—that's the height of the slope— times 1 plus 0.0800—coefficient of friction— divided by <i>tan</i> of the slope angle of 35 degrees and this gives 9.5 meters per second. Now we are expecting some answer that is less than the initial speed so if we had had an answer more than 12, we would have said that doesn't pass the reality test but something less than 12 does. So 9.5 is plausibly the correct answer. | 1,694 | 6,479 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-16 | longest | en | 0.949676 | Question A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in Figure 7.39. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.) Question Image Question by OpenStax is licensed under CC BY 4.0. Final Answer $9.5 \textrm{ m/s}$ Solution Video # OpenStax College Physics Solution, Chapter 7, Problem 24 (Problems & Exercises) (6:50) #### Sign up to view this solution video! View sample solution ## Calculator Screenshots Video Transcript This is College Physics Answers with Shaun Dychko. This skier has some initial velocity of 12.0 meters per second and then they encounter a 35 degree slope. They go up the slope—a total height of 2.5 meters—and then they are skiing horizontally at the top and the question is what speed will they have on this top level assuming that there's a certain amount of friction due to a coefficient of friction of 0.0800 along this slope here. So we are going to figure out how much work is done by the non-conservative force, namely friction, along this slope and then that work is going to be the energy that is taken away from kinetic and potential energy that the skier had initially and then when they are at the top here, they will be left with just some potential and kinetic energy but the total will not be the total initial kinetic and potential energy because some of it will be taken away by the work done by the non-conservative force. There's a plus sign here which suggests that maybe this thing is adding energy but in fact, it's going to be negative because the friction and the displacement are in opposite directions and so this work will be negative. Okay! So the distance that they travel along the slope we can figure that out from this triangle here: we know that <i>sin</i> of this angle <i>Θ</i> is the opposite divided by the hypotenuse so that's <i>h</i> over <i>d</i> and then we can solve for <i>d</i> by multiplying both sides by <i>d</i> over <i>sin Θ</i>. We cancel the <i>sin Θ</i>'s there and cancel the <i>d</i>'s over here and we have <i>d</i> equals <i>h</i> over <i>sin Θ</i>. Okay! This is important because we are gonna substitute in for <i>d</i> here in this formula for the work done by the non-conservative force which is going to be the negative and I put negative there just because, you know, work is force times displacement times <i>cosine</i> of the angle between them and <i>cos</i> of 180 degrees is negative 1 or you could think of it as the displacement and the force are in opposite directions and that's why there's a negative there. Alright. The other factor here is this friction force and it's going to be their coefficient of friction multiplied by the normal force applied by the slope and the normal force is going to equal the component of gravity that is perpendicular to the slope. So gravity is straight down but there's this component here which is the force of gravity times <i>cos</i> of <i>Θ</i> which will be the perpendicular component of gravity and these two have to be equal because the skier is not accelerating perpendicular to the slope. Okay! So we have <i>μ kF N</i> and then substituting in the perpendicular component of gravity in place of normal force and we have <i>μ k</i> is times <i>mgcos Θ</i> is the friction force. So we can substitute for both friction force— <i>μ kmgcos Θ</i>—and substitute for the distance along the slope— which is <i>h</i> over <i>sin Θ</i>— and then we have this formula for the work done by friction. It's gonna be negative times coefficient of friction times <i>mgh</i> divided by <i>tangent Θ</i>— because <i>cos</i> over <i>sin</i>... there's an identity which says that <i>cos Θ</i> divided by <i>sin Θ</i> is the reciprocal of <i>tan Θ</i>. Okay! So all of this gets plugged into the work done by the non-conservative force here in our conservation of energy formula and then we can start plugging in for the other terms as well. So the initial kinetic energy is one-half mass times initial velocity squared plus the initial potential energy and we'll say that that is zero because we'll define this to be our reference level where <i>h</i> equals 0 so <i>mgh</i> would be <i>mg</i> times 0 and then adding to that... that should be a minus or maybe a plus for the formula and then a minus for the substitution but let's just resolve it into a single operation which is minus. Okay! And that equals one-half <i>mv f squared</i> plus <i>mgh</i>— this is the potential energy at the top of the slope and the kinetic energy at the top of the slope and this <i>v f</i> is what we are ultimately trying to find. So we'll make things look a little bit simpler by multiplying everything by 2 over <i>m</i>; it's kind of messy having fractions so we'll get rid of the fractions by multiplying by 2 and then we'll also get rid of the <i>m</i> which is a factor in all of the terms. So we are left with, after you switch the sides around, <i>v f squared</i> plus 2<i>gh</i>—the <i>m</i> is canceled and this 2 is canceled but the 2 appears here because it has to get distributed among both terms into the brackets— and that equals <i>v initial squared</i> minus 2<i>μ kgh</i> over <i>tan Θ</i>. I made a mistake here with the plus sign but nevertheless I corrected it anyway and it's correctly written as a minus here. Okay! So we are gonna subtract 2<i>gh</i> from both sides to solve for <i>v f squared</i> and we have this line and I factored out this common factor 2<i>gh</i> from both of these terms and it becomes minus 2<i>gh</i> times 1 plus <i>μ k</i> over <i>tan Θ</i> and then take the square root of both sides and that's all the algebra that we need to do. So <i>v f</i> is the | square root of <i>v initial squared</i> minus 2<i>gh</i> times 1 plus <i>μ k</i> over <i>tan Θ</i>. So that's the square root of 12.0 meters per second squared minus 2 times 9.80 meters per second squared times 2.5 meters up the slope—that's the height of the slope— times 1 plus 0.0800—coefficient of friction— divided by <i>tan</i> of the slope angle of 35 degrees and this gives 9.5 meters per second. Now we are expecting some answer that is less than the initial speed so if we had had an answer more than 12, we would have said that doesn't pass the reality test but something less than 12 does. So 9.5 is plausibly the correct answer. |
https://www.askiitians.com/forums/Algebra/let-a-min-x-2-4x-5-x-belongs-to-r-and-b_95667.htm | 1,675,829,973,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500671.13/warc/CC-MAIN-20230208024856-20230208054856-00317.warc.gz | 643,636,113 | 32,684 | # Let a = min( x^2 + 4x + 5, x belongs to R) and b= lim??0 1-cos 2?/?^2then find the value ofa-b/a+b .(here ? means theta)
Ajay Verma
8 years ago
solution:
a= min (x^2+ 4x +5)
( using dy/dx= 0 at critical points)
y= x^2+ 4x +5 ; dy/dx = 2x+4 = 0 so x = -2
d2y/dx2 = 2(-2) + 4 = 1 so minimum..
so a = 1
b = lim y --> 0 = (1- cos 2y)/ y2 : 0/0 form so use DL hospi low..
b= lim y-->0 = 2 sin2y / 2y ; lim y-->0 sin 2x/ 2x = 1
soo b= 2
a-b/ a+b = 1-2 / 1+2 = -1/3
Thanks and Regards,
Ajay verma, | 250 | 498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-06 | latest | en | 0.551462 | # Let a = min( x^2 + 4x + 5, x belongs to R) and b= lim??0 1-cos 2?/?^2then find the value ofa-b/a+b .(here ? means theta) Ajay Verma 8 years ago solution: a= min (x^2+ 4x +5) ( using dy/dx= 0 at critical points) y= x^2+ 4x +5 ; dy/dx = 2x+4 = 0 so x = -2 d2y/dx2 = 2(-2) + 4 = 1 so minimum.. so a = 1 b = lim y --> 0 = (1- cos 2y)/ y2 : 0/0 form so use DL hospi low.. b= lim y-->0 = 2 sin2y / 2y ; lim y-->0 sin 2x/ 2x = 1 soo b= 2 | a-b/ a+b = 1-2 / 1+2 = -1/3 Thanks and Regards, Ajay verma, |
http://mathhelpforum.com/higher-math/208596-solve-using-de-moivre-s-theorem-z-3-i-2.html | 1,527,479,466,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870771.86/warc/CC-MAIN-20180528024807-20180528044807-00112.warc.gz | 188,897,937 | 12,660 | # Thread: Solve using de Moivre's theorem: z^3=i
1. ## Re: Solve using de Moivre's theorem: z^3=i
Originally Posted by Prove It
First of all, when dealing with solving exponential equations, the EXPONENTIAL form of the complex number is ALWAYS easiest to use...
\displaystyle \begin{align*} z^3 &= \mathrm{i} \\ z^3 &= \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{3}} \\ z &= \mathrm{ e }^{ \frac{\pi}{6} + \frac{2\pi\,n}{3} } \end{align*}
So in the first cycle, the possible solutions are \displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}}, \mathrm{e}^{\frac{5\pi}{6}} , \mathrm{e}^{\frac{3\pi}{2}} \end{align*}, which in Cartesian form are:
\displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}} &= \cos{ \left( \frac{\pi}{6} \right) } + \mathrm{i}\sin{\left( \frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{5\pi}{6}} &= \cos{ \left( \frac{5\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\pi}{2}} &= \cos{ \left( \frac{3\pi}{2} \right) } + \mathrm{i}\sin{ \left( \frac{3\pi}{2} \right) } \\ &= -\mathrm{i} \end{align*}
I am really bad in this. Sorry guys. I cant really solve it. Can you try solve whole example and like for real bad student ?
For example how did you get from i e on something ?
If someone have time would be nice if he can solve this examples but every step what you do..
I have like 4-5 examples and cant really solve it. and i have exam in 2 days
Best would be way like" Prove it " solved it. We are learning it like that.
z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i
or is there some manual where they describe everything.
Anyway thats for help what you allready solved.
2. ## Re: Solve using de Moivre's theorem: z^3=i
If you can put your equation in "cis" form, then you can put the equation in exponential form. In general \displaystyle \begin{align*} r\,\mathrm{e}^{\mathrm{i}\,\theta} \equiv r \left[ \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \right] \end{align*}
3. ## Re: Solve using de Moivre's theorem: z^3=i
Originally Posted by Martinun
For example how did you get from i e on something ?
If someone have time would be nice if he can solve this examples but every step what you do..
I have like 4-5 examples and cant really solve it. and i have exam in 2 days
Best would be way like" Prove it " solved it. We are learning it like that.
z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i
You are a victim of poor mathematical education practice. There is a good trend to use a more geometric approach to teaching complex analysis.
First you need to understand that any complex has n nth roots. Those n roots are located on a circle centered at the origin of radius the nth root of absolute value of the number. Those n roots divide the circle into n equal arcs.
Let's stop a moment for an example. There are six sixth roots of $\displaystyle 3-4\bf{i}$. They are located on a circle centered at $\displaystyle (0,0)$ and radius $\displaystyle \sqrt[6]5$. On that circle the roots are separated by arcs of $\displaystyle \dfrac{2\pi}{6}$, where $\displaystyle 2\pi$ is the whole circle and six equal parts.
Now give some notation to simplify the way we can list the roots.
Let $\displaystyle \exp(i\theta)=\cos(\theta)+i\sin(\theta)$.
Then for a complex number $\displaystyle z=x+yi$ then $\displaystyle \exp(z)=e^x\left(\exp(iy)\right)=\sqrt{x^2+y^2}\bf {cis}(\theta)$.
Finally we give the notation for the argument of the complex number:
Suppose that $\displaystyle x\cdot y\ne 0$ then $\displaystyle \theta= Arg(x + yi) = \left\{ {\begin{array}{{lr}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.$
Here is a example of the process: $\displaystyle z^5=2-2i$. $\displaystyle \bf{Arg}(2-2i})=\theta=\arctan\left(\frac{-2}{2}\right)$
This is the principle root, $\displaystyle \rho=\sqrt[5]{|2-2i|}\exp\left(\dfrac{i\theta}{5}\right)$, so that $\displaystyle \rho^5=2-2i$
List all the roots. Let $\displaystyle \zeta=\exp\left(\frac{2\pi}{5}~\bf{i}\right)$
the five roots are: $\displaystyle \rho\cdot\zeta^k,~k=0,1,2,3,4$
4. ## Re: Solve using de Moivre's theorem: z^3=i
Originally Posted by Martinun
Can you please tell me how you get 4k+1 from 2kpi ?
MarkFL didn't get 4k+ 1 from $\displaystyle 2k\pi$.
What he had was $\displaystyle \frac{\pi}{2}+ 2k\pi$. Getting the "common denominator", that is $\displaystyle \frac{\pi}{2}+ \frac{4k\pi}{2}$$\displaystyle = \frac{\pi+ 4k\pi}{2}= \frac{\pi}{2}(4k+ 1)$.
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# z^3 i=0
Click on a term to search for related topics. | 1,728 | 4,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-22 | latest | en | 0.717295 | # Thread: Solve using de Moivre's theorem: z^3=i 1. ## Re: Solve using de Moivre's theorem: z^3=i Originally Posted by Prove It First of all, when dealing with solving exponential equations, the EXPONENTIAL form of the complex number is ALWAYS easiest to use... \displaystyle \begin{align*} z^3 &= \mathrm{i} \\ z^3 &= \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \textrm{ where } n \in \mathbf{Z} \\ z &= \left[ \mathrm{e}^{\left( \frac{\pi}{2} + 2\pi\,n \right) \mathrm{i}} \right] ^{\frac{1}{3}} \\ z &= \mathrm{ e }^{ \frac{\pi}{6} + \frac{2\pi\,n}{3} } \end{align*} So in the first cycle, the possible solutions are \displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}}, \mathrm{e}^{\frac{5\pi}{6}} , \mathrm{e}^{\frac{3\pi}{2}} \end{align*}, which in Cartesian form are: \displaystyle \begin{align*} \mathrm{e}^{\frac{\pi}{6}} &= \cos{ \left( \frac{\pi}{6} \right) } + \mathrm{i}\sin{\left( \frac{\pi}{6} \right) } \\ &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{5\pi}{6}} &= \cos{ \left( \frac{5\pi}{6} \right) } + \mathrm{i}\sin{ \left( \frac{5\pi}{6} \right) } \\ &= -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \\ \\ \mathrm{e}^{\frac{3\pi}{2}} &= \cos{ \left( \frac{3\pi}{2} \right) } + \mathrm{i}\sin{ \left( \frac{3\pi}{2} \right) } \\ &= -\mathrm{i} \end{align*} I am really bad in this. Sorry guys. I cant really solve it. Can you try solve whole example and like for real bad student ? For example how did you get from i e on something ? If someone have time would be nice if he can solve this examples but every step what you do.. I have like 4-5 examples and cant really solve it. and i have exam in 2 days Best would be way like" Prove it " solved it. We are learning it like that. z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i or is there some manual where they describe everything. Anyway thats for help what you allready solved. 2. ## Re: Solve using de Moivre's theorem: z^3=i If you can put your equation in "cis" form, then you can put the equation in exponential form. In general \displaystyle \begin{align*} r\,\mathrm{e}^{\mathrm{i}\,\theta} \equiv r \left[ \cos{ \left( \theta \right) } + \mathrm{i}\sin{ \left( \theta \right) } \right] \end{align*} 3. ## Re: Solve using de Moivre's theorem: z^3=i Originally Posted by Martinun For example how did you get from i e on something ? If someone have time would be nice if he can solve this examples but every step what you do.. I have like 4-5 examples and cant really solve it. and i have exam in 2 days Best would be way like" Prove it " solved it. We are learning it like that. z^4 = -16i , z^4 = -1, z^2=1-i√3, z^3=-2+2i, z^5=2+2i You are a victim of poor mathematical education practice. There is a good trend to use a more geometric approach to teaching complex analysis. First you need to understand that any complex has n nth roots. Those n roots are located on a circle centered at the origin of radius the nth root of absolute value of the number. Those n roots divide the circle into n equal arcs. Let's stop a moment for an example. There are six sixth roots of $\displaystyle 3-4\bf{i}$. They are located on a circle centered at $\displaystyle (0,0)$ and radius $\displaystyle \sqrt[6]5$. On that circle the roots are separated by arcs of $\displaystyle \dfrac{2\pi}{6}$, where $\displaystyle 2\pi$ is the whole circle and six equal parts. Now give some notation to simplify the way we can list the roots. Let $\displaystyle \exp(i\theta)=\cos(\theta)+i\sin(\theta)$. Then for a complex number $\displaystyle z=x+yi$ then $\displaystyle \exp(z)=e^x\left(\exp(iy)\right)=\sqrt{x^2+y^2}\bf {cis}(\theta)$. Finally we give the notation for the argument of the complex number: Suppose that $\displaystyle x\cdot y\ne 0$ then $\displaystyle \theta= Arg(x + yi) = \left\{ {\begin{array}{{lr}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right.$ Here is a example of the process: $\displaystyle z^5=2-2i$. $\displaystyle \bf{Arg}(2-2i})=\theta=\arctan\left(\frac{-2}{2}\right)$ This is the principle root, $\displaystyle \rho=\sqrt[5]{|2-2i|}\exp\left(\dfrac{i\theta}{5}\right)$, so that $\displaystyle \rho^5=2-2i$ List all the roots. Let $\displaystyle \zeta=\exp\left(\frac{2\pi}{5}~\bf{i}\right)$ the five roots are: $\displaystyle \rho\cdot\zeta^k,~k=0,1,2,3,4$ 4. ## Re: Solve using de Moivre's theorem: z^3=i Originally Posted | by Martinun Can you please tell me how you get 4k+1 from 2kpi ? MarkFL didn't get 4k+ 1 from $\displaystyle 2k\pi$. What he had was $\displaystyle \frac{\pi}{2}+ 2k\pi$. Getting the "common denominator", that is $\displaystyle \frac{\pi}{2}+ \frac{4k\pi}{2}$$\displaystyle = \frac{\pi+ 4k\pi}{2}= \frac{\pi}{2}(4k+ 1)$. Page 2 of 2 First 12 , , , , , , , , , , , , , , # z^3 i=0 Click on a term to search for related topics. |
https://socratic.org/questions/how-do-you-simplify-3a-2b-9a-2-4b-2 | 1,580,070,561,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251690379.95/warc/CC-MAIN-20200126195918-20200126225918-00529.warc.gz | 659,117,630 | 5,914 | # How do you simplify -3a^2b(9a^2-4b^2)?
Oct 30, 2014
In order to simplify $- 3 {a}^{2} b \left(9 {a}^{2} - 4 {b}^{2}\right)$ you would use the distributive property and multiply $- 3 {a}^{2} b$ by each of the terms in the parenthesis.
$- 3 {a}^{2} b \left(9 {a}^{2}\right)$ - $- 3 {a}^{2} b \left(4 {b}^{2}\right)$
$\left({a}^{2}\right) \left({a}^{2}\right)$ = (a)(a)(a)(a) = ${a}^{4}$ $\left(b\right) \left({b}^{2}\right)$ = (b)(b)(b) = ${b}^{3}$
$- 27 {a}^{4} b - \left(- 12 {a}^{2} {b}^{3}\right)$
Simplify the signs.
$- 27 {a}^{4} b + 12 {a}^{2} {b}^{3}$ | 277 | 565 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-05 | longest | en | 0.543195 | # How do you simplify -3a^2b(9a^2-4b^2)? Oct 30, 2014 In order to simplify $- 3 {a}^{2} b \left(9 {a}^{2} - 4 {b}^{2}\right)$ you would use the distributive property and multiply $- 3 {a}^{2} b$ by each of the terms in the parenthesis. $- 3 {a}^{2} b \left(9 {a}^{2}\right)$ - $- 3 {a}^{2} b \left(4 {b}^{2}\right)$ $\left({a}^{2}\right) \left({a}^{2}\right)$ = (a)(a)(a)(a) = ${a}^{4}$ $\left(b\right) \left({b}^{2}\right)$ = (b)(b)(b) = ${b}^{3}$ $- 27 {a}^{4} b - \left(- 12 {a}^{2} {b}^{3}\right)$ Simplify the | signs. $- 27 {a}^{4} b + 12 {a}^{2} {b}^{3}$ |
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1 5.6 Independent Events YOU WILL NEED calculator EXPLORE The Fortin family has two children. Cam determines the probability that the family has two girls. Rushanna determines the probability that the family has two girls, given that the first child is a girl. How are these probabilities similar, and how are they different? GOAL Understand and solve problems that involve independent events. INVESTIGATE the Math Anne and Abby each have 9 marbles: red and 8 blue. Anne places 7 red marbles and 3 blue marbles in bag. She places the rest of her marbles in bag 2. Abby places all of her marbles in bag 3. Anne then draws one marble from bag and one marble from bag 2. Abby draws two marbles from bag 3, without replacement.? Are both girls equally likely to draw two red marbles? A. Let A represent Anne drawing a red marble from bag. Let B represent Anne drawing a red marble from bag 2. Are A and B independent or dependent events? Explain. B. Let C represent Abby drawing a red marble from bag 3 on her first draw. Let D represent Abby drawing a red marble from bag 3 on her second draw. Are C and D independent or dependent events? Explain. C. Determine PA2 and PB2. D. Determine PA d B2 to three decimal places. E. Determine PC 2 and PD 0 C 2, the probability that Abby will draw a red marble from bag 3 on the second draw, given that the first marble she drew was red. F. Determine PC d D2 to three decimal places. G. Are both girls equally likely to draw two red marbles? Explain. Reflecting H. Explain how you can tell if two events are independent or dependent. I. Suppose that Abby were to draw both of her marbles together, instead of drawing one after the other. Would this affect the probability of Abby drawing two red marbles? Justify your answer. J. Naveen used the following formula to determine the probability of Anne drawing two red marbles: PA d B2 5 PA2 # PB 0 A2 Would this formula give the correct answer? Explain. 354 Chapter 5 Probability NEL
2 APPLY the Math example Determining probabilities of independent events Mokhtar and Chantelle are playing a die and coin game. Each turn consists of rolling a regular die and tossing a coin. Points are awarded for rolling a 6 on the die and/or tossing heads with the coin: point for either outcome 3 points for both outcomes 0 points for neither outcome Players alternate turns. The first player who gets 0 points wins. a) Determine the probability that Mokhtar will get, 3, or 0 points on his first turn. b) Verify your results for part a). Explain what you did. Chantelle s Solution a) Let S represent rolling a 6 on the die. Let H represent tossing heads with the coin. PS PSr PH PHr PSr2 5 5 PHr I created a tree diagram showing the probabilities associated with the events of rolling a six on a die, S, and tossing heads with a coin, H, along with the probabilities of their complements, Srand Hr. Rolling a Die Tossing a Coin Probabilities I defined events S and H. These events are independent. The probability of tossing heads with the coin is not affected by the probability of rolling 6 on the die. I determined the probability that Mokhtar will not get either outcome when rolling the die and tossing the coin. These events are complementary. P(H) 2 heads P(S H) 6. 2 or P(S) 6 six P(H ) 2 tails P(S H ) 6. 2 or I determined the probability of each pair of events occurring on the tree diagram by multiplying. P(S ) 5 6 not six P(H) 2 heads 5 P(S H). or P(H ) 5 tails P(S H ). or NEL 5.6 Independent Events 355
3 P(scores 3 points): PS d H 2 5 P(scores 0 points): PSr d Hr2 5 5 P(scores point): PSr d H 2 c PS d Hr2 5 5 PSr d H 2 c PS d Hr2 5 6 or 2 He has a chance of scoring 3 points, a 2 chance of scoring point, and a 5 chance of scoring 0 points. b) Verify: Total probability Total probability 5 or Mokhtar gets 3 points when he rolls a 6 and tosses heads. He gets 0 points when he does not roll a 6 and does not toss heads. He gets point when he either rolls a 6 or tosses heads I verified my solution by adding the three probabilities. The sum is, so I knew that my answer is correct. Your Turn a) Are you more likely to get points or not get points on each turn? Explain. b) Would the probabilities determined for Mokhtar s first turn change for his next turn? Explain. example 2 Solving a problem that involves independent events using a graphic organizer All 000 tickets for a charity raffle have been sold and placed in a drum. There will be two draws. The first draw will be for the grand prize, and the second draw will be for the consolation prize. After each draw, the winning ticket will be returned to the drum so that it might be drawn again. Max has bought five tickets. Determine the probability, to a tenth of a percent, that he will win at least one prize. Max s Solution: Using complements Let X represent winning the grand prize. Let Y represent winning the consolation prize. Let Z represent winning at least one prize. I defined events X, Y, and Z. 356 Chapter 5 Probability NEL
4 PX or 200 PY or 200 The events are independent. Since the winning ticket will be replaced after the first draw, the probability of winning either prize is the same. In other words, winning the consolation prize is not affected by winning the grand prize. X \ Y X Y Y \ X X Y I drew a Venn diagram to represent the situation. The event of winning at least one prize, Z, is represented by being in any area of the two ovals in the Venn diagram. P(Z ), the probability of winning at least one prize, is the complement of winning no prizes: PZ PX r d Y r2 PXr or PYr or PX r d Y r2 5 PX r2 # PY r2 PXr d Yr # PXr d Yr PZ PXr d Yr PZ PZ PZ or The probability I will win at least one prize is about.0%. I had to determine the probability of winning the grand prize or the consolation prize. The only other possibility is winning no prizes. I determined the probability of not winning the grand prize, the probability of not winning the consolation prize, and the probability of winning neither prize. I determined the probability of winning at least one prize. I rounded to the nearest tenth of a percent. NEL 5.6 Independent Events 357
5 Melissa s Solution: Using a tree diagram Let X represent winning the grand prize. Let Y represent winning the consolation prize. Let Z represent winning at least one prize. PX or PY or PX r2 5 2 PX 2 PY r2 5 2 PY 2 PX r PY r PX r PY r I defined events X, Y, and Z. I determined the probability of winning the grand prize and the probability of winning the consolation prize. I also determined the complement of these events: the probability of not winning the grand prize and the probability of not winning the consolation prize. P(X) P(X ) Grand Prize P(Y) P(Y ) P(Y) Consolation Prize P(X Y) (0.005)(0.005) P(X Y ) (0.005)(0.995) P(X Y) (0.995)(0.005) I drew a tree diagram to list all the possible outcomes. The blue branches represent winning a prize, and the red branches represent not winning a prize. I determined the probability of each pair of events by multiplying the individual probabilities. P(Y ) P(X Y ) (0.995)(0.995) Based on my tree diagram, Max could win two prizes in one way. P(win 2 prizes) 5 PX d Y 2 P(win 2 prizes) Max could win one prize in two ways. P(win prize) 5 PX d Y r2 or PX r d Y 2 P(win prize) 5 PX d Y r2 c PX r d Y 2 P(win prize) P(win prize) These probabilities are equal and can be added together, since the two events (only winning the grand prize or only winning the consolation prize) are mutually exclusive. 358 Chapter 5 Probability NEL
6 Winning at least one prize means that Max would win exactly one prize or both prizes. P(Z ) 5 P(win prize) or P(win 2 prizes) P(Z ) 5 P(win prize) c P(win 2 prizes) P(Z ) 5 P(win prize) P(win 2 prizes) PZ PZ The probability Max will win at least one prize is.0%. Verify: Total probability 5 P(win at least one) P(win none) Total probability Total probability 5 These probabilities are different and can be added together, since the two events (winning one prize and winning two prizes) are mutually exclusive. I rounded to the nearest tenth of a percent. I verified my calculations by adding the probabilities. The sum is, so I knew that my answer is correct. Your Turn Suppose that the rules for the raffle are changed, so the first ticket drawn is not returned to the drum before the second draw. a) Are the events winning the grand prize and winning the consolation prize dependent or independent? b) Do you think Max s probability of winning at least one prize will be greater or less than before? Justify your answer. In Summary Key Ideas If the probability of event B does not depend on the probability of event A occurring, then these events are called independent events. For example, tossing tails with a coin and drawing the ace of spades from a standard deck of 52 playing cards are independent events. The probability that two independent events, A and B, will both occur is the product of their individual probabilities: P(A d B) 5 P(A) # P(B) Need to Know A tree diagram is often useful for modelling problems that involve independent events. Drawing an item and then drawing another item, after replacing the first item, results in a pair of independent events. NEL 5.6 Independent Events 359
7 CHECK Your Understanding. For each situation, classify the events as either independent or dependent. Justify your classification. a) A four-colour spinner is spun, and a die is rolled. The first event is spinning red, and the second event is rolling a 2. b) A red die and a green die are rolled. The first event is rolling a on the red die, and the second event is rolling a 5 on the green die. c) Two cards are drawn, without being replaced, from a standard deck of 52 playing cards. The first event is drawing a king, and the second event is drawing an ace. d) There are 30 cards, numbered to 30, in a box. Two cards are drawn, one at a time, with replacement. The first event is drawing a prime number, and the second event is drawing a number that is a multiple of Celeste goes to the gym five days a week. Each day, she does a cardio workout using either a treadmill, an elliptical walker, or a stationary bike. She follows this with a strength workout using either free weights or the weight machines. Celeste randomly chooses which cardio workout and which strength workout to do each day. a) Are choosing a cardio workout and choosing a strength workout dependent or independent events? Explain. b) Determine the probability that Celeste will use a stationary bike and free weights the next day she goes to the gym. 3. Ian also goes to the gym five days a week, but he does two different cardio workouts each day. His choices include using a treadmill, a stepper, or an elliptical walker, and running the track. a) Are the two cardio workouts that Ian chooses dependent or independent events? b) Determine the probability that the next time Ian goes to the gym he will use the elliptical walker and then run the track. PRACTISING 4. For each situation described in question, determine the probability that both events will occur. 5. a) Suppose that P(A) , P(B) 5 0.4, and P(A d B) Are A and B independent events? Explain. b) Suppose that P(Q) , P(R) , and P(Q d R) Are Q and R independent events? Explain. 360 Chapter 5 Probability NEL
8 6. There are two children in the Angel family. a) Draw a tree diagram that shows all the possible gender combinations for the two children. b) Determine the probability that both children are boys. c) Determine the probability that one child is a boy and the other child is a girl. 7. A particular game uses 40 cards from a standard deck of 52 playing cards: the ace to the 0 from the four suits. One card is dealt to each of two players. Determine the probability that the first card dealt is a club and the second card dealt is a heart. Are these events independent or dependent? 8. A standard die is rolled twice. Determine the probability for the following: a) The first roll is a, and the second roll is a 6. b) The first roll is greater than 3, and the second roll is even. c) The first roll is greater than, and the second roll is less than Jeremiah is going on a cruise up the Nile. According to the travel brochure, the probability that he will see a camel is 4 5, and the probability that he will see an ibis is 3 4. Determine the probability that Jeremiah will see the following: a) A camel and an ibis b) Neither a camel nor an ibis c) Only one of these sights 0. a) Design a spinner so that when you toss a coin and spin the spinner, the probability of getting heads and spinning a 6 is. b) Repeat part a) with a probability of 20.. Recall Anne and Abby, from the beginning of this lesson. They each have 9 marbles: red and 8 blue. Anne places 7 red marbles and 3 blue marbles in bag. She places the rest of her marbles in bag 2. Abby places all of her marbles in bag 3. Anne then draws one marble from bag and one marble from bag 2. Abby draws two marbles from bag 3. a) Are Anne and Abby equally likely to draw two blue marbles from their bags? Explain. b) Determine the probability Anne and Abby will both draw one red marble and one blue marble. Explain what you did. c) Suppose that Anne now has 5 red marbles and 5 blue marbles in each of her two bags, while Abby has 0 red marbles and 0 blue marbles in her one bag. Will Abby still be more likely to draw two red marbles? Explain. NEL 5.6 Independent Events 36
9 . A paper bag contains a mixture of three types of treats: 0 granola bars, 7 fruit bars, and 3 cheese strips. Suppose that you play a game in which a treat is randomly taken from the bag and replaced, and then a second treat is drawn from the bag. You are allowed to keep the second treat only if it was the same type as the treat that was drawn the first time. Determine the probability of each of the following: a) You will be able to keep a granola bar. b) You will be able to keep any treat. c) You will not be able to keep any treat. 3. Tiegan s school is holding a chocolate bar sale. For every case of chocolate bars sold, the seller receives a ticket for a prize draw. Tiegan has sold five cases, so she has five tickets for the draw. At the time of the draw, 00 tickets have been entered. There are two prizes, and the ticket that is drawn for the first prize is returned so it can be drawn for the second prize. a) Determine the probability that Tiegan will win both prizes. b) Determine the probability that she will win no prizes. 4. a) Create a problem that involves determining the probability of two independent events. Give your problem to a classmate to solve. b) Modify the problem you created in part a) so that it now involves two dependent events. Give your problem to a classmate to solve. 5. Two single-digit random numbers (0 to 9 inclusive) are selected independently. Determine the probability that their sum is 0. Closing 6. a) Explain why the formula you would use to calculate P(A d B) would depend on whether A and B are dependent or independent events. b) Give an example of how you would calculate P(A d B) if A and B were independent events. c) Give an example of how you would calculate P(A d B) if A and B were dependent events. Extending 7. A particular machine has 00 parts. Over a year, the probability that each part of the machine will fail is %. If any part fails, the machine will stop. a) Determine the probability that the machine will operate continuously for year. b) Suppose that the probability of each part failing within one year dropped to 0.5%. Determine the probability that the machine would operate continuously for year. c) Suppose that the probability of the machine operating continuously for year must be 90%. What would the probability of not failing need to be for each part? 362 Chapter 5 Probability NEL
### 2. Julie draws a card at random from a standard deck of 52 playing cards. Determine the probability of the card being a diamond.
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Rushanna determines the probability that the family has two girls, given that the first child is a girl. How are these probabilities similar, and how are they different? GOAL Understand and solve problems that involve independent events. INVESTIGATE the Math Anne and Abby each have 9 marbles: red and 8 blue. Anne places 7 red marbles and 3 blue marbles in bag. She places the rest of her marbles in bag 2. Abby places all of her marbles in bag 3. Anne then draws one marble from bag and one marble from bag 2. Abby draws two marbles from bag 3, without replacement.? Are both girls equally likely to draw two red marbles? A. Let A represent Anne drawing a red marble from bag. Let B represent Anne drawing a red marble from bag 2. Are A and B independent or dependent events? Explain. B. Let C represent Abby drawing a red marble from bag 3 on her first draw. Let D represent Abby drawing a red marble from bag 3 on her second draw. Are C and D independent or dependent events? Explain. C. Determine PA2 and PB2. D. Determine PA d B2 to three decimal places. E. Determine PC 2 and PD 0 C 2, the probability that Abby will draw a red marble from bag 3 on the second draw, given that the first marble she drew was red. F. Determine PC d D2 to three decimal places. G. Are both girls equally likely to draw two red marbles? Explain. Reflecting H. Explain how you can tell if two events are independent or dependent. I. Suppose that Abby were to draw both of her marbles together, instead of drawing one after the other. Would this affect the probability of Abby drawing two red marbles? Justify your answer. J. Naveen used the following formula to determine the probability of Anne drawing two red marbles: PA d B2 5 PA2 # PB 0 A2 Would this formula give the correct answer? Explain. 354 Chapter 5 Probability NEL 2 APPLY the Math example Determining probabilities of independent events Mokhtar and Chantelle are playing a die and coin game. Each turn consists of rolling a regular die and tossing a coin. Points are awarded for rolling a 6 on the die and/or tossing heads with the coin: point for either outcome 3 points for both outcomes 0 points for neither outcome Players alternate turns. The first player who gets 0 points wins. a) Determine the probability that Mokhtar will get, 3, or 0 points on his first turn. b) Verify your results for part a). Explain what you did. Chantelle s Solution a) Let S represent rolling a 6 on the die. Let H represent tossing heads with the coin. PS PSr PH PHr PSr2 5 5 PHr I created a tree diagram showing the probabilities associated with the events of rolling a six on a die, S, and tossing heads with a coin, H, along with the probabilities of their complements, Srand Hr. Rolling a Die Tossing a Coin Probabilities I defined events S and H. These events are independent. The probability of tossing heads with the coin is not affected by the probability of rolling 6 on the die. I determined the probability that Mokhtar will not get either outcome when rolling the die and tossing the coin. These events are complementary. P(H) 2 heads P(S H) 6. 2 or P(S) 6 six P(H ) 2 tails P(S H ) 6. 2 or I determined the probability of each pair of events occurring on the tree diagram by multiplying. P(S ) 5 6 not six P(H) 2 heads 5 P(S H). or P(H ) 5 tails P(S H ). or NEL 5.6 Independent Events 355 3 P(scores 3 points): PS d H 2 5 P(scores 0 points): PSr d Hr2 5 5 P(scores point): PSr d H 2 c PS d Hr2 5 5 PSr d H 2 c PS d Hr2 5 6 or 2 He has a chance of scoring 3 points, a 2 chance of scoring point, and a 5 chance of scoring 0 points. b) Verify: Total probability Total probability 5 or Mokhtar gets 3 points when he rolls a 6 and tosses heads. He gets 0 points when he does not roll a 6 and does not toss heads. He gets point when he either rolls a 6 or tosses heads I verified my solution by adding the three probabilities. The sum is, so I knew that my answer is correct. Your Turn a) Are you more likely to get points or not get points on each turn? Explain. b) Would the probabilities determined for Mokhtar s first turn change for his next turn? Explain. example 2 Solving a problem that involves independent events using a graphic organizer All 000 tickets for a charity raffle have been sold and placed in a drum. There will be two draws. The first draw will be for the grand prize, and the second draw will be for the consolation prize. After each draw, the winning ticket will be returned to the drum so that it might be drawn again. Max has bought five tickets. Determine the probability, to a tenth of a percent, that he will win at least one prize. Max s Solution: Using complements Let X represent winning the grand prize. Let Y represent winning the consolation prize. Let Z represent winning at least one prize. I defined events X, Y, and Z. 356 Chapter 5 Probability NEL 4 PX or 200 PY or 200 The events are independent. Since the winning ticket will be replaced after the first draw, the probability of winning either prize is the same. In other words, winning the consolation prize is not affected by winning the grand prize. X \ Y X Y Y \ X X Y I drew a Venn diagram to represent the situation. The event of winning at least one prize, Z, is represented by being in any area of the two ovals in the Venn diagram. P(Z ), the probability of winning at least one prize, is the complement of winning no prizes: PZ PX r d Y r2 PXr or PYr or PX r d Y r2 5 PX r2 # PY r2 PXr d Yr # PXr d Yr PZ PXr d Yr PZ PZ PZ or The probability I will win at least one prize is about.0%. I had to determine the probability of winning the grand prize or the consolation prize. The only other possibility is winning no prizes. I determined the probability of not winning the grand prize, the probability of not winning the consolation prize, and the probability of winning neither prize. I determined the probability of winning at least one prize. I rounded to the nearest tenth of a percent. NEL 5.6 Independent Events 357 5 Melissa s Solution: Using a tree diagram Let X represent winning the grand prize. Let Y represent winning the consolation prize. Let Z represent winning at least one prize. PX or PY or PX r2 5 2 PX 2 PY r2 5 2 PY 2 PX r PY r PX r PY r I defined events X, Y, and Z. I determined the probability of winning the grand prize and the probability of winning the consolation prize. I also determined the complement of these events: the probability of not winning the grand prize and the probability of not winning the consolation prize. P(X) P(X ) Grand Prize P(Y) P(Y ) P(Y) Consolation Prize P(X Y) (0.005)(0.005) P(X Y ) (0.005)(0.995) P(X Y) (0.995)(0.005) I drew a tree diagram to list all the possible outcomes. The blue branches represent winning a prize, and the red branches represent not winning a prize. I determined the probability of each pair of events by multiplying the individual probabilities. P(Y ) P(X Y ) (0.995)(0.995) Based on my tree diagram, Max could win two prizes in one way. P(win 2 prizes) 5 PX d Y 2 P(win 2 prizes) Max could win one prize in two ways. P(win prize) 5 PX d Y r2 or PX r d Y 2 P(win prize) 5 PX d Y r2 c PX r d Y 2 P(win prize) P(win prize) These probabilities are equal and can be added together, since the two events (only winning the grand prize or only winning the consolation prize) are mutually exclusive. 358 Chapter 5 Probability NEL 6 Winning at least one prize means that Max would win exactly one prize or both prizes. P(Z ) 5 P(win prize) or P(win 2 prizes) P(Z ) 5 P(win prize) c P(win 2 prizes) P(Z ) 5 P(win prize) P(win 2 prizes) PZ PZ The probability Max will win at least one prize is.0%. Verify: Total probability 5 P(win at least one) P(win none) Total probability Total probability 5 These probabilities are different and can be added together, since the two events (winning one prize and winning two prizes) are mutually exclusive. I rounded to the nearest tenth of a percent. I verified my calculations by adding the probabilities. The sum is, so I knew that my answer is correct. Your Turn Suppose that the rules for the raffle are changed, so the first ticket drawn is not returned to the drum before the second draw. a) Are the events winning the grand prize and winning the consolation prize dependent or independent? b) Do you think Max s probability of winning at least one prize will be greater or less than before? Justify your answer. In Summary Key Ideas If the probability of event B does not depend on the probability of event A occurring, then these events are called independent events. For example, tossing tails with a coin and drawing the ace of spades from a standard deck of 52 playing cards are independent events. The probability that two independent events, A and B, will both occur is the product of their individual probabilities: P(A d B) 5 P(A) # P(B) Need to Know A tree diagram is often useful for modelling problems that involve independent events. Drawing an item and then drawing another item, after replacing the first item, results in a pair of independent events. NEL 5.6 Independent Events 359 7 CHECK Your Understanding. For each situation, classify the events as either independent or dependent. Justify your classification. a) A four-colour spinner is spun, and a die is rolled. The first event is spinning red, and the second event is rolling a 2. b) A red die and a green die are rolled. The first event is rolling a on the red die, and the second event is rolling a 5 on the green die. c) Two cards are drawn, without being replaced, from a standard deck of 52 playing cards. The first event is drawing a king, and the second event is drawing an ace. d) There are 30 cards, numbered to 30, in a box. Two cards are drawn, one at a time, with replacement. The first event is drawing a prime number, and the second event is drawing a number that is a multiple of Celeste goes to the gym five days a week. Each day, she does a cardio workout using either a treadmill, an elliptical walker, or a stationary bike. She follows this with a strength workout using either free weights or the weight machines. Celeste randomly chooses which cardio workout and which strength workout to do each day. a) Are choosing a cardio workout and choosing a strength workout dependent or independent events? Explain. b) Determine the probability that Celeste will use a stationary bike and free weights the next day she goes to the gym. 3. Ian also goes to the gym five days a week, but he does two different cardio workouts each day. His choices include using a treadmill, a stepper, or an elliptical walker, and running the track. a) Are the two cardio workouts that Ian chooses dependent or independent events? b) Determine the probability that the next time Ian goes to the gym he will use the elliptical walker and then run the track. PRACTISING 4. For each situation described in question, determine the probability that both events will occur. 5. a) Suppose that P(A) , P(B) 5 0.4, and P(A d B) Are A and B independent events? Explain. b) Suppose that P(Q) , P(R) , and P(Q d R) Are Q and R independent events? Explain. 360 Chapter 5 Probability NEL 8 6. There are two children in the Angel family. a) Draw a tree diagram that shows all the possible gender combinations for the two children. b) Determine the probability that both children are boys. c) Determine the probability that one child is a boy and the other child is a girl. 7. A particular game uses 40 cards from a standard deck of 52 playing cards: the ace to the 0 from the four suits. One card is dealt to each of two players. Determine the probability that the first card dealt is a club and the second card dealt is a heart. Are these events independent or dependent? 8. A standard die is rolled twice. Determine the probability for the following: a) The first roll is a, and the second roll is a 6. b) The first roll is greater than 3, and the second roll is even. c) The first roll is greater than, and the second roll is less than Jeremiah is going on a cruise up the Nile. According to the travel brochure, the probability that he will see a camel is 4 5, and the probability that he will see an ibis is 3 4. Determine the probability that Jeremiah will see the following: a) A camel and an ibis b) Neither a camel nor an ibis c) Only one of these sights 0. a) Design a spinner so that when you toss a coin and spin the spinner, the probability of getting heads and spinning a 6 is. b) Repeat part a) with a probability of 20.. Recall Anne and Abby, from the beginning of this lesson. They each have 9 marbles: red and 8 blue. Anne places 7 red marbles and 3 blue marbles in bag. She places the rest of her marbles in bag 2. Abby places all of her marbles in bag 3. Anne then draws one marble from bag and one marble from bag 2. Abby draws two marbles from bag 3. a) Are Anne and Abby equally likely to draw two blue marbles from their bags? Explain. b) Determine the probability Anne and Abby will both draw one red marble and one blue marble. Explain what you did. c) Suppose that Anne now has 5 red marbles and 5 blue marbles in each of her two bags, while Abby has 0 red marbles and 0 blue marbles in her one bag. Will Abby still be more likely to draw two red marbles? Explain. NEL 5.6 Independent Events 36 9 . A paper bag contains a mixture of three types of treats: 0 granola bars, 7 fruit bars, and 3 cheese strips. Suppose that you play a game in which a treat is randomly taken from the bag and replaced, and then a second treat is drawn from the bag. You are allowed to keep the second treat only if it was the same type as the treat that was drawn the first time. Determine the probability of each of the following: a) You will be able to keep a granola bar. b) You will be able to keep any treat. c) You will not be able to keep any treat. 3. Tiegan s school is holding a chocolate bar sale. For every case of chocolate bars sold, the seller receives a ticket for a prize draw. Tiegan has sold five cases, so she has five tickets for the draw. At the time of the draw, 00 tickets have been entered. There are two prizes, and the ticket that is drawn for the first prize is returned so it can be drawn for the second prize. a) Determine the probability that Tiegan will win both prizes. b) Determine the probability that she will win no prizes. 4. a) Create a problem that involves determining the probability of two independent events. Give your problem to a classmate to solve. b) Modify the problem you created in part a) so that it now involves two dependent events. Give your problem to a classmate to solve. 5. Two single-digit random numbers (0 to 9 inclusive) are selected independently. Determine the probability that their sum is 0. Closing 6. a) Explain why the formula you would use to calculate P(A d B) would depend on whether A and B are dependent or independent events. b) Give an example of how you would calculate P(A d B) if A and B were independent events. c) Give an example of how you would calculate P(A d B) if A and B were dependent events. Extending 7. A particular machine has 00 parts. Over a year, the probability that each part of the machine will fail is %. If any part fails, the machine will stop. a) Determine the probability that the machine will operate continuously for year. b) Suppose that the probability of each part failing within one year dropped to 0.5%. Determine the probability that the machine would operate continuously for year. c) Suppose that the probability of the machine operating continuously for year must be 90%. What would the probability of not failing need to be for each part? 362 Chapter 5 Probability NEL ### 2. 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Lecture 10: Randomness and Probability Model The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL STOR 155 Introductory Statistics Lecture 10: Randomness and Probability Model 10/6/09 Lecture 10 1 The Monty Hall Problem Let s Make A Deal: a game show ### Math 1 Unit 4 Mid-Unit Review Chances of Winning Math 1 Unit 4 Mid-Unit Review Chances of Winning Name My child studied for the Unit 4 Mid-Unit Test. I am aware that tests are worth 40% of my child s grade. Parent Signature MM1D1 a. Apply the addition ### Counting Methods and Probability CHAPTER Counting Methods and Probability Many good basketball players can make 90% of their free throws. However, the likelihood of a player making several free throws in a row will be less than 90%. You ### Probability Review before Quiz. Unit 6 Day 6 Probability Probability Review before Quiz Unit 6 Day 6 Probability Warm-up: Day 6 1. A committee is to be formed consisting of 1 freshman, 1 sophomore, 2 juniors, and 2 seniors. How many ways can this committee be ### Key Concepts. Theoretical Probability. Terminology. Lesson 11-1 Key Concepts Theoretical Probability Lesson - Objective Teach students the terminology used in probability theory, and how to make calculations pertaining to experiments where all outcomes are equally ### Review Questions on Ch4 and Ch5 Review Questions on Ch4 and Ch5 1. Find the mean of the distribution shown. x 1 2 P(x) 0.40 0.60 A) 1.60 B) 0.87 C) 1.33 D) 1.09 2. A married couple has three children, find the probability they are all ### Class XII Chapter 13 Probability Maths. Exercise 13.1 Exercise 13.1 Question 1: Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E F) = 0.2, find P (E F) and P(F E). It is given that P(E) = 0.6, P(F) = 0.3, and P(E F) = 0.2 Question 2: ### RANDOM EXPERIMENTS AND EVENTS Random Experiments and Events 18 RANDOM EXPERIMENTS AND EVENTS In day-to-day life we see that before commencement of a cricket match two captains go for a toss. 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# Difference between revisions of "2013 AMC 10A Problems/Problem 18"
## Problem
Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$?
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$
## Solution
First, various area formulas (shoelace, splitting, etc) allow us to find that $[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.
Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$.
Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \frac{27}{8}$.
From this, we know that $E = (\frac{27}{8}, \frac{15}{8})$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$ | 490 | 1,349 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 30, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2022-05 | longest | en | 0.797918 | During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made. # Difference between revisions of "2013 AMC 10A Problems/Problem 18" ## Problem Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $(\frac{p}{q}, \frac{r}{s})$, where these fractions are in lowest terms. What is $p+q+r+s$? $\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$ ## Solution First, various area formulas (shoelace, splitting, etc) allow us to find that $[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$. Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$. Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that | $x = \frac{27}{8}$. From this, we know that $E = (\frac{27}{8}, \frac{15}{8})$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$ |
https://socratic.org/questions/two-charges-of-3-c-and-4-c-are-positioned-on-a-line-at-points-9-and-4-respective | 1,569,093,359,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574662.80/warc/CC-MAIN-20190921190812-20190921212812-00367.warc.gz | 692,030,525 | 6,301 | Two charges of -3 C and -4 C are positioned on a line at points -9 and 4 , respectively. What is the net force on a charge of 4 C at 2 ?
Mar 15, 2016
${F}_{1} = {F}_{2} = 0 , 64 \cdot {10}^{9} \text{ } N$
$\text{The force between two charges is given as:}$
${F}_{1} = {F}_{2} = k \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$
$d = 9 + 4 = 13 \text{ "q_1=-3C" } {q}_{2} = - 4 C$
${F}_{1} = {F}_{2} = 9 \cdot {19}^{9} \frac{\left(- 3\right) \left(- 4\right)}{13} ^ 2$
${F}_{1} = {F}_{2} = \frac{12 \cdot 9 \cdot {10}^{9}}{169}$ | 248 | 526 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 6, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-39 | longest | en | 0.64622 | Two charges of -3 C and -4 C are positioned on a line at points -9 and 4 , respectively. What is the net force on a charge of 4 C at 2 ? Mar 15, 2016 ${F}_{1} = {F}_{2} = 0 , 64 \cdot {10}^{9} \text{ } N$ $\text{The force between two charges is given as:}$ ${F}_{1} = {F}_{2} = k \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$ $d = 9 + 4 = 13 \text{ "q_1=-3C" } {q}_{2} = - 4 C$ ${F}_{1} = {F}_{2} = 9 \cdot {19}^{9} \frac{\left(- 3\right) \left(- 4\right)}{13} | ^ 2$ ${F}_{1} = {F}_{2} = \frac{12 \cdot 9 \cdot {10}^{9}}{169}$ |
http://wpressutexas.net/oldcoursewiki/index.php?title=Travis:_Segment_5 | 1,571,608,039,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00051.warc.gz | 224,283,397 | 5,335 | # Travis: Segment 5
### Problems
#### To Compute
1. You throw a pair of fair dice 10 times and, each time, you record the total number of spots. When you are done, what is the probability that exactly 5 of the 10 recorded totals are prime?
The probability of rolling a prime number with two dice is $\frac{15}{32}$.
Prime numbers in 2-dice roll
+ 1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
The probability of getting exactly n successes in N trials is given by the probability mass function:
$f(n;N,p) = {N\choose n}p^n(1-p)^{N-n}$
So,
$f(5;10,\frac{15}{36}) = {10\choose 5}\bigg(\frac{15}{36}\bigg)^5\bigg(\frac{21}{36}\bigg)^{5} = 0.2138...$
2. If you flip a fair coin one billion times, what is the probability that the number of heads is between 500010000 and 500020000, inclusive? (Give answer to 4 significant figures.) | 333 | 894 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2019-43 | latest | en | 0.83522 | # Travis: Segment 5 ### Problems #### To Compute 1. You throw a pair of fair dice 10 times and, each time, you record the total number of spots. When you are done, what is the probability that exactly 5 of the 10 recorded totals are prime? The probability of rolling a prime number with two dice is $\frac{15}{32}$. Prime numbers in 2-dice roll + 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 The probability of getting exactly n successes in N trials is given by the probability mass function: $f(n;N,p) = {N\choose n}p^n(1-p)^{N-n}$ So, $f(5;10,\frac{15}{36}) = {10\choose 5}\bigg(\frac{15}{36}\bigg)^5\bigg(\frac{21}{36}\bigg)^{5} = 0.2138...$ 2. If you flip a fair coin one billion times, what is the | probability that the number of heads is between 500010000 and 500020000, inclusive? (Give answer to 4 significant figures.) |
https://socratic.org/questions/a-closed-food-jar-has-a-fixed-volume-at-stp-what-would-the-new-pressure-be-at-45 | 1,576,499,825,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540565544.86/warc/CC-MAIN-20191216121204-20191216145204-00171.warc.gz | 538,040,972 | 6,891 | # A closed food jar has a fixed volume at STP. What would the new pressure be at 45°C?
Mar 15, 2016
$318 k P a$
#### Explanation:
Recall that Gay-Lussac's Law states that pressure is directly proportional to temperature , as long as the volume and number of moles of the gas remain constant.
Gay-Lussac's Law can be expressed mathematically with the formula,
$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where:
${P}_{1} =$initial pressure
${T}_{1} =$inital temperature (Kelvin)
${P}_{2} =$final pressure
${T}_{2} =$final temperature (Kelvin)
Determining the Final Pressure
$1$. Start by determining the values for each variable in the formula.
STP conditions (initial):
initial pressure $\textcolor{\mathmr{and} a n \ge}{\left({P}_{1}\right)}$: $\textcolor{\mathmr{and} a n \ge}{101.325 k P a}$
initial temperature $\textcolor{p u r p \le}{\left({T}_{1}\right)}$: $\textcolor{p u r p \le}{273.15 K}$
New conditions (final)
final pressure $\textcolor{t e a l}{\left({P}_{2}\right)}$: $\textcolor{t e a l}{{P}_{2}}$
final temperature $\textcolor{m a \ge n t a}{\left({T}_{2}\right)}$: ${45}^{\circ} C + 273.15 = \textcolor{m a \ge n t a}{318.15 K}$
$2$. Rearrange Gay-Lussac's formula in terms of $\textcolor{t e a l}{{P}_{2}}$.
$\frac{\textcolor{\mathmr{and} a n \ge}{{P}_{1}}}{\textcolor{p u r p \le}{{T}_{1}}} = \frac{\textcolor{t e a l}{{P}_{2}}}{\textcolor{m a \ge n t a}{{T}_{2}}}$
$\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{{T}_{2}}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{{P}_{1}}}{\textcolor{p u r p \le}{{T}_{1}}}\right)$
$3$. Using these values, substitute them into the rearranged formula.
$\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{318.15 K}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{101.325 k P a}}{\textcolor{p u r p \le}{273.15 K}}\right)$
$4$. Solve for $\textcolor{t e a l}{{P}_{2}}$.
$\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{318.15 \textcolor{red}{\cancel{\textcolor{m a \ge n t a}{K}}}}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{101.325 k P a}}{\textcolor{p u r p \le}{273.15 \textcolor{red}{\cancel{\textcolor{p u r p \le}{K}}}}}\right)$
$\textcolor{t e a l}{{P}_{2}} = 118.0177512 k P a$
$\textcolor{t e a l}{{P}_{2}} \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 318 k P a \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\therefore$, the final pressure is $318 k P a$. | 965 | 2,555 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 28, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2019-51 | longest | en | 0.616675 | # A closed food jar has a fixed volume at STP. What would the new pressure be at 45°C? Mar 15, 2016 $318 k P a$ #### Explanation: Recall that Gay-Lussac's Law states that pressure is directly proportional to temperature , as long as the volume and number of moles of the gas remain constant. Gay-Lussac's Law can be expressed mathematically with the formula, $\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}}$ where: ${P}_{1} =$initial pressure ${T}_{1} =$inital temperature (Kelvin) ${P}_{2} =$final pressure ${T}_{2} =$final temperature (Kelvin) Determining the Final Pressure $1$. Start by determining the values for each variable in the formula. STP conditions (initial): initial pressure $\textcolor{\mathmr{and} a n \ge}{\left({P}_{1}\right)}$: $\textcolor{\mathmr{and} a n \ge}{101.325 k P a}$ initial temperature $\textcolor{p u r p \le}{\left({T}_{1}\right)}$: $\textcolor{p u r p \le}{273.15 K}$ New conditions (final) final pressure $\textcolor{t e a l}{\left({P}_{2}\right)}$: $\textcolor{t e a l}{{P}_{2}}$ final temperature $\textcolor{m a \ge n t a}{\left({T}_{2}\right)}$: ${45}^{\circ} C + 273.15 = \textcolor{m a \ge n t a}{318.15 K}$ $2$. Rearrange Gay-Lussac's formula in terms of $\textcolor{t e a l}{{P}_{2}}$. $\frac{\textcolor{\mathmr{and} a n \ge}{{P}_{1}}}{\textcolor{p u r p \le}{{T}_{1}}} = \frac{\textcolor{t e a l}{{P}_{2}}}{\textcolor{m a \ge n t a}{{T}_{2}}}$ $\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{{T}_{2}}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{{P}_{1}}}{\textcolor{p u r p \le}{{T}_{1}}}\right)$ $3$. Using these values, substitute them into the rearranged formula. $\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{318.15 K}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{101.325 k P a}}{\textcolor{p u r p \le}{273.15 K}}\right)$ $4$. Solve for $\textcolor{t e a l}{{P}_{2}}$. $\textcolor{t e a l}{{P}_{2}} = \left(\textcolor{m a \ge n t a}{318.15 \textcolor{red}{\cancel{\textcolor{m a \ge n t a}{K}}}}\right) \left(\frac{\textcolor{\mathmr{and} a n \ge}{101.325 k P a}}{\textcolor{p u r p \le}{273.15 \textcolor{red}{\cancel{\textcolor{p u r p \le}{K}}}}}\right)$ $\textcolor{t e a l}{{P}_{2}} = 118.0177512 k P | a$ $\textcolor{t e a l}{{P}_{2}} \approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 318 k P a \textcolor{w h i t e}{\frac{a}{a}} |}}}$ $\therefore$, the final pressure is $318 k P a$. |
https://cracku.in/blog/data-interpretation-questions-for-cat-pdf/ | 1,721,754,983,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518059.67/warc/CC-MAIN-20240723163815-20240723193815-00772.warc.gz | 160,430,605 | 38,480 | 0
476
# CAT Data Interpretation Important Questions PDF [With Video Solutions]
Data Interpretation is one of the most important topics in the CAT LRDI Section. If you’re new to these questions, you can check out these CAT Data Interpretation Questions from the CAT previous year papers. In this article, we will look into some very important Data Interpretation questions PDF(with solutions) for CAT. You can also download these CAT Data Interpretation questions with detailed solutions, which also include important tricks to solve these questions.
Instructions
The different bars in the diagram above provide information about different orders in various categories (Art, Binders, ….) that were booked in the first two weeks of September of a store for one client. The colour and pattern of a bar denotes the ship mode (First Class / Second Class / Standard Class). The left end point of a bar indicates the booking day of the order, while the right end point indicates the dispatch day of the order. The difference between the dispatch day and the booking day (measured in terms of the number of days) is called the processing time of the order. For the same category, an order is considered for booking only after the previous order of the same category is dispatched. No two consecutive orders of the same category had identical ship mode during this period.
For example, there were only two orders in the furnishing category during this period. The first one was shipped in the Second Class. It was booked on Sep 1 and dispatched on Sep 5. The second order was shipped in the Standard class. It was booked on Sep 5 (although the order might have been placed before that) and dispatched on Sep 12. So the processing times were 4 and 7 days respectively for these orders.
Question 1:Â How many days between Sep 1 and Sep 14 (both inclusive) had no booking from this client considering all the above categories?
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now No booking days from the table are : September 8,9,10,11,12 and 14.
So a total of 6 days .
Question 2:Â What was the average processing time of all orders in the categories which had only one type of ship mode?
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now Envelopes and Accessories has only 1 ship mode i.e Standard class .
So therefore processing days for envelopes = 7-3 =4
and processing days for accessories = 19-1 =18
Therefore average =Â $\frac{\left(18+4\right)}{2}=11$
Question 3:Â The sequence of categories — Art, Binders, Paper and Phones — in decreasing order of average processing time of their orders in this period is:
a)Â Art, Binders, Paper, Phones
b)Â Phones, Art, Binders, Paper
c)Â Phones, Binders, Art, Paper
d)Â Paper, Binders, Art, Phones
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now taking average processing time per order for the above mentioned categories we get :
Art =$\frac{2+8+2+1+7}{5}\ \ =4$
Binders =$\frac{1+1+11+2}{4}\ \ =3.75$
Papers =Â $\frac{3+2+5}{3}\ \ =3.33$
Phones =$\frac{2+12+1}{3}\ \ =5$
So in decreasing order we get Phones , Art ,Binder , Paper.
Question 4:Â Approximately what percentage of orders had a processing time of one day during the period Sep 1 to Sep 22 (both dates inclusive)?
a)Â 22%
b)Â 16%
c)Â 20%
d)Â 25%
Solution:
Accumulating all the data :
We get the following table :
Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched .
Now from the table we observe that the total number of orders are 35 and 7 orders have a processing time of 1 unit
The 7 orders are : Arts Standard class,Binders First class and standard class, Phones First class, Bookcases second class ( 2 orders)Â and Chairs standard class.
So the percentage = $\frac{7}{35}\times\ 100\ =\ 20$
Instructions
The figure above shows the schedule of four employees – Abani, Bahni, Danni, and Tinni – whom Dhoni supervised in 2020. Altogether there were five projects which started and concluded in 2020 in which they were involved. For each of these projects and for each employee, the starting day was at the beginning of a month and the concluding day was the end of a month, and these are indicated by the left and right end points of the corresponding horizontal bars. The number within each bar indicates the percentage of assigned work completed by the employee for that project, as assessed by Dhoni.
For each employee, his/her total project-month (in 2020) is the sum of the number of months (s)he worked across the five projects, while his/her annual completion index is the weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects. For each project, the total employee-month is the sum of the number of months four employees worked in this project, while its completion index is the weightage average of the completion percentage assigned for the employees who worked in this project, with the weights being the corresponding number of months they worked in this project.
Question 5:Â Which of the following statements is/are true?
I: The total project-month was the same for the four employees.
II: The total employee-month was the same for the five projects.
a)Â Only II
b)Â Both I and II
c)Â Neither I nor II
d)Â Only I
Solution:
The total project month is the number of months Abani, Bahni, Danni, and Tinni individually worked for all the projects combined :
Abani – 2+2+5 = 9 months
Bahni – 2+4+3 = 9 months
Danni – 3+3+2+1 = 9 months
Tinni – 2+2+3+2 = 9 months.
The total employee month for all the five projects is the sum of the total employee-month is the sum of the number of months four employees worked in this project.
Project -1 = 2+2+2 = 6 months
Project -2 = 3+2 = 5 months
Project – 3 = 2+4+3 = 9 months.
Project – 4 = 5+2+3 = 10 months.
Project – 5 = 3+1+2 = 6 months.
Only statement 1 is true.
Question 6:Â Which employees did not work in multiple projects for any of the months in 2020?
a)Â Only Abani, Bahni and Danni
b)Â Only Abani and Bahni
c)Â All four of them
d)Â Only Tinni
Solution:
Abani, Banni, and Danni did not work on multiple projects simultaneously in a month
Tinni was the only person who worked on multiple projects which are project 4 and project 5 in the month of september.
Question 7:Â The project duration, measured in terms of the number of months, is the time during which at least one employee worked in the project. Which of the following pairs of the projects had the same duration?
a)Â Project 1, Project 5
b)Â Project 4, Project 5
c)Â Project 3, Project 5
d)Â Project 3, Project 4
Solution:
Considering the information provided :
For project 1 : 3 months.
Project – 2: 3 months.
Project – 3: 5 months.
Project – 4: 5 months.
Project – 5: 4 months.
Among the given options option D is true which is project 3, project 4.
Question 8:Â The list of employees in decreasing order of annual completion index is:
a)Â Danni, Tinni, Bahni, Abani
b)Â Bahni, Abani, Tinni, Danni
c)Â Danni, Tinni, Abani, Bahni
d)Â Tinni, Danni, Abani, Bahni
Solution:
The annual completion index for different people is :
The weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects.
For Abani :
$\frac{\left(\left(100\cdot2\right)+\left(100\cdot2\right)+\left(80\cdot5\right)\right)}{2+2+5}=\ \frac{800}{9}$
For Bahni :
$\frac{\left(\left(100\cdot2\right)+\left(75\cdot4\right)+\left(90\cdot3\right)\right)}{2+3+4}=\ \frac{770}{9}$
For Danni :
$\frac{\left(\left(90\cdot3\right)+\left(100\cdot3\right)+\left(100\cdot2\right)+\left(100\cdot1\right)\right)}{2+3+2+1}=\ \frac{870}{9}$
For Tinni :
$\frac{\left(\left(80\cdot2\right)+\left(100\cdot2\right)+\left(100\cdot3\right)+\left(100\cdot2\right)\right)}{2+2+3+2}=\ \frac{860}{9}$
The descending order for the four people is :
Danni, Tinni, Abani, Bahni.
Instructions
DIRECTIONS for the following four questions:
A low-cost airline company connects ten India cities, A to J. The table below gives the distance between a pair of airports and the corresponding price charged by the company. Travel is permitted only from a departure airport to an arrival airport. The customers do not travel by a route where they have to stop at more than two intermediate airports.
<img “=”” alt=”” class=”img-responsive” src=”https://cracku.in/media/questionGroup/DI_6_3.png”/>
Question 9:Â What is the lowest possible fare, in rupees, from A to J?
a)Â 2275
b)Â 2850
c)Â 2890
d)Â 2930
e)Â 3340
Solution:
From the table we can see that, the lowest price would be from A to H and H to J.
The cost of travel from A to H = Rs 1850
The cost of travel from H to J = Rs 425
Total cost = 1850 + 425 = Rs 2275.
Question 10:Â The company plans to introduce a direct flight between A and J. The market research results indicate that all its existing passengers travelling between A and J will use this direct flight if it is priced 5% below the minimum price that they pay at present. What should the company charge approximately, in rupees, for this direct flight?
a)Â 1991
b)Â 2161
c)Â 2707
d)Â 2745
e)Â 2783
Solution:
From the table we can see that, the lowest price would be from A to H and H to J.
The cost of travel from A to H = Rs 1850
The cost of travel from H to JÂ = Rs 425
Total cost = 1850 + 425 = Rs 2275
Lowest price = Rs 2275
95% of 2275 = Rs 2161
Question 11:Â If the airports C, D and H are closed down owing to security reasons, what would be the minimum price, in rupees, to be paid by a passenger travelling from A to J?
a)Â 2275
b)Â 2615
c)Â 2850
d)Â 2945
e)Â 3190
Solution:
If the airports C, D and H are closed down  the minimum price to be paid by a passenger travelling from A to J would be by first travelling to F and then from F to J.
The cost of travel from A to FÂ = Rs 1700
The cost of travel from FÂ to JÂ = Rs 1150
Total cost = 1700 + 1150 = Rs 2850
Question 12:Â If the prices include a margin of 10% over the total cost that the company incurs, what is the minimum cost per kilometer that the company incurs in flying from A to J?
a)Â 0.77
b)Â 0.88
c)Â 0.99
d)Â 1.06
e)Â 1.08
Solution:
The minimum cost from A to J we know is 2275.
Let the CP to company be C
Since 10% over actual CP is the total price i.e. $\text{CP}\times1.1 = 2275 \rightarrow CP = \frac{2275}{1.1}$
The total distance is 1950+1400=2350 Km.
Cost per Km = $\dfrac{\frac{2275}{1.1}}{2350}$ = Rs 0.88/Km
Question 13:Â If the prices include a margin of 15% over the total cost that the company incurs, which among the following is the distance to be covered in flying from A to J that minimizes the total cost of travel for the company?
a)Â 2170
b)Â 2180
c)Â 2315
d)Â 2350
e) 2390 | 3,150 | 11,309 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-30 | latest | en | 0.952784 | 0 476 # CAT Data Interpretation Important Questions PDF [With Video Solutions] Data Interpretation is one of the most important topics in the CAT LRDI Section. If you’re new to these questions, you can check out these CAT Data Interpretation Questions from the CAT previous year papers. In this article, we will look into some very important Data Interpretation questions PDF(with solutions) for CAT. You can also download these CAT Data Interpretation questions with detailed solutions, which also include important tricks to solve these questions. Instructions The different bars in the diagram above provide information about different orders in various categories (Art, Binders, ….) that were booked in the first two weeks of September of a store for one client. The colour and pattern of a bar denotes the ship mode (First Class / Second Class / Standard Class). The left end point of a bar indicates the booking day of the order, while the right end point indicates the dispatch day of the order. The difference between the dispatch day and the booking day (measured in terms of the number of days) is called the processing time of the order. For the same category, an order is considered for booking only after the previous order of the same category is dispatched. No two consecutive orders of the same category had identical ship mode during this period. For example, there were only two orders in the furnishing category during this period. The first one was shipped in the Second Class. It was booked on Sep 1 and dispatched on Sep 5. The second order was shipped in the Standard class. It was booked on Sep 5 (although the order might have been placed before that) and dispatched on Sep 12. So the processing times were 4 and 7 days respectively for these orders. Question 1: How many days between Sep 1 and Sep 14 (both inclusive) had no booking from this client considering all the above categories? Solution: Accumulating all the data : We get the following table : Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched . Now No booking days from the table are : September 8,9,10,11,12 and 14. So a total of 6 days . Question 2: What was the average processing time of all orders in the categories which had only one type of ship mode? Solution: Accumulating all the data : We get the following table : Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched . Now Envelopes and Accessories has only 1 ship mode i.e Standard class . So therefore processing days for envelopes = 7-3 =4 and processing days for accessories = 19-1 =18 Therefore average = $\frac{\left(18+4\right)}{2}=11$ Question 3: The sequence of categories — Art, Binders, Paper and Phones — in decreasing order of average processing time of their orders in this period is: a) Art, Binders, Paper, Phones b) Phones, Art, Binders, Paper c) Phones, Binders, Art, Paper d) Paper, Binders, Art, Phones Solution: Accumulating all the data : We get the following table : Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched . Now taking average processing time per order for the above mentioned categories we get : Art =$\frac{2+8+2+1+7}{5}\ \ =4$ Binders =$\frac{1+1+11+2}{4}\ \ =3.75$ Papers = $\frac{3+2+5}{3}\ \ =3.33$ Phones =$\frac{2+12+1}{3}\ \ =5$ So in decreasing order we get Phones , Art ,Binder , Paper. Question 4: Approximately what percentage of orders had a processing time of one day during the period Sep 1 to Sep 22 (both dates inclusive)? a) 22% b) 16% c) 20% d) 25% Solution: Accumulating all the data : We get the following table : Note a-b : represents the duration where a is the day when order is booked and b is the day when it is dispatched . Now from the table we observe that the total number of orders are 35 and 7 orders have a processing time of 1 unit The 7 orders are : Arts Standard class,Binders First class and standard class, Phones First class, Bookcases second class ( 2 orders) and Chairs standard class. So the percentage = $\frac{7}{35}\times\ 100\ =\ 20$ Instructions The figure above shows the schedule of four employees – Abani, Bahni, Danni, and Tinni – whom Dhoni supervised in 2020. Altogether there were five projects which started and concluded in 2020 in which they were involved. For each of these projects and for each employee, the starting day was at the beginning of a month and the concluding day was the end of a month, and these are indicated by the left and right end points of the corresponding horizontal bars. The number within each bar indicates the percentage of assigned work completed by the employee for that project, as assessed by Dhoni. For each employee, his/her total project-month (in 2020) is the sum of the number of months (s)he worked across the five projects, while his/her annual completion index is the weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects. For each project, the total employee-month is the sum of the number of months four employees worked in this project, while its completion index is the weightage average of the completion percentage assigned for the employees who worked in this project, with the weights being the corresponding number of months they worked in this project. Question 5: Which of the following statements is/are true? I: The total project-month was the same for the four employees. II: The total employee-month was the same for the five projects. a) Only II b) Both I and II c) Neither I nor II d) Only I Solution: The total project month is the number of months Abani, Bahni, Danni, and Tinni individually worked for all the projects combined : Abani – 2+2+5 = 9 months Bahni – 2+4+3 = 9 months Danni – 3+3+2+1 = 9 months Tinni – 2+2+3+2 = 9 months. The total employee month for all the five projects is the sum of the total employee-month is the sum of the number of months four employees worked in this project. Project -1 = 2+2+2 = 6 months Project -2 = 3+2 = 5 months Project – 3 = 2+4+3 = 9 months. Project – 4 = 5+2+3 = 10 months. Project – 5 = 3+1+2 = 6 months. Only statement 1 is true. Question 6: Which employees did not work in multiple projects for any of the months in 2020? a) Only Abani, Bahni and Danni b) Only Abani and Bahni c) All four of them d) Only Tinni Solution: Abani, Banni, and Danni did not work on multiple projects simultaneously in a month Tinni was the only person who worked on multiple projects which are project 4 and project 5 in the month of september. Question 7: The project duration, measured in terms of the number of months, is the time during which at least one employee worked in the project. Which of the following pairs of the projects had the same duration? a) Project 1, Project 5 b) Project 4, Project 5 c) Project 3, Project 5 d) Project 3, Project 4 Solution: Considering the information provided : For project 1 : 3 months. Project – 2: 3 months. Project – 3: 5 months. Project – 4: 5 months. Project – 5: 4 months. Among the given options option D is true which is project 3, project 4. Question 8: The list of employees in decreasing order of annual completion index is: a) Danni, Tinni, Bahni, Abani b) Bahni, Abani, Tinni, Danni c) Danni, Tinni, Abani, Bahni d) Tinni, Danni, Abani, Bahni Solution: The annual completion index for different people is : The weightage average of the completion percentage assigned from the different projects, with the weights being the corresponding number of months (s)he worked in these projects. For Abani : $\frac{\left(\left(100\cdot2\right)+\left(100\cdot2\right)+\left(80\cdot5\right)\right)}{2+2+5}=\ \frac{800}{9}$ For Bahni : $\frac{\left(\left(100\cdot2\right)+\left(75\cdot4\right)+\left(90\cdot3\right)\right)}{2+3+4}=\ \frac{770}{9}$ For Danni : $\frac{\left(\left(90\cdot3\right)+\left(100\cdot3\right)+\left(100\cdot2\right)+\left(100\cdot1\right)\right)}{2+3+2+1}=\ \frac{870}{9}$ For Tinni : $\frac{\left(\left(80\cdot2\right)+\left(100\cdot2\right)+\left(100\cdot3\right)+\left(100\cdot2\right)\right)}{2+2+3+2}=\ \frac{860}{9}$ The descending order for the four people is : Danni, Tinni, Abani, Bahni. Instructions DIRECTIONS for the following four questions: A low-cost airline company connects ten India cities, A to J. The table below gives the distance between a pair of airports and the corresponding price charged by the company. Travel is permitted only from a departure airport to an arrival airport. The customers do not travel by a route where they have to stop at more than two intermediate airports. <img “=”” alt=”” class=”img-responsive” src=”https://cracku.in/media/questionGroup/DI_6_3.png”/> Question 9: What is the lowest possible fare, in rupees, from A to J? a) 2275 b) 2850 c) 2890 d) 2930 e) 3340 Solution: From the table we can see that, the lowest price would be from A to H and H to J. The cost of travel from A to H = Rs 1850 The cost of travel from H to J = Rs 425 Total cost = 1850 + 425 = Rs 2275. Question 10: The company plans to introduce a direct flight between A and J. The market research results indicate that all its existing passengers travelling between A and J will use this direct flight if it is priced 5% below the minimum price that they pay at present. What should the company charge approximately, in rupees, for this direct flight? a) 1991 b) 2161 c) 2707 d) 2745 e) 2783 Solution: From the table we can see that, the lowest price would be from A to H and H to J. The cost of travel from A to H = Rs 1850 The cost of travel from H to J = Rs 425 Total cost = 1850 + 425 = Rs 2275 Lowest price = Rs 2275 95% of 2275 = Rs 2161 Question 11: If the airports C, D and H are closed down owing to security reasons, what would be the minimum price, in rupees, to be paid by a passenger travelling from A to J? a) 2275 b) 2615 c) 2850 d) 2945 e) 3190 Solution: If the airports C, D and H are closed down  the minimum price to be paid by a passenger travelling from A to | J would be by first travelling to F and then from F to J. The cost of travel from A to F = Rs 1700 The cost of travel from F to J = Rs 1150 Total cost = 1700 + 1150 = Rs 2850 Question 12: If the prices include a margin of 10% over the total cost that the company incurs, what is the minimum cost per kilometer that the company incurs in flying from A to J? a) 0.77 b) 0.88 c) 0.99 d) 1.06 e) 1.08 Solution: The minimum cost from A to J we know is 2275. Let the CP to company be C Since 10% over actual CP is the total price i.e. $\text{CP}\times1.1 = 2275 \rightarrow CP = \frac{2275}{1.1}$ The total distance is 1950+1400=2350 Km. Cost per Km = $\dfrac{\frac{2275}{1.1}}{2350}$ = Rs 0.88/Km Question 13: If the prices include a margin of 15% over the total cost that the company incurs, which among the following is the distance to be covered in flying from A to J that minimizes the total cost of travel for the company? a) 2170 b) 2180 c) 2315 d) 2350 e) 2390 |
https://en.m.wikipedia.org/wiki/Pushforward_(differential) | 1,610,832,595,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00033.warc.gz | 334,747,498 | 15,406 | Pushforward (differential)
In differential geometry, pushforward is a linear approximation of smooth maps on tangent spaces. Suppose that φ : MN is a smooth map between smooth manifolds; then the differential of φ at a point x is, in some sense, the best linear approximation of φ near x. It can be viewed as a generalization of the total derivative of ordinary calculus. Explicitly, it is a linear map from the tangent space of M at x to the tangent space of N at φ(x). Hence it can be used to push tangent vectors on M forward to tangent vectors on N. The differential of a map φ is also called, by various authors, the derivative or total derivative of φ.
If a map, φ, carries every point on manifold M to manifold N then the pushforward of φ carries vectors in the tangent space at every point in M to a tangent space at every point in N.
Motivation
Let φ : UV be a smooth map from an open subset U of Rm to an open subset V of Rn. For any point x in U, the Jacobian of φ at x (with respect to the standard coordinates) is the matrix representation of the total derivative of φ at x, which is a linear map
${\displaystyle d\varphi _{x}\colon \mathbf {R} ^{m}\to \mathbf {R} ^{n}\ .}$
We wish to generalize this to the case that φ is a smooth function between any smooth manifolds M and N.
The differential of a smooth map
Let φ : MN be a smooth map of smooth manifolds. Given some xM, the differential of φ at x is a linear map
${\displaystyle d\varphi _{x}:T_{x}M\to T_{\varphi (x)}N\,}$
from the tangent space of M at x to the tangent space of N at φ(x). The application of x to a tangent vector X is sometimes called the pushforward of X by φ. The exact definition of this pushforward depends on the definition one uses for tangent vectors (for the various definitions see tangent space).
If one defines tangent vectors as equivalence classes of curves through x then the differential is given by
${\displaystyle d\varphi _{x}\left(\gamma ^{\prime }(0)\right)=(\varphi \circ \gamma )^{\prime }(0).}$
Here γ is a curve in M with γ(0) = x. In other words, the pushforward of the tangent vector to the curve γ at 0 is just the tangent vector to the curve φγ at 0.
Alternatively, if tangent vectors are defined as derivations acting on smooth real-valued functions, then the differential is given by
${\displaystyle d\varphi _{x}(X)(f)=X(f\circ \varphi ).}$
Here XTxM, therefore X is a derivation defined on M and f is a smooth real-valued function on N. By definition, the pushforward of X at a given x in M is in Tφ(x)N and therefore itself is a derivation.
After choosing charts around x and φ(x), φ is locally determined by a smooth map
${\displaystyle {\widehat {\varphi }}:U\to V}$
between open sets of Rm and Rn, and x has representation (at x)
${\displaystyle d\varphi _{x}\left({\frac {\partial }{\partial u^{a}}}\right)={\frac {\partial {\widehat {\varphi }}^{b}}{\partial u^{a}}}{\frac {\partial }{\partial v^{b}}},}$
in the Einstein summation notation, where the partial derivatives are evaluated at the point in U corresponding to x in the given chart.
Extending by linearity gives the following matrix
${\displaystyle \left(d\varphi _{x}\right)_{a}^{\;b}={\frac {\partial {\widehat {\varphi }}^{b}}{\partial u^{a}}}.}$
Thus the differential is a linear transformation, between tangent spaces, associated to the smooth map φ at each point. Therefore, in some chosen local coordinates, it is represented by the Jacobian matrix of the corresponding smooth map from Rm to Rn. In general the differential need not be invertible. If φ is a local diffeomorphism, then the pushforward at x is invertible and its inverse gives the pullback of Tφ(x)N.
The differential is frequently expressed using a variety of other notations such as
${\displaystyle D\varphi _{x},\;\left(\varphi _{*}\right)_{x},\;\varphi '(x),\;T_{x}\varphi .}$
It follows from the definition that the differential of a composite is the composite of the differentials (i.e., functorial behaviour). This is the chain rule for smooth maps.
Also, the differential of a local diffeomorphism is a linear isomorphism of tangent spaces.
The differential on the tangent bundle
The differential of a smooth map φ induces, in an obvious manner, a bundle map (in fact a vector bundle homomorphism) from the tangent bundle of M to the tangent bundle of N, denoted by or φ, which fits into the following commutative diagram:
where πM and πN denote the bundle projections of the tangent bundles of M and N respectively.
${\displaystyle \operatorname {d} \!\varphi }$ induces a bundle map from TM to the pullback bundle φTN over M via
${\displaystyle (m,v_{m})\mapsto (m,\operatorname {d} \!\varphi (m,v_{m})),}$
where ${\displaystyle m\in M}$ and ${\displaystyle v_{m}\in T_{m}M.}$ The latter map may in turn be viewed as a section of the vector bundle Hom(TM, φTN) over M. The bundle map is also denoted by and called the tangent map. In this way, T is a functor.
Pushforward of vector fields
Given a smooth map φ : MN and a vector field X on M, it is not usually possible to identify a pushforward of X by φ with some vector field Y on N. For example, if the map φ is not surjective, there is no natural way to define such a pushforward outside of the image of φ. Also, if φ is not injective there may be more than one choice of pushforward at a given point. Nevertheless, one can make this difficulty precise, using the notion of a vector field along a map.
A section of φTN over M is called a vector field along φ. For example, if M is a submanifold of N and φ is the inclusion, then a vector field along φ is just a section of the tangent bundle of N along M; in particular, a vector field on M defines such a section via the inclusion of TM inside TN. This idea generalizes to arbitrary smooth maps.
Suppose that X is a vector field on M, i.e., a section of TM. Then, ${\displaystyle \operatorname {d} \!\varphi \circ X}$ yields, in the above sense, the pushforward φX, which is a vector field along φ, i.e., a section of φTN over M.
Any vector field Y on N defines a pullback section φY of φTN with (φY)x = Yφ(x). A vector field X on M and a vector field Y on N are said to be φ-related if φX = φY as vector fields along φ. In other words, for all x in M, x(X) = Yφ(x).
In some situations, given a X vector field on M, there is a unique vector field Y on N which is φ-related to X. This is true in particular when φ is a diffeomorphism. In this case, the pushforward defines a vector field Y on N, given by
${\displaystyle Y_{y}=\varphi _{*}\left(X_{\varphi ^{-1}(y)}\right).}$
A more general situation arises when φ is surjective (for example the bundle projection of a fiber bundle). Then a vector field X on M is said to be projectable if for all y in N, x(Xx) is independent of the choice of x in φ−1({y}). This is precisely the condition that guarantees that a pushforward of X, as a vector field on N, is well defined.
References
• Lee, John M. (2003). Introduction to Smooth Manifolds. Springer Graduate Texts in Mathematics. 218.
• Jost, Jürgen (2002). Riemannian Geometry and Geometric Analysis. Berlin: Springer-Verlag. ISBN 3-540-42627-2. See section 1.6.
• Abraham, Ralph; Marsden, Jerrold E. (1978). Foundations of Mechanics. London: Benjamin-Cummings. ISBN 0-8053-0102-X. See section 1.7 and 2.3. | 1,935 | 7,328 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 14, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2021-04 | latest | en | 0.876963 | Pushforward (differential) In differential geometry, pushforward is a linear approximation of smooth maps on tangent spaces. Suppose that φ : MN is a smooth map between smooth manifolds; then the differential of φ at a point x is, in some sense, the best linear approximation of φ near x. It can be viewed as a generalization of the total derivative of ordinary calculus. Explicitly, it is a linear map from the tangent space of M at x to the tangent space of N at φ(x). Hence it can be used to push tangent vectors on M forward to tangent vectors on N. The differential of a map φ is also called, by various authors, the derivative or total derivative of φ. If a map, φ, carries every point on manifold M to manifold N then the pushforward of φ carries vectors in the tangent space at every point in M to a tangent space at every point in N. Motivation Let φ : UV be a smooth map from an open subset U of Rm to an open subset V of Rn. For any point x in U, the Jacobian of φ at x (with respect to the standard coordinates) is the matrix representation of the total derivative of φ at x, which is a linear map ${\displaystyle d\varphi _{x}\colon \mathbf {R} ^{m}\to \mathbf {R} ^{n}\ .}$ We wish to generalize this to the case that φ is a smooth function between any smooth manifolds M and N. The differential of a smooth map Let φ : MN be a smooth map of smooth manifolds. Given some xM, the differential of φ at x is a linear map ${\displaystyle d\varphi _{x}:T_{x}M\to T_{\varphi (x)}N\,}$ from the tangent space of M at x to the tangent space of N at φ(x). The application of x to a tangent vector X is sometimes called the pushforward of X by φ. The exact definition of this pushforward depends on the definition one uses for tangent vectors (for the various definitions see tangent space). If one defines tangent vectors as equivalence classes of curves through x then the differential is given by ${\displaystyle d\varphi _{x}\left(\gamma ^{\prime }(0)\right)=(\varphi \circ \gamma )^{\prime }(0).}$ Here γ is a curve in M with γ(0) = x. In other words, the pushforward of the tangent vector to the curve γ at 0 is just the tangent vector to the curve φγ at 0. Alternatively, if tangent vectors are defined as derivations acting on smooth real-valued functions, then the differential is given by ${\displaystyle d\varphi _{x}(X)(f)=X(f\circ \varphi ).}$ Here XTxM, therefore X is a derivation defined on M and f is a smooth real-valued function on N. By definition, the pushforward of X at a given x in M is in Tφ(x)N and therefore itself is a derivation. After choosing charts around x and φ(x), φ is locally determined by a smooth map ${\displaystyle {\widehat {\varphi }}:U\to V}$ between open sets of Rm and Rn, and x has representation (at x) ${\displaystyle d\varphi _{x}\left({\frac {\partial }{\partial u^{a}}}\right)={\frac {\partial {\widehat {\varphi }}^{b}}{\partial u^{a}}}{\frac {\partial }{\partial v^{b}}},}$ in the Einstein summation notation, where the partial derivatives are evaluated at the point in U corresponding to x in the given chart. Extending by linearity gives the following matrix ${\displaystyle \left(d\varphi _{x}\right)_{a}^{\;b}={\frac {\partial {\widehat {\varphi }}^{b}}{\partial u^{a}}}.}$ Thus the differential is a linear transformation, between tangent spaces, associated to the smooth map φ at each point. Therefore, in some chosen local coordinates, it is represented by the Jacobian matrix of the corresponding smooth map from Rm to Rn. In general the differential need not be invertible. If φ is a local diffeomorphism, then the pushforward at x is invertible and its inverse gives the pullback of Tφ(x)N. The differential is frequently expressed using a variety of other notations such as ${\displaystyle D\varphi _{x},\;\left(\varphi _{*}\right)_{x},\;\varphi '(x),\;T_{x}\varphi .}$ It follows from the definition that the differential of a composite is the composite of the differentials (i.e., functorial behaviour). This is the chain rule for smooth maps. Also, the differential of a local diffeomorphism is a linear isomorphism of tangent spaces. The differential on the tangent bundle The differential of a smooth map φ induces, in an obvious manner, a bundle map (in fact a vector bundle homomorphism) from the tangent bundle of M to the tangent bundle of N, denoted by or φ, which fits into the following commutative diagram: where πM and πN denote the bundle projections of the tangent bundles of M and N respectively. ${\displaystyle \operatorname {d} \!\varphi }$ induces a bundle map from TM to the pullback bundle φTN over M via ${\displaystyle (m,v_{m})\mapsto (m,\operatorname {d} \!\varphi (m,v_{m})),}$ where ${\displaystyle m\in M}$ and ${\displaystyle v_{m}\in T_{m}M.}$ The latter map may in turn be viewed as a section of the vector bundle Hom(TM, φTN) over M. The bundle map is also denoted by and called the tangent map. In this way, T is a functor. Pushforward of vector fields Given a smooth map φ : MN and a vector field X on M, it is not usually possible to identify a pushforward of X by φ with some vector field Y on N. For example, if the map φ is not surjective, there is no natural way to define such a pushforward outside of the image of φ. Also, if φ is not injective there may be more than one choice of pushforward at a given point. Nevertheless, one can make this difficulty precise, using the notion of a vector field along a map. A section of φTN over M is called a vector field along φ. For example, if M is a submanifold of N and φ is the inclusion, then a vector field along φ is just a section of the tangent bundle of N along M; in particular, a vector field on M defines such a section via the inclusion of TM inside TN. This idea generalizes to arbitrary smooth maps. Suppose that X is a vector field on M, i.e., a section of TM. Then, ${\displaystyle \operatorname {d} \!\varphi \circ X}$ yields, in the above sense, the pushforward φX, which is a vector field along φ, i.e., a section of φTN over M. Any vector field Y on N defines a pullback section φY of φTN with (φY)x = Yφ(x). A vector field X on M and a vector field Y on N are said to be φ-related if φX = φY as vector fields along φ. In other words, for all x in M, x(X) = Yφ(x). In some situations, given a X vector field on M, there is a unique vector field Y on N which is φ-related to X. This is true in particular when φ is a diffeomorphism. In this case, the pushforward defines a vector field Y | on N, given by ${\displaystyle Y_{y}=\varphi _{*}\left(X_{\varphi ^{-1}(y)}\right).}$ A more general situation arises when φ is surjective (for example the bundle projection of a fiber bundle). Then a vector field X on M is said to be projectable if for all y in N, x(Xx) is independent of the choice of x in φ−1({y}). This is precisely the condition that guarantees that a pushforward of X, as a vector field on N, is well defined. References • Lee, John M. (2003). Introduction to Smooth Manifolds. Springer Graduate Texts in Mathematics. 218. • Jost, Jürgen (2002). Riemannian Geometry and Geometric Analysis. Berlin: Springer-Verlag. ISBN 3-540-42627-2. See section 1.6. • Abraham, Ralph; Marsden, Jerrold E. (1978). Foundations of Mechanics. London: Benjamin-Cummings. ISBN 0-8053-0102-X. See section 1.7 and 2.3. |
https://www.coursehero.com/file/15526/570-homework6-sol/ | 1,490,807,610,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190754.6/warc/CC-MAIN-20170322212950-00327-ip-10-233-31-227.ec2.internal.warc.gz | 879,492,543 | 21,613 | 570 homework6 sol
# 570 homework6 sol - HW6 SOLUTION 1 Chapter 7 Problem 3 Part(a ν f = X e out of s f e = 6 3 1 = 10 Constructing the residual graph of the current
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Unformatted text preview: HW6 SOLUTION 1. Chapter 7 Problem 3: Part (a) ν ( f ) = X e out of s f ( e ) = 6 + 3 + 1 = 10 Constructing the residual graph of the current flow network by the Ford-Fulkerson algorithm in the text book, there is a s-t path( s → a → c → b → d → t ) in the residual graph of current flow network, thus the current net flow is not the maximum of the network Part (b) By the Ford-Fulkerson algorithm, we can get the max-flow and the max-flow is 7 + 3 + 1 = 11. And the minimum s- t cut is A = { s,a,b,c } ,B = { d,t } since C ( A,B ) = X e out of A c ( e ) = 5 + 5 + 1 = 11 Problem 5: The statement is false. We take Figure (7.3)(a) in Page 341 as the counter example by changing the capacity c u,v = 9 . 5. Clearly the minimum cut is A = { s,u } and B = { v,t } . After we add 1 to every capacity, ( A,B ) is not a minimum cut any more as the value of max flow is 32 now while the cut between A and B is 32 . 5. Problem 7: We first construct a bipartite graph G whose node are the customers c i (0 < i ≤ n ) and the base station b j (0 < j ≤ k ). We add edge e ij = ( c i ,b j ) in the graph G if and only if the distance of client c i and the base station b j is within r (the coordinates of both are given) and set c e ij = 1. then we add a source s and sink t in G . For each customer c i , we add e = ( s,c i ) in the graph G and set c e = 1.For each base station b j , we add e = ( b j ,t ) in the graph G and set c e = L . After building the graph G for the original problem, we find the maximum s-t-flow value v in graph G by Fulkerson algorithm. If v = n , then every client can be connected simultaneously to a base station, otherwise it can’t. The correctness of the algorithm comes from the following lemma. Lemma 1: Every client can be connected simultaneous to a base station if and only the value of the maximum value of an s- t- flow in G is n . Proof: First if every client can be connected simultaneous to a base station, by the con- struction of the graph G , it’s easy to see that this flow meets all capacity constraints and has value n . For the converse direction, assume that there is an s- t- flow of value n . The construction given in the algorithm makes sure that the base station is within the Date : Oct 29, 2006. 1 2 HW6 SOLUTION distance r of the customers since flow goes from c i to b j . So we only need to make sure that all customers are connected to some base station. Notice that { s } ,V-{ s } is a minimum s- t cut(it has value n ), so the flow must saturate all edges crossing the cut. Therefore, all customers must be connected to some base stations. This completes the correctness proof. Running time: The time complexity of constructing the graph is O ( nk ) since we need to see if the distance of each ( b i ,c j ) is below the range parameter r or not....
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## This note was uploaded on 02/27/2008 for the course CSCI 570 taught by Professor Shamsian during the Fall '06 term at USC.
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570 homework6 sol - HW6 SOLUTION 1 Chapter 7 Problem 3 Part(a ν f = X e out of s f e = 6 3 1 = 10 Constructing the residual graph of the current
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Ask a homework question - tutors are online | 1,007 | 3,686 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-13 | longest | en | 0.887108 | 570 homework6 sol # 570 homework6 sol - HW6 SOLUTION 1 Chapter 7 Problem 3 Part(a ν f = X e out of s f e = 6 3 1 = 10 Constructing the residual graph of the current This preview shows pages 1–3. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: HW6 SOLUTION 1. Chapter 7 Problem 3: Part (a) ν ( f ) = X e out of s f ( e ) = 6 + 3 + 1 = 10 Constructing the residual graph of the current flow network by the Ford-Fulkerson algorithm in the text book, there is a s-t path( s → a → c → b → d → t ) in the residual graph of current flow network, thus the current net flow is not the maximum of the network Part (b) By the Ford-Fulkerson algorithm, we can get the max-flow and the max-flow is 7 + 3 + 1 = 11. And the minimum s- t cut is A = { s,a,b,c } ,B = { d,t } since C ( A,B ) = X e out of A c ( e ) = 5 + 5 + 1 = 11 Problem 5: The statement is false. We take Figure (7.3)(a) in Page 341 as the counter example by changing the capacity c u,v = 9 . 5. Clearly the minimum cut is A = { s,u } and B = { v,t } . After we add 1 to every capacity, ( A,B ) is not a minimum cut any more as the value of max flow is 32 now while the cut between A and B is 32 . 5. Problem 7: We first construct a bipartite graph G whose node are the customers c i (0 < i ≤ n ) and the base station b j (0 < j ≤ k ). We add edge e ij = ( c i ,b j ) in the graph G if and only if the distance of client c i and the base station b j is within r (the coordinates of both are given) and set c e ij = 1. then we add a source s and sink t in G . For each customer c i , we add e = ( s,c i ) in the graph G and set c e = 1.For each base station b j , we add e = ( b j ,t ) in the graph G and set c e = L . After building the graph G for the original problem, we find the maximum s-t-flow value v in graph G by Fulkerson algorithm. If v = n , then every client can be connected simultaneously to a base station, otherwise it can’t. The correctness of the algorithm comes from the following lemma. Lemma 1: Every client can be connected simultaneous to a base station if and only the value of the maximum value of an s- t- flow in G is n . Proof: First if every client can be connected simultaneous to a base station, by the con- struction of the graph G , it’s easy to see that this flow meets all capacity constraints and has value n . For the converse direction, assume that there is an s- t- flow of value n . The construction given in the algorithm makes sure that the base station is within the Date : Oct 29, 2006. 1 2 HW6 SOLUTION distance r of the customers since flow goes from c i to b j . So we only need to make sure that all customers are connected to some base station. Notice that { s } ,V-{ s } is a minimum s- t cut(it has value n ), so the flow must saturate all edges crossing the cut. Therefore, all customers must be connected to some base stations. This completes the correctness proof. Running time: The time complexity of constructing the graph is O ( nk ) since we need to see if the distance of each ( b i ,c j ) is below the range parameter r or not.... View Full Document ## This note was uploaded | on 02/27/2008 for the course CSCI 570 taught by Professor Shamsian during the Fall '06 term at USC. ### Page1 / 6 570 homework6 sol - HW6 SOLUTION 1 Chapter 7 Problem 3 Part(a ν f = X e out of s f e = 6 3 1 = 10 Constructing the residual graph of the current This preview shows document pages 1 - 3. Sign up to view the full document. View Full Document Ask a homework question - tutors are online |
https://www.jiskha.com/questions/458485/jill-has-4-pints-of-water-in-a-4-pint-bucket-jack-has-an-empty-1-pint-bucket-and-an-empty | 1,580,145,033,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700988.64/warc/CC-MAIN-20200127143516-20200127173516-00078.warc.gz | 915,010,382 | 6,296 | # MATH - PLZ NEED HELP
Jill has 4 pints of water in a 4-pint bucket. Jack has an empty 1-pint bucket and an empty 3-pint bucket. None of the buckets has measurement markings. How can they divide the water so each has exactly 2 pints?
1. 👍 0
2. 👎 0
3. 👁 155
1. You can't. The 1 pint bucket will not hold 2 pints.
Also, there are 3 buckets. 3x2=6.
The question is horribly flawed.
1. 👍 0
2. 👎 0
2. From the 4L can, fill the 1L can
Condidtion:
4L can : has 3 L
1L can is full
3L can is emptyl
pour the remaining 3L from the 4L can into the empty 3 L can
Condition:
4L can is empty
1L can is full
3L can is full
Pour the 1L can back into the now empty 4L can, and fill the 1L from the 3L can
Conditon:
4L can holds 1 L
1L can has 1 L
3L can has 2 L
pour the 1L can into the 4L can
Final condition
4L can has 2 L
1L can is empty
3L can has 2 L
so Jill has 2 L in her 4L can, and Jack has 2 L in his 3L can.
1. 👍 0
2. 👎 0
posted by Reiny
3. But then the 1 liter bucket has 0 liters of water.
1. 👍 0
2. 👎 0
4. I interpreted the "they" in "How can they divide the water so each has exactly 2 pints?" as referring to Jack and Jill.
Jack and Jill were the subjects in the first two sentences.
1. 👍 0
2. 👎 0
posted by Reiny
5. Badly written question, then, right? The word "each" needs to be clarified.
1. 👍 0
2. 👎 0
6. I agree.
1. 👍 0
2. 👎 0
posted by Reiny
## Similar Questions
1. ### Math
Jill has 4 pints of water in a 4-pint bucket. Jack has an empty 1-pint bucket and an empty 3-pint bucket. None of the buckets has measurement markings. How can they divide the water so each has exactly 2 pints?
asked by Roni on December 1, 2010
Water is poured into a bucket according to the rate F(t)=(t+7)/(2+t), and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/(t+2), with both F(t) and E(t) measured in pints per minute. How much
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Water is poured into a bucket according to the rate F(t)=(t+7)/(t+2) , and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/t+2 , with both F(t) and E(t) measured in pints per minute. How much
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How do I figure out 6c=__?__pt showing work See my other response. =) Brooke, You REALLY need to learn to do these systematically. Learn to watch the units, make the ones you don't want cancel, and keep the units you want for the
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how do you measure and get 4 gallons of water only having an unmarked 5 gallon bucket and an unmarked 3 gallon bucket and a large container of water ok this is simple, u just have to think out of the bucket...heh.. so here goes
asked by john on June 1, 2007
8. ### math
6c=___pt Can someone show me steps please? You look up the conversion factor. There are two ways it can look; i.e., 2 cups = 1 pint OR 1 pint = 2 cups. Here is the way to do ALL conversions. ??pts = what you are given x conversion
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Imagine that you fill a bucket to the top with water and then place the bucket outside in freezing weather. Which of the following describes what will happen when the water in the bucket freezes? A. The level of the water and the
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asked by Jenny on June 10, 2017
More Similar Questions | 1,267 | 4,075 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-05 | latest | en | 0.961108 | # MATH - PLZ NEED HELP Jill has 4 pints of water in a 4-pint bucket. Jack has an empty 1-pint bucket and an empty 3-pint bucket. None of the buckets has measurement markings. How can they divide the water so each has exactly 2 pints? 1. 👍 0 2. 👎 0 3. 👁 155 1. You can't. The 1 pint bucket will not hold 2 pints. Also, there are 3 buckets. 3x2=6. The question is horribly flawed. 1. 👍 0 2. 👎 0 2. From the 4L can, fill the 1L can Condidtion: 4L can : has 3 L 1L can is full 3L can is emptyl pour the remaining 3L from the 4L can into the empty 3 L can Condition: 4L can is empty 1L can is full 3L can is full Pour the 1L can back into the now empty 4L can, and fill the 1L from the 3L can Conditon: 4L can holds 1 L 1L can has 1 L 3L can has 2 L pour the 1L can into the 4L can Final condition 4L can has 2 L 1L can is empty 3L can has 2 L so Jill has 2 L in her 4L can, and Jack has 2 L in his 3L can. 1. 👍 0 2. 👎 0 posted by Reiny 3. But then the 1 liter bucket has 0 liters of water. 1. 👍 0 2. 👎 0 4. I interpreted the "they" in "How can they divide the water so each has exactly 2 pints?" as referring to Jack and Jill. Jack and Jill were the subjects in the first two sentences. 1. 👍 0 2. 👎 0 posted by Reiny 5. Badly written question, then, right? The word "each" needs to be clarified. 1. 👍 0 2. 👎 0 6. I agree. 1. 👍 0 2. 👎 0 posted by Reiny ## Similar Questions 1. ### Math Jill has 4 pints of water in a 4-pint bucket. Jack has an empty 1-pint bucket and an empty 3-pint bucket. None of the buckets has measurement markings. How can they divide the water so each has exactly 2 pints? asked by Roni on December 1, 2010 Water is poured into a bucket according to the rate F(t)=(t+7)/(2+t), and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/(t+2), with both F(t) and E(t) measured in pints per minute. How much asked by Ke\$ha on May 22, 2017 Water is poured into a bucket according to the rate F(t)=(t+7)/(t+2) , and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/t+2 , with both F(t) and E(t) measured in pints per minute. How much asked by Hannah on April 27, 2017 4. ### Machine Jack & Jill went to a well to fetch a bucket of water. The full bucket was 500N. Jack weighed 50kg and Jill weighed 40kg. How would you get bucket from the well asked by Paul on January 30, 2016 5. ### Math George has a 5 pint pitcher and a 3 pint pitcher that are both empty. Each move consists of either pouring water into a pitcher, out of a pitcher, or from one pitcher to the other. What is the least number of moves it would take asked by PT on January 16, 2019 6. ### math How do I figure out 6c=__?__pt showing work See my other response. =) Brooke, You REALLY need to learn to do these systematically. Learn to watch the units, make the ones you don't want cancel, and keep the units you want for the asked by Brooke on November 28, 2006 7. ### math how do you measure and get 4 gallons of water only having an unmarked 5 gallon bucket and an unmarked 3 gallon bucket and a large container of water ok this is simple, u just have to think out of the bucket...heh.. so here goes asked by john on June 1, 2007 8. ### math 6c=___pt Can someone show me steps please? You look up the conversion factor. There are two ways it can look; i.e., 2 cups = 1 pint OR 1 pint = 2 cups. Here is the way to do ALL conversions. ??pts = what you are given x conversion asked by brooke on November 30, 2006 9. ### science Imagine that you fill a bucket to the top with water and then place the bucket outside in freezing weather. | Which of the following describes what will happen when the water in the bucket freezes? A. The level of the water and the asked by Batman on February 27, 2013 10. ### Math using a certain garden hose, liza can fill her fancy new bucket with water in 45 seconds. Unfortunately, Henry accidently makes a hole in the bucket, so if initially full, the bucket will now become empty in 1 minute. How long asked by Jenny on June 10, 2017 More Similar Questions |
https://oeis.org/A244279 | 1,726,818,507,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652138.47/warc/CC-MAIN-20240920054402-20240920084402-00310.warc.gz | 396,875,570 | 5,276 | The OEIS is supported by the many generous donors to the OEIS Foundation.
A244279
Numerators of the n-th iteration of the alternating continued fraction of the positive integers, initiated with (1 + ...).
4
1, 1, 7, 17, 127, 547, 5111, 31865, 358781, 2938437, 38808271, 394282041, 5982064475, 72608885159, 1245025688399, 17581129642961, 336297031232409, 5417081623572649, 114375064174857015, 2069902867431592833, 47819312187294567447, 960634689914268797707
OFFSET
1,3
COMMENTS
As n-->inf, a(n)/A244280(n) converges to 0.628736607098954801603428... ; this number has a surprisingly elegant standard continued fraction representation of [0; 1, 1, 1, 2, 3, 1, 4, 5, 1, 6, 7, 1, 8, 9...].
LINKS
FORMULA
This is the result of taking the numerator of a continued fraction with alternating signs a(n) = 1/(1+1/(2-1/(3+1/(4-...1/(n +/- 1))))), where addition follows an odd number and subtraction follows an even number.
EXAMPLE
a(1) = 1/(1+1) = 1/2;
a(2) = 1/(1+1/(2-1)) = 1/2;
a(3) = 1/(1+1/(2-1/(3+1))) = 7/11;
a(4) = 1/(1+1/(2-1/(3+1/(4-1)))) = 17/27.
MAPLE
seq(numer(numtheory:-cfrac([0, [1, 1], seq([(-1)^j, j], j=2..n), [(-1)^(n+1), 1]])), n = 1..40); # Robert Israel, Jan 17 2016
PROG
(PARI) a(n) = if(n%2==0, s=-1, s=1); t=1; while(n>0, t=n+s/t; n--; s=-s); numerator(t=1/t)
vector(30, n, a(n)) \\ Colin Barker, Jul 20 2014
CROSSREFS
Cf. A244280 (Denominators).
Sequence in context: A266382 A118108 A227506 * A364703 A325584 A375426
KEYWORD
nonn,frac
AUTHOR
Mohamed Sabba, Jun 24 2014
EXTENSIONS
More terms from Colin Barker, Jul 20 2014
STATUS
approved
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Last modified September 20 03:37 EDT 2024. Contains 376016 sequences. (Running on oeis4.) | 732 | 1,884 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2024-38 | latest | en | 0.633857 | The OEIS is supported by the many generous donors to the OEIS Foundation. A244279 Numerators of the n-th iteration of the alternating continued fraction of the positive integers, initiated with (1 + ...). 4 1, 1, 7, 17, 127, 547, 5111, 31865, 358781, 2938437, 38808271, 394282041, 5982064475, 72608885159, 1245025688399, 17581129642961, 336297031232409, 5417081623572649, 114375064174857015, 2069902867431592833, 47819312187294567447, 960634689914268797707 OFFSET 1,3 COMMENTS As n-->inf, a(n)/A244280(n) converges to 0.628736607098954801603428... ; this number has a surprisingly elegant standard continued fraction representation of [0; 1, 1, 1, 2, 3, 1, 4, 5, 1, 6, 7, 1, 8, 9...]. LINKS FORMULA This is the result of taking the numerator of a continued fraction with alternating signs a(n) = 1/(1+1/(2-1/(3+1/(4-...1/(n +/- 1))))), where addition follows an odd number and subtraction follows an even number. EXAMPLE a(1) = 1/(1+1) = 1/2; a(2) = 1/(1+1/(2-1)) = 1/2; a(3) = 1/(1+1/(2-1/(3+1))) = 7/11; a(4) = 1/(1+1/(2-1/(3+1/(4-1)))) = 17/27. MAPLE seq(numer(numtheory:-cfrac([0, [1, 1], seq([(-1)^j, j], j=2..n), [(-1)^(n+1), 1]])), n = 1..40); # Robert Israel, Jan 17 2016 PROG (PARI) a(n) = if(n%2==0, s=-1, s=1); t=1; while(n>0, t=n+s/t; n--; s=-s); numerator(t=1/t) vector(30, n, a(n)) \\ Colin Barker, Jul 20 2014 CROSSREFS Cf. A244280 (Denominators). Sequence in context: A266382 A118108 A227506 * A364703 A325584 A375426 KEYWORD nonn,frac AUTHOR Mohamed Sabba, Jun 24 2014 EXTENSIONS More terms from Colin Barker, Jul 20 2014 STATUS approved Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | WebCam Contribute new seq. or comment | Format | Style Sheet | | Transforms | Superseeker | Recents The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified September 20 03:37 EDT 2024. Contains 376016 sequences. (Running on oeis4.) |
https://www.gradesaver.com/textbooks/math/algebra/elementary-and-intermediate-algebra-concepts-and-applications-6th-edition/chapter-9-inequalities-and-problem-solving-9-1-inequalities-and-applications-9-1-exercise-set-page-580/6 | 1,534,246,642,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221209021.21/warc/CC-MAIN-20180814101420-20180814121420-00206.warc.gz | 1,130,809,063 | 13,534 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Using the Distributive Property, which is given by $a(b+c)=ab+ac,$ the expression , $2(4x+1),$ is equivalent to \begin{array}{l}\require{cancel} 2(4x+1) \\\\= 2(4x)+2(1) \\\\= 8x+2 .\end{array} Since this is the same as the second given expression, then the given are $\text{ equivalent expressions .}$ | 116 | 381 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2018-34 | longest | en | 0.854368 | ## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) Using the Distributive Property, which is given by $a(b+c)=ab+ac,$ the expression , $2(4x+1),$ is equivalent to \begin{array}{l}\require{cancel} 2(4x+1) \\\\= 2(4x)+2(1) \\\\= 8x+2 .\end{array} Since this is the same as the second given expression, then the | given are $\text{ equivalent expressions .}$ |
https://stacks.math.columbia.edu/tag/01OJ | 1,679,947,919,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00035.warc.gz | 568,643,250 | 7,308 | ## 28.3 Integral, irreducible, and reduced schemes
Definition 28.3.1. Let $X$ be a scheme. We say $X$ is integral if it is nonempty and for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is an integral domain.
Lemma 28.3.2. Let $X$ be a scheme. The following are equivalent.
1. The scheme $X$ is reduced, see Schemes, Definition 26.12.1.
2. There exists an affine open covering $X = \bigcup U_ i$ such that each $\Gamma (U_ i, \mathcal{O}_ X)$ is reduced.
3. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.
4. For every open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced.
Lemma 28.3.3. Let $X$ be a scheme. The following are equivalent.
1. The scheme $X$ is irreducible.
2. There exists an affine open covering $X = \bigcup _{i \in I} U_ i$ such that $I$ is not empty, $U_ i$ is irreducible for all $i \in I$, and $U_ i \cap U_ j \not= \emptyset$ for all $i, j \in I$.
3. The scheme $X$ is nonempty and every nonempty affine open $U \subset X$ is irreducible.
Proof. Assume (1). By Schemes, Lemma 26.11.1 we see that $X$ has a unique generic point $\eta$. Then $X = \overline{\{ \eta \} }$. Hence $\eta$ is an element of every nonempty affine open $U \subset X$. This implies that $\eta \in U$ is dense hence $U$ is irreducible. It also implies any two nonempty affines meet. Thus (1) implies both (2) and (3).
Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets. For every $i$ we see that either $U_ i \subset Z_1$ or $U_ i \subset Z_2$. Pick some $i \in I$ and assume $U_ i \subset Z_1$ (possibly after renumbering $Z_1$, $Z_2$). For any $j \in I$ the open subset $U_ i \cap U_ j$ is dense in $U_ j$ and contained in the closed subset $Z_1 \cap U_ j$. We conclude that also $U_ j \subset Z_1$. Thus $X = Z_1$ as desired.
Assume (3). Choose an affine open covering $X = \bigcup _{i \in I} U_ i$. We may assume that each $U_ i$ is nonempty. Since $X$ is nonempty we see that $I$ is not empty. By assumption each $U_ i$ is irreducible. Suppose $U_ i \cap U_ j = \emptyset$ for some pair $i, j \in I$. Then the open $U_ i \amalg U_ j = U_ i \cup U_ j$ is affine, see Schemes, Lemma 26.6.8. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved. $\square$
Lemma 28.3.4. A scheme $X$ is integral if and only if it is reduced and irreducible.
Proof. If $X$ is irreducible, then every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma 28.3.2 above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain.
If $X$ is integral, then for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma 28.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 28.3.3. $\square$
In Examples, Section 109.6 we construct a connected affine scheme all of whose local rings are domains, but which is not integral.
Comment #4228 by Lin on
Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?
Comment #4229 by Lin on
Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ?
Comment #4409 by on
Hmm... I think you are saying the same thing as the proof says.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). | 1,268 | 3,952 | {"found_math": true, "script_math_tex": 10, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-14 | latest | en | 0.794391 | ## 28.3 Integral, irreducible, and reduced schemes Definition 28.3.1. Let $X$ be a scheme. We say $X$ is integral if it is nonempty and for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is an integral domain. Lemma 28.3.2. Let $X$ be a scheme. The following are equivalent. 1. The scheme $X$ is reduced, see Schemes, Definition 26.12.1. 2. There exists an affine open covering $X = \bigcup U_ i$ such that each $\Gamma (U_ i, \mathcal{O}_ X)$ is reduced. 3. For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced. 4. For every open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is reduced. Lemma 28.3.3. Let $X$ be a scheme. The following are equivalent. 1. The scheme $X$ is irreducible. 2. There exists an affine open covering $X = \bigcup _{i \in I} U_ i$ such that $I$ is not empty, $U_ i$ is irreducible for all $i \in I$, and $U_ i \cap U_ j \not= \emptyset$ for all $i, j \in I$. 3. The scheme $X$ is nonempty and every nonempty affine open $U \subset X$ is irreducible. Proof. Assume (1). By Schemes, Lemma 26.11.1 we see that $X$ has a unique generic point $\eta$. Then $X = \overline{\{ \eta \} }$. Hence $\eta$ is an element of every nonempty affine open $U \subset X$. This implies that $\eta \in U$ is dense hence $U$ is irreducible. It also implies any two nonempty affines meet. Thus (1) implies both (2) and (3). Assume (2). Suppose $X = Z_1 \cup Z_2$ is a union of two closed subsets. For every $i$ we see that either $U_ i \subset Z_1$ or $U_ i \subset Z_2$. Pick some $i \in I$ and assume $U_ i \subset Z_1$ (possibly after renumbering $Z_1$, $Z_2$). For any $j \in I$ the open subset $U_ i \cap U_ j$ is dense in $U_ j$ and contained in the closed subset $Z_1 \cap U_ j$. We conclude that also $U_ j \subset Z_1$. Thus $X = Z_1$ as desired. Assume (3). Choose an affine open covering $X = \bigcup _{i \in I} U_ i$. We may assume that each $U_ i$ is nonempty. Since $X$ is nonempty we see that $I$ is not empty. By assumption each $U_ i$ is irreducible. Suppose $U_ i \cap U_ j = \emptyset$ for some pair $i, j \in I$. Then the open $U_ i \amalg U_ j = U_ i \cup U_ j$ is affine, see Schemes, Lemma 26.6.8. Hence it is irreducible by assumption which is absurd. We conclude that (3) implies (2). The lemma is proved. $\square$ Lemma 28.3.4. A scheme $X$ is integral if and only if it is reduced and irreducible. Proof. If $X$ is irreducible, then every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma 28.3.2 above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain. If $X$ is integral, then for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma 28.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 28.3.3. $\square$ In Examples, Section 109.6 we construct a connected affine scheme all of whose local rings are domains, but which is not integral. Comment #4228 by Lin on Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ? Comment #4229 by Lin on Correct me If am wrong, it seems that (2) to (1) of lemma [01OM], 2nd paragraph, follows directly from the first line. If $U_j \cap U_i \not= \emptyset$, then $U_j$ cannot | contain a point in $Z_2$, hence $U_j \subseteq Z_1$, ( otherwise the $Z_1, Z_2$ forms a decomposition. ) ? Comment #4409 by on Hmm... I think you are saying the same thing as the proof says. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). |
https://www.sarthaks.com/1386191/solubility-product-constant-cro-agbr-1xx10-0xx10-respectively-calculate-ratio-molarit | 1,685,648,227,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648000.54/warc/CC-MAIN-20230601175345-20230601205345-00542.warc.gz | 1,082,908,683 | 15,395 | # The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarit
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The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarities of their saturated solutions.
by (65.9k points)
selected
Ag_(2)CrO_(4) dissolves in water and the equilibrium in the saturated solution is
Ag_(2)CrO_(4)(s)= 2Ag^(o+)(aq)+CrO_(4)^(2-)(aq)
Let the solubolty of AgBr be S_(1)
[Ag^(o+)(aq)]=2S_(1)
and [CrO_(4)^(2-)(aq)=S_(1)]
K_(sp)=[Ag^(o+)(aq)]^(2)[CrO_(4)^(2-)(aq)]=(2S_(1))^(2)xxS=4S_(1)^(3)
or S_(1)=(K_(sp)/4)^(1//3)=((1.1xx10^(-12))/4)^(1//3)=0.65xx10^(-14) M
Similarly AgBr=Ag^(o+)(aq)+Br^(Θ)(aq)
Let the solubility of AgBr be S_(2)
[Ag^(o+)(aq)]=S_(2) and Br^(Θ)(aq)]=S_(2)
K_(sp)=[Ag^(o+)(aq)][Br^(Θ)(aq)]
S_(2)xxS_(2)=S_(2)^(2)
or S_(2)=(K_(sp))^(1//2)=(5.0xx10^(-13))^(1//2)
=(0.5xx10^(-12))^(1//2)=0.707xx10^(-6) mol L^(-1)
Ratio of the molarities of silver chromate to silver bromide comes to =(0.65xx10^(-4))/(0.707xx10^(-6))=91.9 silver chromate is more soluble. | 506 | 1,135 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2023-23 | latest | en | 0.716339 | # The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarit 26 views closed The solubility product constant of Ag_(2)CrO_(4) and AgBr are 1.1xx10^(-12) and 5.0xx10^(-13) respectively. Calculate the ratio of the molarities of their saturated solutions. by (65.9k points) selected Ag_(2)CrO_(4) dissolves in water and the equilibrium in the saturated solution is Ag_(2)CrO_(4)(s)= 2Ag^(o+)(aq)+CrO_(4)^(2-)(aq) Let the solubolty of AgBr be S_(1) [Ag^(o+)(aq)]=2S_(1) and [CrO_(4)^(2-)(aq)=S_(1)] K_(sp)=[Ag^(o+)(aq)]^(2)[CrO_(4)^(2-)(aq)]=(2S_(1))^(2)xxS=4S_(1)^(3) or S_(1)=(K_(sp)/4)^(1//3)=((1.1xx10^(-12))/4)^(1//3)=0.65xx10^(-14) M Similarly AgBr=Ag^(o+)(aq)+Br^(Θ)(aq) Let the solubility of AgBr be S_(2) [Ag^(o+)(aq)]=S_(2) and Br^(Θ)(aq)]=S_(2) K_(sp)=[Ag^(o+)(aq)][Br^(Θ)(aq)] S_(2)xxS_(2)=S_(2)^(2) or S_(2)=(K_(sp))^(1//2)=(5.0xx10^(-13))^(1//2) =(0.5xx10^(-12))^(1//2)=0.707xx10^(-6) mol L^(-1) Ratio of the molarities of silver | chromate to silver bromide comes to =(0.65xx10^(-4))/(0.707xx10^(-6))=91.9 silver chromate is more soluble. |
https://samuelsonmathxp.com/ | 1,500,726,607,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424060.49/warc/CC-MAIN-20170722122816-20170722142816-00679.warc.gz | 700,984,133 | 53,832 | ## First Order Differential Equations
As noted above, the subject of this entry is first order differential equations. Before getting directly to that, however, a few examples of more basic equations representing rates of change will be reflected on.
Too often, students are given rules and algorithms to follow in “solving” problems. While becoming fluent with these procedures is not unimportant, students relying on these strategies alone will eventually find themselves swimming in a pool of confusion when faced with unfamiliar looking problems; if problems not recognizable, they become impossible to resolve. In an effort to minimize such predicaments, an intuitive perspective can often be helpful. Directly below are two examples of solutions to basic differential equations; the first resulting in a quadratic function, the second being exponential. The first two frames below show an “intuitive” approach to solving these problems while the third frame shows a more typical procedure.
Exponential Function
Procedure using Leibniz Notation
An intuitive feel is supplementary to the procedures shown above and can add understanding to the process. Equally important is developing the ability to create these mathematical models from information conveyed in the written word.
A huge stumbling block for many students is knowing why a specific model works for one scenario and not the next; the deciding factor is sometimes a seemingly small detail that is easily overlooked early in the learning process. The example below draws attention to this and will serve as a bridge between simple separable differential equations and the more complex first order linear differential equations.
Mixing Problems and Solution Concentrations
This portion of our journey will focus on the changing concentration of a solution in a container as the ratio of solute to solvent increases or decreases. One such example is provided below.
A tank initially contains 50 gallons of brine in which 5 lb of salt is dissolved. Brine containing 1 lb of salt per gallon flows into the tank at a constant rate of 4 gal/min. The concentration of the brine in the tank is kept uniform throughout the tank by stirring. While the brine is entering the tank, brine from the tank is also being drawn off at a rate of 2 gal/min. Find the amount of salt in the tank after 25 minutes.
(Fraleigh, p.903)
In this scenario, the concentration of brine being added to the tank differs from that originally present with the new mixture being drained off simultaneously; this is the “small detail” making the problem here a more complex one than the other examples cited earlier. The two frames immediately below illustrate simplified versions of this and will serve as an introduction to mixing problems; both allow for a straight forward model involving separable differential equations.
In the scenario directly below, the concentration of solute initially contained in the tank has been altered to match that which is being added (1 lb/gal). That being the case, the concentration of solution being drained away will be a constant 1 lb/gal as well; the resulting function will therefore be linear.
Separable Differential Equation (Equal Concentration)
This second example sees the concentration initially contained in the tank revert to 0.1 lb/gal with inflow of 1 lb/gal. This time, however, no solution will be drawn from the container, once again resulting in a linear function.
Separable Differential Equation (Outflow Zero)
First Order Differential Equations
It is now time to tackle the problem posed by Fraleigh. To reiterate, the concentration of solution flowing into the container differs from that initially present AND this new mixture is being drawn off simultaneously, thereby requiring a more complex model. Since the concentration of solution being extracted is constantly changing, that quantity will be expressed as a ratio of salt (in lbs) to number of gallons of brine in the container at any time “t”. Since the quantity of salt in the tank is changing continuously, a variable will be assigned to represent that quantity; we’ll use “y” to denote pounds of salt.
Inflow: 4 gal/min at 1 lb/gal → 4 lb/min Outflow: 2 gal/min at y lb/(50+2t) gal → [2/(50+2t)]y lb/min
From this, we can easily represent the rate of change of salt (in lbs) with respect to time as follows:
dy/dt=4-[2/(50+2t)]y (lb/min)
This expression is referred to as “Item 2” in the frame directly below. The “item” preceding that is a reminder of the FTC that is called upon in solving for what is referred to as an “Integrating Factor”. Its purpose is also discovered below and is denoted by “I”.
Integrating Factor
Purpose of the Integrating Factor (clarification)
Given Iy’+Ipy=qI
If (Iy)’=Iy’+Ipy , then (Iy)’=qI which leads to Iy=∫qIdt
The integrating factor ultimately allows us to express “y” as a function of “t” (which is our objective).
While first order differential equations are not required in cases where the concentration of inflow equals that already in the container, initial conditions for that scenario are nevertheless included below. The solution to Fraleigh’s problem is shown as well (bottom right) for comparison purposes.
Solution to Fraleigh’s Problem
To reiterate: If the concentration of solution flowing into the container is equal to that already present OR outflow is zero, the function describing quantity of solute will be linear in nature. The following three frames serves as examples of this important detail.
Linear Function: Constant Concentration (1 lb/gal)
Linear Function: Constant Concentration (2 lb/gal)
Linear Function: Zero Outflow
The frame that follows is a variation of the problem posed by Fraleigh, this time with an inflow of 5 gal/min. A graph of the resulting non-linear function is also included.
Non-Linear Function
For further comparison and analysis, I’ve included two additional scenarios directly below, each with an interactive link.
Inflow Exceeds Outflow
Click on the link provided here to explore scenarios where inflow exceeds outflow.
Outflow Exceeds Inflow
Click on the link provided here to explore scenarios where outflow exceeds inflow.
The following image illustrates the function shown directly above on a different scale. The lack of symmetry here creates additional opportunities for discussion with respect to the rate at which the quantity of solute is changing over time. For example, at t=11.16 minutes it can be stated that the quantity of solute flowing in equals that which is flowing out. What is happening before and after that point in time and why? Is this a reasonable conclusion and does the graph of our function reflect that?
Nearly the end………….
I had initially planned on ending this entry with the previous frame; my students, however, wanted one more day to explore mixing problems. I’ve decided to summarize that final day’s sequence here.
As always, this mixing problem began as follows:
y’=q-py , where y’ = rate of change of solute (lb/min) q = inflow of solute (lb/min) py = outflow of solute (lb/min)
We then decided initial conditions, concentration of solute being added as well as rates of inflow and outflow. The first set of values chosen produced what we thought was a reasonable solution until that function’s stationary point was determined; it occurred at (31.53,-1.96). These values caused some concern so our work was double checked. Since the math seemed to be fine, we decided to graph our function and interpret that. Once everything was taken into consideration, this result was perfectly logical. The tank initially contained 150 lbs of solute (3 lbs/gal); we were adding only 0.2 lb/gal at a rate of 6 gal/min and draining away the diluted mixture at a rate of 8 gal/min. Upon reflection, our result seemed perfectly reasonable. This scenario and resulting function are shown immediately below.
Rapid Dilution
Following our satisfactory explanation of the dilution scenario above, the discussion focused on what would need to change so that the function would produce a maximum value greater that 150 lb (the initial content). It was decided that increasing the concentration of solute flowing in would produce the desired result. A concentration of 4 lb/gal was agreed upon and the function reworked. Upon graphing this new function, the location of its maximum was revealed. The question immediately asked was ‘Why is the maximum located at t=0?’. The derivative was determined and set equal to zero, quickly satisfying everyone. The frame immediately below illustrates this scenario.
Local Maximum at t=0
To further solidify our understanding of this concept, it was decided that the vertex of our graph should move to the right if the concentration of solute flowing in was further increased. This final iteration of our mixing problem is shown in the following frame.
Local Maximum at t>0
Click on the link provided here to further explore mixing problems.
The end.
References
Kline, Morris. (1998). Calculus: An Intuitive and Physical Approach. Mineola, NY: Dover Publications.
Quote — Posted: April 22, 2016 in Calculus: An Introduction
## Projectile Motion….including Baseball!
Posted: March 31, 2016 in Calculus: An Introduction, Differential Calculus
Tags:
Projectile motion is a natural fit and provides an interesting application in the introduction of calculus at the high school level. A previous post focused on calculating the horizontal velocity of a ball rolling off the end of a table; this entry takes things a bit further by launching projectiles at various angles to determine the maximum horizontal travel.
Before lighting the fuse on our launching device, some important theory should first be dealt with. Since our launch angle will be somewhere between 0° and 90°, the projectile’s travel will be directed both vertically and horizontally. Since these values will be dependent on the angle with which the projectile is launched, expressions for distance traveled in terms of that angle will be required. These calculations are shown directly below.
Projections of Velocity
The maximum of the horizontal component of a projectile’s motion occurs when its vertical component has been fully depleted. Since these two conditions occur simultaneously, our work will be straight forward. A new function representing the projectile’s height at any position “x” can be easily determined; x-intercepts of this new function will then lead us to the desired solution.
Analytical Solution
The following is a geometric representation of the analytical solution shown above. The graph on the left shows the projectile’s height as a function of its horizontal position “x”. The second graph measures height over time.
Maximum Horizontal Travel
Click on the link provided here to interact with projectile motion and discover maximum horizontal travel.
The scenario directly below is supplementary to a previous entry describing a ball rolling off a table. It is included here with the intent of having students further explore the behavior of projectiles launched horizontally from various heights.
Horizontally Launched Projectile
Click on the link provided here to explore horizontally fired projectiles with variable height and velocity.
The image and link directly below show how far a baseball would travel with zero drag from air-resistance. The initial launch angle and height have been set to 35° and 1 m respectively. The launch angle can vary greatly and still constitute a baseball scenario; the same cannot be said for the launch height. I decided to leave that variable visible to provide another option for further exploration.
Theoretical Flight of a Baseball (and more)
Click on the link provided here to explore the path of a baseball assuming zero drag from air-resistance.
The following links verify the accuracy of the model used above and also provide additional insights into the flight of a baseball.
Baseball Trajectory
The Science of Baseball: What Is The Farthest Home Run?
Baseball Physics: Anatomy of a Home Run
Conclusion
Each year, Major League Baseball provides many satisfying projectile launches; the most gratifying (for me) occurred in the 1988 post-season. For a reminder of this moment in time, click on the link provided here to witness what I consider to be the greatest projectile launch of all time……….I love baseball.
## Physics Problem & Differential Equations
Posted: March 29, 2016 in Calculus: An Introduction
Our school’s physics teacher and I were chatting recently about an experiment he likes to conduct with his Grade 11 students and a related Mythbuster’s episode. The approach in Physics 20 requires students to use the displacement and velocity formulas provided there. This entry revisits the same problem and brings simple separable differential equations to the table.
View the Mythbuster’s video here first: Bullet Fired vs Bullet Dropped
Separable Differential Equations
Click on the link provided here to adjust horizontal velocity of the ball.
Thanks for looking.
## Centers of Mass
Posted: March 27, 2016 in Calculus: An Introduction, Centers of Mass, Integral Calculus
Tags: ,
The power of integral calculus is once again exploited in this entry, this time to determine the centers of mass of one- and two-dimensional objects. Before getting to that, however, some preliminary “discourse” will be engaged in to set the stage.
Everyone (of my age at least) can relate to the scenario involving two children playing on a see-saw. If the children have equal mass and are sitting an equal distance from the fulcrum, they can achieve perfect balance; the fulcrum in this scene is located at the center of mass. If, however, two differing masses are involved (all other things remaining equal), the side containing the greater mass will rotate downward on the fulcrum. This brings us to a very important term for this and other concepts involving rotational motion. This term is called the MOMENT of a force and is defined below:
The moment of a force is a measure of its tendency to rotate about an axis or a point. The moment can be influenced by two quantities: the object’s mass and its distance from the axis or point of rotation; this distance can be referred to as the moment arm.
Moment=Force x Distance , with units measured in ft-lb, kgf-m, etc.
In our example, when two children of the same mass are positioned an equal distance on opposing sides of the fulcrum, a state of equilibrium is achieved. If a bird suddenly joins in the fun and lands on the head of one of the participants, the mass on that end is increased and begins to rotate downward; this increase in mass has created a moment. A moment can also be created if one participant increases his/her position relative to the fulcrum. The moment arm on that side has now been lengthened, thereby creating a moment.
When finding the center of mass in one dimension, the same principles apply; this is a very straight forward procedure if the object has uniform mass density over its entire length. Complications arise, however, if the object’s mass density is not uniform throughout. To address this issue, the object is analyzed as a collection of very small points, each having its own mass and unique position (moment arm) within the main object; each of these will be referred to as “point-mass”. As in the playground scenario described earlier, the further each point-mass is situated from the axes (or point) of rotation, the greater its contribution will be to the moment and, thereby, its influence on the location of the center of mass. The calculation for center of mass is built upon the concept of weighted averages; while the most frequently occurring outcomes have a significant say with respect to the overall average, the extreme outliers can also have a measurable impact.
Before weighted averages can be referenced in this context, the notation and the underlying concept that will be utilized throughout must be introduced. This is initiated below in the Mean Value Theorem of integral calculus.
Mean Value Theorem of Integral Calculus
NOTE: In the calculations that follow, x- and y-components of the moment appear. Since centers of mass occur at a point of equilibrium, force due to gravity is ignored and omitted from the units chosen to represent those quantities. I wanted clarification on this item here since moments are once again called upon when dealing with torque. In that application, force is included in the units of measurement when describing moments of inertia.
A reference was made earlier to “point-masses” and their relative position within the object containing them. A direct connection between this and weighted averages exists and is presented below.
Weighted Average
The end result in the first two examples in the following image are common sense and serve as a “trial run” on the theory developed above; all three can be related directly to our playground scenario described earlier.
One Dimensional Center of Mass
While limiting ourselves here to one-dimension would be silly, attempting to illustrate centers of mass in three dimensions on a 2-D surface could be considered reckless. For this reason, two-dimensions will be the extent to which this topic is pursued here.
Two Dimensions: x-component
Two Dimensions: y-component
In the examples that follow, centers of mass are determined using the theory developed above. Interactive links also provide the opportunity to change one or more parameters in these examples to observe variations in the various integrals involved.
Constant Function
Click on the link provided here to interact with centers of mass on one dimension.
Linear Function
One Image from the exploration that follows……
Click on the link provided here to explore centers of mass defined by linear functions.
Click on the link provided here to explore centers of mass defined by quadratic functions.
The examples that follow have mass density increasing exponentially along the x-axis. With exponentials in the mix, the need for a new method of integration (by parts) emerges; the power of WolframAlpha is also introduced to do the “heavy lifting”.
Constant Function: Exponential Increase in Mass Density
The x-component of the center of mass in the example above can be calculated manually using integration by parts; this procedure is included here.
Integration by Parts
Click on the link provided here to explore centers of mass resulting from an exponential increase in mass density.
While integration by parts can be exploited to evaluate all integrals of this form, the process can become a time-consuming one. The following example is one result obtained from the exploration directly above; it is included here with the intent of introducing students to the power of WolframAlpha. The x-component of the center of mass is shown in a screenshot of the WolframAlpha app available on any device. Interested students have the option of verifying this and other results manually using integration by parts.
Quadratic Function: Exponential Increase in Mass Density
WolframAlpha App
I’ve included a link here to the web-based version of the app shown above. To verify the y-component of the center of mass in the final example shown above, click on WolframAlpha.
References
Courant, Richard., John, Fritz (1999). Introduction to Calculus and Analytics: Classics of Mathematics. New York, NY: Springer-Verlag Berlin Heidelberg.
Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath.
## Fluid Pressure & Force
Posted: March 20, 2016 in Calculus: An Introduction, Fluid Force, Integral Calculus, Trigonometric Substitution
Tags: ,
Pressure is a force per unit of area exerted over the surface of an object (as in 35 psi in the tires on your car). When an object is immersed in water, or some other liquid medium, the fluid pressure exerted on that object varies with the depth at which it is submerged. For example, the volume of water pressing down on an object submerged 10 feet is twice that of an object submerged half that depth.
Fluid Pressure (force per unit area) can therefore be defined as follows:
p=wh ,where w = weight density of the fluid h = depth at which the object is submerged
Fluid Force (total force exerted on object) is therefore given by
F=pA ,where A=total area of surface object in question F=whA
According to Pascal’s Law (principle), an object submerged in a fluid is subjected to equal pressure in all directions (at any given depth). For a sheet of metal submerged in water and resting horizontally at a given depth, the fluid force is constant over its entire surface. If, however, the submerged sheet is resting vertically, the entire force exerted over this object by the water varies with depth; the bottom of the sheet will experience more fluid force than its top. In order to determine the total force acting on this vertically oriented sheet, the force exerted on each rectangular cross-section of infinitesimal width (Δy) will be calculated and summed over the object’s entire vertical span; enter integration. With a well-chosen location for the y-axis in our model, the length of each rectangular cross-section can be easily expressed as some variation of f(y).
Rectangular Plate
The example above was relatively simple since f(y) was a constant throughout its vertical span. With shapes whose widths are not constant, the mathematical model can once again vary depending on the perspective chosen. In the examples below, circular plates have been introduced since they provide opportunities for multiple forms of substitution in the integration process, thereby maximizing learning opportunities for students.
The fluid force acting on the ends of a cylindrical water tank is the subject below. I felt that this would be more interesting than imagining the force exerted on a circular plate submerged in a body of water. To simplify matters here, atmospheric pressure and other factors such as sliding forces have been ignored.
Cylindrical Tank: Half-full (Perspective 1)
Cylindrical Tank: Half-full (Perspective 2)
Cylindrical Tank: Filled to Capacity
It is worthwhile drawing attention to the forces acting on the ends of top half of the tank filled to capacity (451.34 lb) and the bottom half filled to half capacity (332.8 lb). This type of thoughtful comparison can add to the students’ understanding of this topic.
The two images directly below once again show the two perspectives of fluid forces acting on the ends of our water tank. They are included here to illustrate the contents of the interactive link that follows.
Tank Centered at (0,0)
Tank Centered at (0,-2)
Click on the link provided here to interact with fluid force on the ends of a cylindrical tank.
In the image and link below, the circular end has been removed from the tank and submerged on its own. Once again, various mathematical models could be used to describe this scenario; the one chosen here has placed the origin at the circle’s center.
Click on the link provided here to interact with fluid force on a submerged vertical plate with center at origin.
The following links will be of interest to some:
Reference
Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath.
## Related Rates: Ball Drop & Shadow
Posted: March 13, 2016 in Calculus: An Introduction, Differential Calculus, Differential Equations, Related Rates
Tags: ,
Here’s problem involving a falling object and the speed at which its shadow travels along the ground. As usual, in related rates, once a relationship between the variables involved has been established, the calculus required to reach its conclusion is very straight forward.
In order to make efficient use of time, these problems provide students the opportunity to practice simple differentiation procedures. In addition, the graphs provided below open the door to a discussion on the Mean Value Theorem of differential calculus, serving to either introduce or reinforce that concept.
Falling Ball
Click on the link provided here to interact with the falling ball and its shadow.
The ball’s displacement from its release point was provided in the image above. As a review (since integral calculus has already been introduced), that displacement formula is once again derived through basic differential equations; this is shown directly below.
Acceleration, Velocity and Displacement
I’ve included solutions for t=1 and t=2 below. In keeping with my belief that students can learn effectively through comparison and contrast, three varied methods are shown.
Solutions
The main subject of this entry was originally planned as an optimization problem involving differential calculus only; its been slightly modified. This more interesting approach provides the derivative up front, presenting students with three separate tasks to pursue from that point. As a consequence, students are reintroduced to differential equations and curve sketching.
A talking point emerges as well: Is there a difference between derivatives and differential equations?
Inscribed Triangle of Maximum Area
Click on the link provided here to explore area of the inscribed triangle.
Thanks for looking. | 5,238 | 25,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2017-30 | longest | en | 0.939901 | ## First Order Differential Equations As noted above, the subject of this entry is first order differential equations. Before getting directly to that, however, a few examples of more basic equations representing rates of change will be reflected on. Too often, students are given rules and algorithms to follow in “solving” problems. While becoming fluent with these procedures is not unimportant, students relying on these strategies alone will eventually find themselves swimming in a pool of confusion when faced with unfamiliar looking problems; if problems not recognizable, they become impossible to resolve. In an effort to minimize such predicaments, an intuitive perspective can often be helpful. Directly below are two examples of solutions to basic differential equations; the first resulting in a quadratic function, the second being exponential. The first two frames below show an “intuitive” approach to solving these problems while the third frame shows a more typical procedure. Exponential Function Procedure using Leibniz Notation An intuitive feel is supplementary to the procedures shown above and can add understanding to the process. Equally important is developing the ability to create these mathematical models from information conveyed in the written word. A huge stumbling block for many students is knowing why a specific model works for one scenario and not the next; the deciding factor is sometimes a seemingly small detail that is easily overlooked early in the learning process. The example below draws attention to this and will serve as a bridge between simple separable differential equations and the more complex first order linear differential equations. Mixing Problems and Solution Concentrations This portion of our journey will focus on the changing concentration of a solution in a container as the ratio of solute to solvent increases or decreases. One such example is provided below. A tank initially contains 50 gallons of brine in which 5 lb of salt is dissolved. Brine containing 1 lb of salt per gallon flows into the tank at a constant rate of 4 gal/min. The concentration of the brine in the tank is kept uniform throughout the tank by stirring. While the brine is entering the tank, brine from the tank is also being drawn off at a rate of 2 gal/min. Find the amount of salt in the tank after 25 minutes. (Fraleigh, p.903) In this scenario, the concentration of brine being added to the tank differs from that originally present with the new mixture being drained off simultaneously; this is the “small detail” making the problem here a more complex one than the other examples cited earlier. The two frames immediately below illustrate simplified versions of this and will serve as an introduction to mixing problems; both allow for a straight forward model involving separable differential equations. In the scenario directly below, the concentration of solute initially contained in the tank has been altered to match that which is being added (1 lb/gal). That being the case, the concentration of solution being drained away will be a constant 1 lb/gal as well; the resulting function will therefore be linear. Separable Differential Equation (Equal Concentration) This second example sees the concentration initially contained in the tank revert to 0.1 lb/gal with inflow of 1 lb/gal. This time, however, no solution will be drawn from the container, once again resulting in a linear function. Separable Differential Equation (Outflow Zero) First Order Differential Equations It is now time to tackle the problem posed by Fraleigh. To reiterate, the concentration of solution flowing into the container differs from that initially present AND this new mixture is being drawn off simultaneously, thereby requiring a more complex model. Since the concentration of solution being extracted is constantly changing, that quantity will be expressed as a ratio of salt (in lbs) to number of gallons of brine in the container at any time “t”. Since the quantity of salt in the tank is changing continuously, a variable will be assigned to represent that quantity; we’ll use “y” to denote pounds of salt. Inflow: 4 gal/min at 1 lb/gal → 4 lb/min Outflow: 2 gal/min at y lb/(50+2t) gal → [2/(50+2t)]y lb/min From this, we can easily represent the rate of change of salt (in lbs) with respect to time as follows: dy/dt=4-[2/(50+2t)]y (lb/min) This expression is referred to as “Item 2” in the frame directly below. The “item” preceding that is a reminder of the FTC that is called upon in solving for what is referred to as an “Integrating Factor”. Its purpose is also discovered below and is denoted by “I”. Integrating Factor Purpose of the Integrating Factor (clarification) Given Iy’+Ipy=qI If (Iy)’=Iy’+Ipy , then (Iy)’=qI which leads to Iy=∫qIdt The integrating factor ultimately allows us to express “y” as a function of “t” (which is our objective). While first order differential equations are not required in cases where the concentration of inflow equals that already in the container, initial conditions for that scenario are nevertheless included below. The solution to Fraleigh’s problem is shown as well (bottom right) for comparison purposes. Solution to Fraleigh’s Problem To reiterate: If the concentration of solution flowing into the container is equal to that already present OR outflow is zero, the function describing quantity of solute will be linear in nature. The following three frames serves as examples of this important detail. Linear Function: Constant Concentration (1 lb/gal) Linear Function: Constant Concentration (2 lb/gal) Linear Function: Zero Outflow The frame that follows is a variation of the problem posed by Fraleigh, this time with an inflow of 5 gal/min. A graph of the resulting non-linear function is also included. Non-Linear Function For further comparison and analysis, I’ve included two additional scenarios directly below, each with an interactive link. Inflow Exceeds Outflow Click on the link provided here to explore scenarios where inflow exceeds outflow. Outflow Exceeds Inflow Click on the link provided here to explore scenarios where outflow exceeds inflow. The following image illustrates the function shown directly above on a different scale. The lack of symmetry here creates additional opportunities for discussion with respect to the rate at which the quantity of solute is changing over time. For example, at t=11.16 minutes it can be stated that the quantity of solute flowing in equals that which is flowing out. What is happening before and after that point in time and why? Is this a reasonable conclusion and does the graph of our function reflect that? Nearly the end…………. I had initially planned on ending this entry with the previous frame; my students, however, wanted one more day to explore mixing problems. I’ve decided to summarize that final day’s sequence here. As always, this mixing problem began as follows: y’=q-py , where y’ = rate of change of solute (lb/min) q = inflow of solute (lb/min) py = outflow of solute (lb/min) We then decided initial conditions, concentration of solute being added as well as rates of inflow and outflow. The first set of values chosen produced what we thought was a reasonable solution until that function’s stationary point was determined; it occurred at (31.53,-1.96). These values caused some concern so our work was double checked. Since the math seemed to be fine, we decided to graph our function and interpret that. Once everything was taken into consideration, this result was perfectly logical. The tank initially contained 150 lbs of solute (3 lbs/gal); we were adding only 0.2 lb/gal at a rate of 6 gal/min and draining away the diluted mixture at a rate of 8 gal/min. Upon reflection, our result seemed perfectly reasonable. This scenario and resulting function are shown immediately below. Rapid Dilution Following our satisfactory explanation of the dilution scenario above, the discussion focused on what would need to change so that the function would produce a maximum value greater that 150 lb (the initial content). It was decided that increasing the concentration of solute flowing in would produce the desired result. A concentration of 4 lb/gal was agreed upon and the function reworked. Upon graphing this new function, the location of its maximum was revealed. The question immediately asked was ‘Why is the maximum located at t=0?’. The derivative was determined and set equal to zero, quickly satisfying everyone. The frame immediately below illustrates this scenario. Local Maximum at t=0 To further solidify our understanding of this concept, it was decided that the vertex of our graph should move to the right if the concentration of solute flowing in was further increased. This final iteration of our mixing problem is shown in the following frame. Local Maximum at t>0 Click on the link provided here to further explore mixing problems. The end. References Kline, Morris. (1998). Calculus: An Intuitive and Physical Approach. Mineola, NY: Dover Publications. Quote — Posted: April 22, 2016 in Calculus: An Introduction ## Projectile Motion….including Baseball! Posted: March 31, 2016 in Calculus: An Introduction, Differential Calculus Tags: Projectile motion is a natural fit and provides an interesting application in the introduction of calculus at the high school level. A previous post focused on calculating the horizontal velocity of a ball rolling off the end of a table; this entry takes things a bit further by launching projectiles at various angles to determine the maximum horizontal travel. Before lighting the fuse on our launching device, some important theory should first be dealt with. Since our launch angle will be somewhere between 0° and 90°, the projectile’s travel will be directed both vertically and horizontally. Since these values will be dependent on the angle with which the projectile is launched, expressions for distance traveled in terms of that angle will be required. These calculations are shown directly below. Projections of Velocity The maximum of the horizontal component of a projectile’s motion occurs when its vertical component has been fully depleted. Since these two conditions occur simultaneously, our work will be straight forward. A new function representing the projectile’s height at any position “x” can be easily determined; x-intercepts of this new function will then lead us to the desired solution. Analytical Solution The following is a geometric representation of the analytical solution shown above. The graph on the left shows the projectile’s height as a function of its horizontal position “x”. The second graph measures height over time. Maximum Horizontal Travel Click on the link provided here to interact with projectile motion and discover maximum horizontal travel. The scenario directly below is supplementary to a previous entry describing a ball rolling off a table. It is included here with the intent of having students further explore the behavior of projectiles launched horizontally from various heights. Horizontally Launched Projectile Click on the link provided here to explore horizontally fired projectiles with variable height and velocity. The image and link directly below show how far a baseball would travel with zero drag from air-resistance. The initial launch angle and height have been set to 35° and 1 m respectively. The launch angle can vary greatly and still constitute a baseball scenario; the same cannot be said for the launch height. I decided to leave that variable visible to provide another option for further exploration. Theoretical Flight of a Baseball (and more) Click on the link provided here to explore the path of a baseball assuming zero drag from air-resistance. The following links verify the accuracy of the model used above and also provide additional insights into the flight of a baseball. Baseball Trajectory The Science of Baseball: What Is The Farthest Home Run? Baseball Physics: Anatomy of a Home Run Conclusion Each year, Major League Baseball provides many satisfying projectile launches; the most gratifying (for me) occurred in the 1988 post-season. For a reminder of this moment in time, click on the link provided here to witness what I consider to be the greatest projectile launch of all time……….I love baseball. ## Physics Problem & Differential Equations Posted: March 29, 2016 in Calculus: An Introduction Our school’s physics teacher and I were chatting recently about an experiment he likes to conduct with his Grade 11 students and a related Mythbuster’s episode. The approach in Physics 20 requires students to use the displacement and velocity formulas provided there. This entry revisits the same problem and brings simple separable differential equations to the table. View the Mythbuster’s video here first: Bullet Fired vs Bullet Dropped Separable Differential Equations Click on the link provided here to adjust horizontal velocity of the ball. Thanks for looking. ## Centers of Mass Posted: March 27, 2016 in Calculus: An Introduction, Centers of Mass, Integral Calculus Tags: , The power of integral calculus is once again exploited in this entry, this time to determine the centers of mass of one- and two-dimensional objects. Before getting to that, however, some preliminary “discourse” will be engaged in to set the stage. Everyone (of my age at least) can relate to the scenario involving two children playing on a see-saw. If the children have equal mass and are sitting an equal distance from the fulcrum, they can achieve perfect balance; the fulcrum in this scene is located at the center of mass. If, however, two differing masses are involved (all other things remaining equal), the side containing the greater mass will rotate downward on the fulcrum. This brings us to a very important term for this and other concepts involving rotational motion. This term is called the MOMENT of a force and is defined below: The moment of a force is a measure of its tendency to rotate about an axis or a point. The moment can be influenced by two quantities: the object’s mass and its distance from the axis or point of rotation; this distance can be referred to as the moment arm. Moment=Force x Distance , with units measured in ft-lb, kgf-m, etc. In our example, when two children of the same mass are positioned an equal distance on opposing sides of the fulcrum, a state of equilibrium is achieved. If a bird suddenly joins in the fun and lands on the head of one of the participants, the mass on that end is increased and begins to rotate downward; this increase in mass has created a moment. A moment can also be created if one participant increases his/her position relative to the fulcrum. The moment arm on that side has now been lengthened, thereby creating a moment. When finding the center of mass in one dimension, the same principles apply; this is a very straight forward procedure if the object has uniform mass density over its entire length. Complications arise, however, if the object’s mass density is not uniform throughout. To address this issue, the object is analyzed as a collection of very small points, each having its own mass and unique position (moment arm) within the main object; each of these will be referred to as “point-mass”. As in the playground scenario described earlier, the further each point-mass is situated from the axes (or point) of rotation, the greater its contribution will be to the moment and, thereby, its influence on the location of the center of mass. The calculation for center of mass is built upon the concept of weighted averages; while the most frequently occurring outcomes have a significant say with respect to the overall average, the extreme outliers can also have a measurable impact. Before weighted averages can be referenced in this context, the notation and the underlying concept that will be utilized throughout must be introduced. This is initiated below in the Mean Value Theorem of integral calculus. Mean Value Theorem of Integral Calculus NOTE: In the calculations that follow, x- and y-components of the moment appear. Since centers of mass occur at a point of equilibrium, force due to gravity is ignored and omitted from the units chosen to represent those quantities. I wanted clarification on this item here since moments are once again called upon when dealing with torque. In that application, force is included in the units of measurement when describing moments of inertia. A reference was made earlier to “point-masses” and their relative position within the object containing them. A direct connection between this and weighted averages exists and is presented below. Weighted Average The end result in the first two examples in the following image are common sense and serve as a “trial run” on the theory developed above; all three can be related directly to our playground scenario described earlier. One Dimensional Center of Mass While limiting ourselves here to one-dimension would be silly, attempting to illustrate centers of mass in three dimensions on a 2-D surface could be considered reckless. For this reason, two-dimensions will be the extent to which this topic is pursued here. Two Dimensions: x-component Two Dimensions: y-component In the examples that follow, centers of mass are determined using the theory developed above. Interactive links also provide the opportunity to change one or more parameters in these examples to observe variations in the various integrals involved. Constant Function Click on the link provided here to interact with centers of mass on one dimension. Linear Function One Image from the exploration that follows…… Click on the link provided here to explore centers of mass defined by linear functions. Click on the link provided here to explore centers of mass defined by quadratic functions. The examples that follow have mass density increasing exponentially along the x-axis. With exponentials in the mix, the need for a new method of integration (by parts) emerges; the power of WolframAlpha is also introduced to do the “heavy lifting”. Constant Function: Exponential Increase in Mass Density The x-component of the center of mass in the example above can be calculated manually using integration by parts; this procedure is included here. Integration by Parts Click on the link provided here to explore centers of mass resulting from an exponential increase in mass density. While integration by parts can be exploited to evaluate all integrals of this form, the process can become a time-consuming one. The following example is one result obtained from the exploration directly above; it is included here with the intent of introducing students to the power of WolframAlpha. The x-component of the center of mass is shown in a screenshot of the WolframAlpha app available on any device. Interested students have the option of verifying this and other results manually using integration by parts. Quadratic Function: Exponential Increase in Mass Density WolframAlpha App I’ve included a link here to the web-based version of the app shown above. To verify the y-component of the center of mass in the final example shown above, click on WolframAlpha. References Courant, Richard., John, Fritz (1999). Introduction to Calculus and Analytics: Classics of Mathematics. New York, NY: Springer-Verlag Berlin Heidelberg. Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath. ## Fluid Pressure & Force Posted: March 20, 2016 in Calculus: An Introduction, Fluid Force, Integral Calculus, Trigonometric Substitution Tags: , Pressure is a force per unit of area exerted over the surface of an object (as in 35 psi in the tires on your car). When an object is immersed in water, or some other liquid medium, the fluid pressure exerted on that object varies with the depth at which it is submerged. For example, the volume of water pressing down on an object submerged 10 feet is twice that of an object submerged half that depth. Fluid Pressure (force per unit area) can therefore be defined as follows: p=wh ,where w = weight density of the fluid h = depth at which the object is submerged Fluid Force (total force exerted on object) is therefore given by F=pA ,where A=total area of surface object in question F=whA According to Pascal’s Law (principle), an object submerged in a fluid is subjected to equal pressure in all directions (at any given depth). For a sheet of metal submerged in water and resting horizontally at a given depth, the fluid force is constant over its entire surface. If, however, the submerged sheet is resting vertically, the entire force exerted over this object by the water varies with depth; the bottom of the sheet will experience more fluid force than its top. In order to determine the total force acting on this vertically oriented sheet, the force exerted on each rectangular cross-section of infinitesimal width (Δy) will be calculated and summed over the object’s entire vertical span; enter integration. With a well-chosen location for the y-axis in our model, the length of each rectangular cross-section can be easily expressed as some variation of f(y). Rectangular Plate The example above was relatively simple since f(y) was a constant throughout its vertical span. With shapes whose widths are not constant, the mathematical model can once again vary depending on the perspective chosen. In the examples below, circular plates have been introduced since they provide opportunities for multiple forms of substitution in the integration process, thereby maximizing learning opportunities for students. The fluid force acting on the ends of a cylindrical water tank is the subject below. I felt that this would be more interesting than imagining the force exerted on a circular plate submerged in a body of water. To simplify matters here, atmospheric pressure and other factors such as sliding forces have been ignored. Cylindrical Tank: Half-full (Perspective 1) Cylindrical Tank: Half-full (Perspective 2) Cylindrical Tank: Filled to Capacity It is worthwhile drawing attention to the forces acting on the ends of top half of the tank filled to capacity (451.34 lb) and the bottom half filled to half capacity (332.8 lb). This type of thoughtful comparison can add to the students’ understanding of this topic. The two images directly below once again show the two perspectives of fluid forces acting on the ends of our water tank. They are included here to illustrate the contents of the interactive link that follows. Tank Centered at (0,0) Tank Centered | at (0,-2) Click on the link provided here to interact with fluid force on the ends of a cylindrical tank. In the image and link below, the circular end has been removed from the tank and submerged on its own. Once again, various mathematical models could be used to describe this scenario; the one chosen here has placed the origin at the circle’s center. Click on the link provided here to interact with fluid force on a submerged vertical plate with center at origin. The following links will be of interest to some: Reference Larson, R., Hostetler, R. P., & Edwards, B. H. (1995). Calculus of a Single Variable: Early Transcendental Functions. Lexington, MA: D.C. Heath. ## Related Rates: Ball Drop & Shadow Posted: March 13, 2016 in Calculus: An Introduction, Differential Calculus, Differential Equations, Related Rates Tags: , Here’s problem involving a falling object and the speed at which its shadow travels along the ground. As usual, in related rates, once a relationship between the variables involved has been established, the calculus required to reach its conclusion is very straight forward. In order to make efficient use of time, these problems provide students the opportunity to practice simple differentiation procedures. In addition, the graphs provided below open the door to a discussion on the Mean Value Theorem of differential calculus, serving to either introduce or reinforce that concept. Falling Ball Click on the link provided here to interact with the falling ball and its shadow. The ball’s displacement from its release point was provided in the image above. As a review (since integral calculus has already been introduced), that displacement formula is once again derived through basic differential equations; this is shown directly below. Acceleration, Velocity and Displacement I’ve included solutions for t=1 and t=2 below. In keeping with my belief that students can learn effectively through comparison and contrast, three varied methods are shown. Solutions The main subject of this entry was originally planned as an optimization problem involving differential calculus only; its been slightly modified. This more interesting approach provides the derivative up front, presenting students with three separate tasks to pursue from that point. As a consequence, students are reintroduced to differential equations and curve sketching. A talking point emerges as well: Is there a difference between derivatives and differential equations? Inscribed Triangle of Maximum Area Click on the link provided here to explore area of the inscribed triangle. Thanks for looking. |
https://ask.sagemath.org/question/8717/series-solutions-of-higher-order-odes/?answer=37709 | 1,723,202,596,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640763425.51/warc/CC-MAIN-20240809110814-20240809140814-00799.warc.gz | 87,166,598 | 17,073 | # series solutions of higher order ODEs
I'm trying to use Sage to find the general series solution to $y^{(4)}=\frac{y'y''}{1+x}$. So far my best efforts to derive the coefficient recurrence relations, inspired by a good book draft, have been along these lines:
R10 = QQ['a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10']
a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10 = R10.gens()
R.<x> = PowerSeriesRing(R10)
y = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + \
a6*x^6 + a7*x^7 + a8*x^8 + a9*x^9 + a10*x^10 + O(x^11)
y1 = y.derivative()
y2 = y1.derivative()
y3 = y2.derivative()
y4 = y3.derivative()
f = (1+x)*y4-y1*y2
i = ideal(f)
# g = i.groebner_fan(); g.reduced_groebner_bases() # a wish
# q = R.quotient(i) # works, but not so useful by itself
and my other approaches ended in tracebacks:
x = var('x'); y = function('y', x)
desolve(diff(y,x,4)-diff(y,x)*diff(y,x,2)/(1 + x), y, contrib_ode=True)
# NotImplementedError: Maxima was unable to solve this ODE.
desolve_laplace(diff(y,x,4) - diff(y,x)*diff(y,x,2)/(1 + x), y)
# TypeError: unable to make sense of Maxima expression
I would like to at least solve that and determine the radius of convergence. Ideally (more generally), it would be nice to have a good bag of tricks for working with series DEs such as I imagine others have already created. For this, I would like to find or develop techniques to incorporate:
• more convenient coefficients, e.g. from this thread
• a way to derive the recurrence relations using Python's lambda operator or this nice trick
• a solver for higher order ODEs such as above
Any hints, links, references or suggestions would be appreciated.
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The substitution $z=y'$ is natural, so let us please then forget about the posted $y$.
We will not use $z$. Let us use $y$ "back" instead of $z$ in the sequel. So we solve (instead) around $x=0$: $$y'' ' = y\ y' \ /\ (1+x)\ .$$
Reusing the above code...
R10.<a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10> = \
QQ['a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10']
R.<x> = PowerSeriesRing(R10)
y = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + \
a6*x^6 + a7*x^7 + a8*x^8 + a9*x^9 + a10*x^10 + O(x^11)
y1 = y .derivative()
y2 = y1.derivative()
y3 = y2.derivative()
f = (1+x)*y3 - y*y1
# J = ideal(f) # --> leads to an error
f.list()
... we get the coefficients of $f$:
[-a0*a1 + 6*a3,
-a1^2 - 2*a0*a2 + 6*a3 + 24*a4,
-3*a1*a2 - 3*a0*a3 + 24*a4 + 60*a5,
-2*a2^2 - 4*a1*a3 - 4*a0*a4 + 60*a5 + 120*a6,
-5*a2*a3 - 5*a1*a4 - 5*a0*a5 + 120*a6 + 210*a7,
-3*a3^2 - 6*a2*a4 - 6*a1*a5 - 6*a0*a6 + 210*a7 + 336*a8,
-7*a3*a4 - 7*a2*a5 - 7*a1*a6 - 7*a0*a7 + 336*a8 + 504*a9,
-4*a4^2 - 8*a3*a5 - 8*a2*a6 - 8*a1*a7 - 8*a0*a8 + 504*a9 + 720*a10]
We give or consider a0, a1, a2 as given.
Solving by the Cauchy-Kovalevskaya "procedure", we have to determine recursively a3, a4, a5, a6, a7, a8, a9, a10, and so on, starting from a0, a1, a2, by making each line evaluate to $0$. We get manually $$a_3 = \frac 16 a_0a_1\ ,$$ $$a_4 = \frac 1{24}( a_1^2 + 2a_0a_2 + a_0a_1 )\ ,$$ and so on.
I think it is not a good idea to isolate in this way all these equations, then start solving. Instead, let us write the given DE in the form $$y'' ' = y\ y'\ (1-x+x^2-x^3+\dots)\ ,$$ and consider it formally in the ring $\mathbb Q[[x]]$. Then the equations in $a_3,a_4,\dots$ come already separated, and we can simply write the code to get the first some $100$ coefficients.
(I am not willing to write optimal code, time...)
To have a precise situation, let us better use values instead of the symbols $a_0,a_1, a_2$. For instance $a_0=0$, $a_1=1$, $a_2=2$. Then...
a = [0,1,2] # use other initial conditions, next time
PSR.<x> = PowerSeriesRing( QQ, 100 )
C = 1 / PSR(1+x)
for k in [ 3..99 ]:
print k,
# trying to get a(k) from y = sum( [ a(k) * x^k for k in [0..ooh] ] )
yk = sum( [ a[j] * x^j for j in range(k) ] ) + O(x^k)
ak = 1/(k*(k-1)*(k-2)) * ( yk * diff( yk, x ) * ( C+O(x^k) ) ).list()[k-3]
print ak
a.append( ak )
y = PSR( sum( [ a[k] * x^k for k in [0..99] ] ) )
This computes some values. For instance, the first some $30$ entries are:
sage: a[:30]
[0,
1,
2,
0,
1/24,
1/12,
1/40,
-67/5040,
13/1152,
-7/1440,
4621/1209600,
-38399/13305600,
11287/4838400,
-12737/6988800,
2351057/1614412800,
-171374993/145297152000,
1692465871/1743565824000,
-115101113/142502976000,
32202455561/47424990412800,
-4213094929/7313918976000,
22201339257697/45053740892160000,
-267905698350809/630752372490240000,
5115211318062343/13876552194785280000,
-8562348061276901/26596725040005120000,
928302494123858617/3282795776366346240000,
-7613558036179561/30490363247984640000,
1985785680757233642821/8962032469480125235200000,
-590031815878576041977/2987344156493375078400000,
33273819697787429364371/188202681859082629939200000,
-3468510498746975345435917/21831511095653585072947200000]
(These are exactly computed.) Let us check the solution modulo $O(x^{100})$:
# : diff( y,x,3 ) - y * diff(y,x) / (1+x)
# the above is a mess with terms in degrees 97, 98, 99, but..
sage: diff( y,x,3 ) - y * diff(y,x) / (1+x) + O( x^97 )
O(x^97)
The radius of convergence is in this case probably one, one can try...
for k in [1..99]:
ak = a[k]
print ( "a[%2d] ~ %13.10f and a[%2d]^(1/%2d) ~ %8.6f"
% ( k, RR(ak), k, k, abs(RR(ak))^(1/k) ) )
and the first and last lines shown are:
a[ 1] ~ 1.0000000000 and a[ 1]^(1/ 1) ~ 1.000000
a[ 2] ~ 2.0000000000 and a[ 2]^(1/ 2) ~ 1.414214
a[ 3] ~ 0.0000000000 and a[ 3]^(1/ 3) ~ 0.000000
a[ 4] ~ 0.0416666667 and a[ 4]^(1/ 4) ~ 0.451801
a[ 5] ~ 0.0833333333 and a[ 5]^(1/ 5) ~ 0.608364
a[ 6] ~ 0.0250000000 and a[ 6]^(1/ 6) ~ 0.540742
a[ 7] ~ -0.0132936508 and a[ 7]^(1/ 7) ~ 0.539447
a[ 8] ~ 0.0112847222 and a[ 8]^(1/ 8) ~ 0.570902
a[ 9] ~ -0.0048611111 and a[ 9]^(1/ 9) ~ 0.553313
:::::::::::::::::::::::::::::::::::::::::::::::::
a[90] ~ 0.0000051223 and a[90]^(1/90) ~ 0.873406
a[91] ~ -0.0000049542 and a[91]^(1/91) ~ 0.874386
a[92] ~ 0.0000047934 and a[92]^(1/92) ~ 0.875348
a[93] ~ -0.0000046394 and a[93]^(1/93) ~ 0.876295
a[94] ~ 0.0000044920 and a[94]^(1/94) ~ 0.877225
a[95] ~ -0.0000043507 and a[95]^(1/95) ~ 0.878140
a[96] ~ 0.0000042154 and a[96]^(1/96) ~ 0.879040
a[97] ~ -0.0000040855 and a[97]^(1/97) ~ 0.879925
a[98] ~ 0.0000039610 and a[98]^(1/98) ~ 0.880796
a[99] ~ -0.0000038414 and a[99]^(1/99) ~ 0.881653
I have maybe to stop here with sage sample code.
About the convergence radius... The recurrence relation is thus of the shape $a_k=$ a sum of monomials of order two in the previous variables, with coefficients that can be recursively "controlled". With some work, one may (??) maybe show that the radius of convergence is one. (The one is my quick guess, based on the fact that the limit $r=\lim |a_k|^{1/k}$ may / should exist, and then we expect $r\sim r^2$. Minoration / majoration for the "polynomial part in $k$" is constricted by the power $1/k$. But it is just a guess.)
Note: The use of solve_laplace really wants to perform a Laplace transform. But which is the first step when we try to apply it on the right side of the given DE ?!
Note: The answer is not given mot-a-mot to the posted (relatively general) question, but in the sense of the progress. (Which is subjective.) I think, sample code serves good as an answer.
more | 2,893 | 7,267 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2024-33 | latest | en | 0.852136 | # series solutions of higher order ODEs I'm trying to use Sage to find the general series solution to $y^{(4)}=\frac{y'y''}{1+x}$. So far my best efforts to derive the coefficient recurrence relations, inspired by a good book draft, have been along these lines: R10 = QQ['a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10'] a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10 = R10.gens() R.<x> = PowerSeriesRing(R10) y = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + \ a6*x^6 + a7*x^7 + a8*x^8 + a9*x^9 + a10*x^10 + O(x^11) y1 = y.derivative() y2 = y1.derivative() y3 = y2.derivative() y4 = y3.derivative() f = (1+x)*y4-y1*y2 i = ideal(f) # g = i.groebner_fan(); g.reduced_groebner_bases() # a wish # q = R.quotient(i) # works, but not so useful by itself and my other approaches ended in tracebacks: x = var('x'); y = function('y', x) desolve(diff(y,x,4)-diff(y,x)*diff(y,x,2)/(1 + x), y, contrib_ode=True) # NotImplementedError: Maxima was unable to solve this ODE. desolve_laplace(diff(y,x,4) - diff(y,x)*diff(y,x,2)/(1 + x), y) # TypeError: unable to make sense of Maxima expression I would like to at least solve that and determine the radius of convergence. Ideally (more generally), it would be nice to have a good bag of tricks for working with series DEs such as I imagine others have already created. For this, I would like to find or develop techniques to incorporate: • more convenient coefficients, e.g. from this thread • a way to derive the recurrence relations using Python's lambda operator or this nice trick • a solver for higher order ODEs such as above Any hints, links, references or suggestions would be appreciated. edit retag close merge delete Sort by ยป oldest newest most voted The substitution $z=y'$ is natural, so let us please then forget about the posted $y$. We will not use $z$. Let us use $y$ "back" instead of $z$ in the sequel. So we solve (instead) around $x=0$: $$y'' ' = y\ y' \ /\ (1+x)\ .$$ Reusing the above code... R10.<a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10> = \ QQ['a0,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10'] R.<x> = PowerSeriesRing(R10) y = a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4 + a5*x^5 + \ a6*x^6 + a7*x^7 + a8*x^8 + a9*x^9 + a10*x^10 + O(x^11) y1 = y .derivative() y2 = y1.derivative() y3 = y2.derivative() f = (1+x)*y3 - y*y1 # J = ideal(f) # --> leads to an error f.list() ... we get the coefficients of $f$: [-a0*a1 + 6*a3, -a1^2 - 2*a0*a2 + 6*a3 + 24*a4, -3*a1*a2 - 3*a0*a3 + 24*a4 + 60*a5, -2*a2^2 - 4*a1*a3 - 4*a0*a4 + 60*a5 + 120*a6, -5*a2*a3 - 5*a1*a4 - 5*a0*a5 + 120*a6 + 210*a7, -3*a3^2 - 6*a2*a4 - 6*a1*a5 - 6*a0*a6 + 210*a7 + 336*a8, -7*a3*a4 - 7*a2*a5 - 7*a1*a6 - 7*a0*a7 + 336*a8 + 504*a9, -4*a4^2 - 8*a3*a5 - 8*a2*a6 - 8*a1*a7 - 8*a0*a8 + 504*a9 + 720*a10] We give or consider a0, a1, a2 as given. Solving by the Cauchy-Kovalevskaya "procedure", we have to determine recursively a3, a4, a5, a6, a7, a8, a9, a10, and so on, starting from a0, a1, a2, by making each line evaluate to $0$. We get manually $$a_3 = \frac 16 a_0a_1\ ,$$ $$a_4 = \frac 1{24}( a_1^2 + 2a_0a_2 + a_0a_1 )\ ,$$ and so on. I think it is not a good idea to isolate in this way all these equations, then start solving. Instead, let us write the given DE in the form $$y'' ' = y\ y'\ (1-x+x^2-x^3+\dots)\ ,$$ and consider it formally in the ring $\mathbb Q[[x]]$. Then the equations in $a_3,a_4,\dots$ come already separated, and we can simply write the code to get the first some $100$ coefficients. (I am not willing to write optimal code, time...) To have a precise situation, let us better use values instead of the symbols $a_0,a_1, a_2$. For instance $a_0=0$, $a_1=1$, $a_2=2$. Then... a = [0,1,2] # use other initial conditions, next time PSR.<x> = PowerSeriesRing( QQ, 100 ) C = 1 / PSR(1+x) for k in [ 3..99 ]: print k, # trying to get a(k) from y = sum( [ a(k) * x^k for k in [0..ooh] ] ) yk = sum( [ a[j] * x^j for j in range(k) ] ) + O(x^k) ak = 1/(k*(k-1)*(k-2)) * ( yk * diff( yk, x ) * ( C+O(x^k) ) ).list()[k-3] print ak a.append( ak ) y = PSR( sum( [ a[k] * x^k for k in [0..99] ] ) ) This computes some values. For instance, the first some $30$ entries are: sage: a[:30] [0, 1, 2, 0, 1/24, 1/12, 1/40, -67/5040, 13/1152, -7/1440, 4621/1209600, -38399/13305600, 11287/4838400, -12737/6988800, 2351057/1614412800, -171374993/145297152000, 1692465871/1743565824000, -115101113/142502976000, 32202455561/47424990412800, -4213094929/7313918976000, 22201339257697/45053740892160000, -267905698350809/630752372490240000, 5115211318062343/13876552194785280000, -8562348061276901/26596725040005120000, 928302494123858617/3282795776366346240000, -7613558036179561/30490363247984640000, 1985785680757233642821/8962032469480125235200000, -590031815878576041977/2987344156493375078400000, 33273819697787429364371/188202681859082629939200000, -3468510498746975345435917/21831511095653585072947200000] (These are exactly computed.) Let us check the solution modulo $O(x^{100})$: # : diff( y,x,3 ) - y * diff(y,x) / (1+x) # the above is a mess with terms in degrees 97, 98, 99, but.. sage: diff( y,x,3 ) - y * diff(y,x) / (1+x) + O( x^97 ) O(x^97) The radius of convergence is in this case probably one, one can try... for k in [1..99]: ak = a[k] print ( "a[%2d] ~ %13.10f and a[%2d]^(1/%2d) ~ %8.6f" % ( k, RR(ak), k, k, abs(RR(ak))^(1/k) ) ) and the first and last lines shown are: a[ 1] ~ 1.0000000000 and a[ 1]^(1/ 1) ~ 1.000000 a[ 2] ~ 2.0000000000 and a[ 2]^(1/ 2) ~ 1.414214 a[ 3] ~ 0.0000000000 and a[ 3]^(1/ 3) ~ 0.000000 a[ 4] ~ 0.0416666667 and a[ 4]^(1/ 4) ~ 0.451801 a[ 5] ~ 0.0833333333 and a[ 5]^(1/ 5) ~ 0.608364 a[ 6] ~ 0.0250000000 and a[ 6]^(1/ 6) ~ 0.540742 a[ 7] ~ -0.0132936508 and a[ 7]^(1/ 7) ~ 0.539447 a[ 8] ~ 0.0112847222 and a[ 8]^(1/ 8) ~ 0.570902 a[ 9] ~ -0.0048611111 and a[ 9]^(1/ 9) ~ 0.553313 ::::::::::::::::::::::::::::::::::::::::::::::::: a[90] ~ 0.0000051223 and a[90]^(1/90) ~ 0.873406 a[91] ~ -0.0000049542 and a[91]^(1/91) ~ 0.874386 a[92] ~ 0.0000047934 and a[92]^(1/92) ~ 0.875348 a[93] ~ -0.0000046394 and a[93]^(1/93) ~ 0.876295 a[94] ~ 0.0000044920 and a[94]^(1/94) ~ 0.877225 a[95] ~ -0.0000043507 and a[95]^(1/95) ~ 0.878140 a[96] ~ 0.0000042154 and a[96]^(1/96) ~ 0.879040 a[97] ~ -0.0000040855 and a[97]^(1/97) ~ 0.879925 a[98] ~ 0.0000039610 and a[98]^(1/98) ~ 0.880796 a[99] ~ -0.0000038414 and a[99]^(1/99) ~ 0.881653 I have maybe to stop here with sage sample code. About the convergence radius... The recurrence relation is thus of the shape $a_k=$ a sum of monomials of order two in the previous variables, with coefficients that can be recursively "controlled". With some work, one may (??) maybe show | that the radius of convergence is one. (The one is my quick guess, based on the fact that the limit $r=\lim |a_k|^{1/k}$ may / should exist, and then we expect $r\sim r^2$. Minoration / majoration for the "polynomial part in $k$" is constricted by the power $1/k$. But it is just a guess.) Note: The use of solve_laplace really wants to perform a Laplace transform. But which is the first step when we try to apply it on the right side of the given DE ?! Note: The answer is not given mot-a-mot to the posted (relatively general) question, but in the sense of the progress. (Which is subjective.) I think, sample code serves good as an answer. more |
http://epomme.com/tommy/index.php/2018/04/17/can-sin-1-0-have-a-finite-value/ | 1,524,143,760,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125936969.10/warc/CC-MAIN-20180419130550-20180419150550-00304.warc.gz | 105,190,248 | 22,528 | # Can Sin (1/0) have a finite value?
The expression $\sin(\dfrac{1}{0})$ is equivalent to $\sin(\dfrac{orange}{mango})$, i.e has no meaning.
On the other hand,
$\displaystyle \lim_\limits{x \to 0} \sin(\dfrac{1}{x})$does have a finite value, between $-1$ and $1$.
## 13 Replies to “Can Sin (1/0) have a finite value?”
1. Vaibhav Rastogi says:
Sin(1/0) can be written as sin(1/x) where
X is tending towards 0 or we can write it
Lim x-0(1/x)
It is clear from given function is defined for all values of x except 0 bcz at 0 it's value will be infinity .
So the value of sin(1/0) doesn't exist.
2. Rudresh Pandey says:
no,it can not be.
because the range of sin function lies between -1 to +1.
so for any value of domain, function (sinX) value can not be more then +1 or less then -1.
hope it will make you understand.
thank you.
3. Satya says:
Yes, the value is finite. Explanation – as the value of sine function is always between -1 to +1, it'll not depend on the angel for being infinite. It's not possible to find the exact value but it must be difinite between -1 and +1.
4. Azim Javed says:
First of all sin(1/0) was not possible bcz the maximum value of sin is 1 and minimum is -1.
Sin(1/0) =0
5. Nirbhay Thacker says:
It is just a finite value between 1 and -1. But, one can't tell the exact value. Whatever you put in sine and cosine function……they just lie from -1 to 1.
6. Vaibhav Rastogi says:
Ofcourse it is a finite value indeed but you cannot determine the value. Sin x is always between +-1 .
7. Satya says:
Yes because sin function domain between [1,-1] therefore the value is limited between 1&-1 is finite value
8. Abhishek Singh says:
No,it has infinite values
9. Abhishek Singh says:
Yes, sin(1/0) has a finite value bcoz
Sinx belongs to [-1,1] for every value of x.
Hope it helps!!
Regards!
10. Nirbhay Thacker says:
Yes which is lies between -1 to +1
11. Ritesh Arora says:
answer —— it will be an oscillating number between [-1,1].
12. Ritesh Arora says:
Yes and it lies between -1 and +1, it doesn't matter how large angle gets
13. Abhishek Singh says:
Yep. It’s close to my estimate of your IQ, somewhere between -1 and 1, but not exactly sure what. | 658 | 2,214 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-17 | latest | en | 0.913192 | # Can Sin (1/0) have a finite value? The expression $\sin(\dfrac{1}{0})$ is equivalent to $\sin(\dfrac{orange}{mango})$, i.e has no meaning. On the other hand, $\displaystyle \lim_\limits{x \to 0} \sin(\dfrac{1}{x})$does have a finite value, between $-1$ and $1$. ## 13 Replies to “Can Sin (1/0) have a finite value?” 1. Vaibhav Rastogi says: Sin(1/0) can be written as sin(1/x) where X is tending towards 0 or we can write it Lim x-0(1/x) It is clear from given function is defined for all values of x except 0 bcz at 0 it's value will be infinity . So the value of sin(1/0) doesn't exist. 2. Rudresh Pandey says: no,it can not be. because the range of sin function lies between -1 to +1. so for any value of domain, function (sinX) value can not be more then +1 or less then -1. hope it will make you understand. thank you. 3. Satya says: Yes, the value is finite. Explanation – as the value of sine function is always between -1 to +1, it'll not depend on the angel for being infinite. It's not possible to find the exact value but it must be difinite between -1 and +1. 4. Azim Javed says: First of all sin(1/0) was not possible bcz the maximum value of sin is 1 and minimum is -1. Sin(1/0) =0 5. Nirbhay Thacker says: It is just a finite value between 1 and -1. But, one can't tell the exact value. Whatever you put in sine and cosine function……they just lie from -1 to 1. 6. Vaibhav Rastogi says: Ofcourse it is a finite value indeed but you cannot determine the value. Sin x is always between +-1 . 7. Satya says: Yes because sin function domain between [1,-1] therefore the value is limited between 1&-1 is finite value 8. Abhishek Singh says: No,it has infinite values 9. Abhishek Singh says: Yes, sin(1/0) has a finite value bcoz Sinx belongs to [-1,1] for every value of x. Hope it helps!! Regards! 10. Nirbhay Thacker says: Yes which is lies between -1 to +1 11. Ritesh Arora says: answer —— it will be an oscillating number between [-1,1]. 12. Ritesh | Arora says: Yes and it lies between -1 and +1, it doesn't matter how large angle gets 13. Abhishek Singh says: Yep. It’s close to my estimate of your IQ, somewhere between -1 and 1, but not exactly sure what. |
https://www.hunker.com/13415369/how-to-calculate-a-walls-surface-area | 1,642,940,842,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304261.85/warc/CC-MAIN-20220123111431-20220123141431-00568.warc.gz | 832,630,966 | 100,656 | # How to Calculate a Wall's Surface Area
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• Tape measure
• Calculator
• Pencil
• Paper
#### Tip
Sketch the surface area to be measured and record the measurements and results as you go. Simplify your math by rounding inches to feet unless absolutely necessary.
Because most walls are square or rectangular, you can determine their surface area by multiplying the width and height measurements and then subtracting any door or window space. Atypical architecture and design, however, can result in having triangular, circular or trapezoidal shapes combined with the rectangle. The key to calculating the surface area of a wall accurately is using the basic formulas for finding the area of two-dimensional shapes.
## Step 1
Measure the vertical height and horizontal width of the wall.
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## Step 2
Multiply width and height measurements.
## Step 3
Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area.
## Step 1
Calculate any rectangular area and write down the result.
## Step 2
Locate the triangular shape. Measure the width along its base and the height from the base to the highest point.
## Step 3
Multiply the base and height measurements then divide the result by 2. Add this to the result you recorded in Step 1.
## Step 4
Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area.
## Step 1
Calculate any rectangular area and write down the result.
## Step 2
Measure the width of the bottom, the width of the top and the height. Write these numbers down.
## Step 3
Add the width measurements together and divide by 2. Multiply this number by the height measurement. Add this result to solution in Step 1.
## Step 4
Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area.
## Step 1
Calculate any rectangular area and write down the result.
## Step 2
Measure from the middle of the circle to the edge, which is known as the radius.
## Step 3
Multiply the measurement of the radius by itself, which is also known as squaring.
## Step 4
Multiply the result of squaring the radius by 3.14.
## Step 5
Divide the result in step 4 by 2, because you most likely need the area of only a portion a circle. Add this result to the area that you calculated in Step 1.
## Step 6
Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area. | 581 | 2,711 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-05 | longest | en | 0.873154 | # How to Calculate a Wall's Surface Area Hunker may earn compensation through affiliate links in this story. • Tape measure • Calculator • Pencil • Paper #### Tip Sketch the surface area to be measured and record the measurements and results as you go. Simplify your math by rounding inches to feet unless absolutely necessary. Because most walls are square or rectangular, you can determine their surface area by multiplying the width and height measurements and then subtracting any door or window space. Atypical architecture and design, however, can result in having triangular, circular or trapezoidal shapes combined with the rectangle. The key to calculating the surface area of a wall accurately is using the basic formulas for finding the area of two-dimensional shapes. ## Step 1 Measure the vertical height and horizontal width of the wall. Video of the Day ## Step 2 Multiply width and height measurements. ## Step 3 Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area. ## Step 1 Calculate any rectangular area and write down the result. ## Step 2 Locate the triangular shape. Measure the width along its base and the height from the base to the highest point. ## Step 3 Multiply the base and height measurements then divide the result by 2. Add this to the result you recorded in Step 1. ## Step 4 Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area. ## Step 1 Calculate any rectangular area and write down the result. ## Step 2 Measure the width of the bottom, the width of the top and the height. Write these numbers down. ## Step 3 Add the width measurements together and divide by 2. Multiply this number by the height measurement. Add this result to solution in Step 1. ## Step 4 Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area. ## Step 1 Calculate any rectangular area and write down the result. ## Step 2 Measure from the middle of the circle to the edge, which is known as the radius. ## Step 3 Multiply the measurement of the radius by itself, which is also known as squaring. ## Step 4 Multiply the result of squaring the radius by 3.14. ## Step 5 Divide the result in step 4 by 2, because you most likely | need the area of only a portion a circle. Add this result to the area that you calculated in Step 1. ## Step 6 Calculate the areas for doors and windows, if any. Subtract that amount from your overall measurements to determine the actual wall surface area. |
http://slideplayer.com/slide/5806065/ | 1,695,563,361,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506646.94/warc/CC-MAIN-20230924123403-20230924153403-00134.warc.gz | 40,679,556 | 28,458 | # DERIVING LINEAR REGRESSION COEFFICIENTS
## Presentation on theme: "DERIVING LINEAR REGRESSION COEFFICIENTS"— Presentation transcript:
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y X This sequence shows how the regression coefficients for a simple regression model are derived, using the least squares criterion (OLS, for ordinary least squares) 1
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y X We will start with a numerical example with just three observations: (1,3), (2,5), and (3,6). 2
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X Writing the fitted regression as Y = b1 + b2X, we will determine the values of b1 and b2 that minimize RSS, the sum of the squares of the residuals. ^ 3
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X Given our choice of b1 and b2, the residuals are as shown. 4
DERIVING LINEAR REGRESSION COEFFICIENTS
The sum of the squares of the residuals is thus as shown above. 5
DERIVING LINEAR REGRESSION COEFFICIENTS
The quadratics have been expanded. 6
DERIVING LINEAR REGRESSION COEFFICIENTS
Like terms have been added together. 7
DERIVING LINEAR REGRESSION COEFFICIENTS
For a minimum, the partial derivatives of RSS with respect to b1 and b2 should be zero. (We should also check a second-order condition.) 8
DERIVING LINEAR REGRESSION COEFFICIENTS
The first-order conditions give us two equations in two unknowns. 9
DERIVING LINEAR REGRESSION COEFFICIENTS
Solving them, we find that RSS is minimized when b1 and b2 are equal to 1.67 and 1.50, respectively. 10
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X Here is the scatter diagram again. 11
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X The fitted line and the fitted values of Y are as shown. 12
DERIVING LINEAR REGRESSION COEFFICIENTS
Before we move on to the general case, it is as well to make a small but important mathematical point. 13
DERIVING LINEAR REGRESSION COEFFICIENTS
When we establish the expression for RSS, we do so as a function of b1 and b2. At this stage, b1 and b2 are not specific values. Our task is to determine the particular values that minimize RSS. 14
DERIVING LINEAR REGRESSION COEFFICIENTS
We should give these values special names, to differentiate them from the rest. 15
DERIVING LINEAR REGRESSION COEFFICIENTS
Obvious names would be b1OLS and b2OLS, OLS standing for Ordinary Least Squares and meaning that these are the values that minimize RSS. We have re-written the first-order conditions and their solution accordingly. 16
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y X1 Xn X Now we will proceed to the general case with n observations. 17
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X Given our choice of b1 and b2, we will obtain a fitted line as shown. 18
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X The residual for the first observation is defined. 19
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X Similarly we define the residuals for the remaining observations. That for the last one is marked. 20
DERIVING LINEAR REGRESSION COEFFICIENTS
RSS, the sum of the squares of the residuals, is defined for the general case. The data for the numerical example are shown for comparison.. 21
DERIVING LINEAR REGRESSION COEFFICIENTS
DERIVING LINEAR REGRESSION COEFFICIENTS
Like terms are added together. 23
DERIVING LINEAR REGRESSION COEFFICIENTS
} Note that in this equation the observations on X and Y are just data that determine the coefficients in the expression for RSS. 24
DERIVING LINEAR REGRESSION COEFFICIENTS
} The choice variables in the expression are b1 and b2. This may seem a bit strange because in elementary calculus courses b1 and b2 are usually constants and X and Y are variables. 25
DERIVING LINEAR REGRESSION COEFFICIENTS
} However, if you have any doubts, compare what we are doing in the general case with what we did in the numerical example. 26
DERIVING LINEAR REGRESSION COEFFICIENTS
} The first derivative with respect to b1. 27
DERIVING LINEAR REGRESSION COEFFICIENTS
} With some simple manipulation we obtain a tidy expression for b1 . 28
DERIVING LINEAR REGRESSION COEFFICIENTS
} The first derivative with respect to b2. 29
DERIVING LINEAR REGRESSION COEFFICIENTS
Divide through by 2. 30
DERIVING LINEAR REGRESSION COEFFICIENTS
We now substitute for b1 using the expression obtained for it and we thus obtain an equation that contains b2 only. 31
DERIVING LINEAR REGRESSION COEFFICIENTS
The definition of the sample mean has been used. 32
DERIVING LINEAR REGRESSION COEFFICIENTS
The last two terms have been disentangled. 33
DERIVING LINEAR REGRESSION COEFFICIENTS
Terms not involving b2 have been transferred to the right side. 34
DERIVING LINEAR REGRESSION COEFFICIENTS
To create space, the equation is shifted to the top of the slide. 35
DERIVING LINEAR REGRESSION COEFFICIENTS
Hence we obtain an expression for b2. 36
DERIVING LINEAR REGRESSION COEFFICIENTS
In practice, we shall use an alternative expression. We will demonstrate that it is equivalent. 37
DERIVING LINEAR REGRESSION COEFFICIENTS
Expanding the numerator, we obtain the terms shown. 38
DERIVING LINEAR REGRESSION COEFFICIENTS
In the second term the mean value of Y is a common factor. In the third, the mean value of X is a common factor. The last term is the same for all i. 39
DERIVING LINEAR REGRESSION COEFFICIENTS
We use the definitions of the sample means to simplify the expression. 40
DERIVING LINEAR REGRESSION COEFFICIENTS
Hence we have shown that the numerators of the two expressions are the same. 41
DERIVING LINEAR REGRESSION COEFFICIENTS
The denominator is mathematically a special case of the numerator, replacing Y by X. Hence the expressions are quivalent. 42
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X The scatter diagram is shown again. We will summarize what we have done. We hypothesized that the true model is as shown, we obtained some data, and we fitted a line. 43
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X We chose the parameters of the fitted line so as to minimize the sum of the squares of the residuals. As a result, we derived the expressions for b1 and b2. 44
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X Again, we should make the mathematical point discussed in the context of the numerical example. These are the particular values of b1 and b2 that minimize RSS, and we should differentiate them from the rest by giving them special names, for example b1OLS and b2OLS. 45
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Y Fitted model b2 b1 X1 Xn X However, for the next few chapters, we shall mostly be concerned with the OLS estimators, and so the superscript 'OLS' is not really necessary. It will be dropped, to simplify the notation. 46
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Fitted model Typically, an intercept should be included in the regression specification. Occasionally, however, one may have reason to fit the regression without an intercept. In the case of a simple regression model, the true and fitted models become as shown. 47
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Fitted model We will derive the expression for b2 from first principles using the least squares criterion. The residual in observation i is ei = Yi – b2Xi. 48
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Fitted model With this, we obtain the expression for the sum of the squares of the residuals. 49
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Fitted model We differentiate with respect to b2. The OLS estimator is the value that makes this slope equal to zero (the first-order condition for a minimum). Note that we have differentiated properly between the general b2 and the specific b2OLS. 50
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Fitted model Hence, we obtain the OLS estimator of b2 for this model. 51
DERIVING LINEAR REGRESSION COEFFICIENTS
True model Fitted model The second derivative is positive, confirming that we have found a minimum. 52 | 2,035 | 8,246 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-40 | latest | en | 0.821159 | # DERIVING LINEAR REGRESSION COEFFICIENTS ## Presentation on theme: "DERIVING LINEAR REGRESSION COEFFICIENTS"— Presentation transcript: DERIVING LINEAR REGRESSION COEFFICIENTS True model Y X This sequence shows how the regression coefficients for a simple regression model are derived, using the least squares criterion (OLS, for ordinary least squares) 1 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y X We will start with a numerical example with just three observations: (1,3), (2,5), and (3,6). 2 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X Writing the fitted regression as Y = b1 + b2X, we will determine the values of b1 and b2 that minimize RSS, the sum of the squares of the residuals. ^ 3 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X Given our choice of b1 and b2, the residuals are as shown. 4 DERIVING LINEAR REGRESSION COEFFICIENTS The sum of the squares of the residuals is thus as shown above. 5 DERIVING LINEAR REGRESSION COEFFICIENTS The quadratics have been expanded. 6 DERIVING LINEAR REGRESSION COEFFICIENTS Like terms have been added together. 7 DERIVING LINEAR REGRESSION COEFFICIENTS For a minimum, the partial derivatives of RSS with respect to b1 and b2 should be zero. (We should also check a second-order condition.) 8 DERIVING LINEAR REGRESSION COEFFICIENTS The first-order conditions give us two equations in two unknowns. 9 DERIVING LINEAR REGRESSION COEFFICIENTS Solving them, we find that RSS is minimized when b1 and b2 are equal to 1.67 and 1.50, respectively. 10 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X Here is the scatter diagram again. 11 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X The fitted line and the fitted values of Y are as shown. 12 DERIVING LINEAR REGRESSION COEFFICIENTS Before we move on to the general case, it is as well to make a small but important mathematical point. 13 DERIVING LINEAR REGRESSION COEFFICIENTS When we establish the expression for RSS, we do so as a function of b1 and b2. At this stage, b1 and b2 are not specific values. Our task is to determine the particular values that minimize RSS. 14 DERIVING LINEAR REGRESSION COEFFICIENTS We should give these values special names, to differentiate them from the rest. 15 DERIVING LINEAR REGRESSION COEFFICIENTS Obvious names would be b1OLS and b2OLS, OLS standing for Ordinary Least Squares and meaning that these are the values that minimize RSS. We have re-written the first-order conditions and their solution accordingly. 16 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y X1 Xn X Now we will proceed to the general case with n observations. 17 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X Given our choice of b1 and b2, we will obtain a fitted line as shown. 18 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X The residual for the first observation is defined. 19 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X Similarly we define the residuals for the remaining observations. That for the last one is marked. 20 DERIVING LINEAR REGRESSION COEFFICIENTS RSS, the sum of the squares of the residuals, is defined for the general case. The data for the numerical example are shown for comparison.. 21 DERIVING LINEAR REGRESSION COEFFICIENTS DERIVING LINEAR REGRESSION COEFFICIENTS Like terms are added together. 23 DERIVING LINEAR REGRESSION COEFFICIENTS } Note that in this equation the observations on X and Y are just data that determine the coefficients in the expression for RSS. 24 DERIVING LINEAR REGRESSION COEFFICIENTS } The choice variables in the expression are b1 and b2. This may seem a bit strange because in elementary calculus courses b1 and b2 are usually constants and X and Y are variables. 25 DERIVING LINEAR REGRESSION COEFFICIENTS } However, if you have any doubts, compare what we are doing in the general case with what we did in the numerical example. 26 DERIVING LINEAR REGRESSION COEFFICIENTS } The first derivative with respect to b1. 27 DERIVING LINEAR REGRESSION COEFFICIENTS } With some simple manipulation we obtain a tidy expression for b1 . 28 DERIVING LINEAR REGRESSION COEFFICIENTS } The first derivative with respect to b2. 29 DERIVING LINEAR REGRESSION COEFFICIENTS Divide through by 2. 30 DERIVING LINEAR REGRESSION COEFFICIENTS We now substitute for b1 using the expression obtained for it and we thus obtain an equation that contains b2 only. 31 DERIVING LINEAR REGRESSION COEFFICIENTS The definition of the sample mean has been used. 32 DERIVING LINEAR REGRESSION COEFFICIENTS The last two terms have been disentangled. 33 DERIVING LINEAR REGRESSION COEFFICIENTS Terms not involving b2 have been transferred to the right side. 34 DERIVING LINEAR REGRESSION COEFFICIENTS To create space, the equation is shifted to the top of the slide. 35 DERIVING LINEAR REGRESSION COEFFICIENTS Hence we obtain an expression for b2. 36 DERIVING LINEAR REGRESSION COEFFICIENTS In practice, we shall use an alternative expression. We will demonstrate that it is equivalent. 37 DERIVING LINEAR REGRESSION COEFFICIENTS Expanding the numerator, we obtain the terms shown. 38 DERIVING LINEAR REGRESSION COEFFICIENTS In the second term the mean value of Y is a common factor. In the third, the mean value of X is a common factor. The last term is the same for all i. 39 DERIVING LINEAR REGRESSION COEFFICIENTS We use the definitions of the sample means to simplify the expression. 40 DERIVING LINEAR REGRESSION COEFFICIENTS Hence we have shown that the numerators of the two expressions are the same. 41 DERIVING LINEAR REGRESSION COEFFICIENTS The denominator is mathematically a special case of the numerator, replacing Y by X. Hence the expressions are quivalent. 42 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X The scatter diagram is shown again. We will summarize what we have done. We hypothesized that the true model is as shown, we obtained some data, and we fitted a line. 43 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X We chose the parameters of the fitted line so as to minimize the sum of the squares of the residuals. As a result, we derived the expressions for b1 and b2. 44 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X Again, we should make the mathematical point discussed in the context of the numerical example. These are the particular values of b1 and b2 that minimize RSS, and we should differentiate them from the rest by giving them special names, for example b1OLS and b2OLS. 45 DERIVING LINEAR REGRESSION COEFFICIENTS True model Y Fitted model b2 b1 X1 Xn X However, for the next few chapters, we shall mostly be concerned with the OLS estimators, and so the superscript 'OLS' is not really necessary. It will be dropped, to simplify the notation. 46 DERIVING LINEAR REGRESSION COEFFICIENTS True model Fitted model Typically, an intercept should be included in the regression specification. Occasionally, however, one may have reason to fit the regression without an intercept. In the case of a simple regression model, the true and fitted models become as shown. 47 DERIVING LINEAR REGRESSION COEFFICIENTS True model Fitted model We will derive the expression | for b2 from first principles using the least squares criterion. The residual in observation i is ei = Yi – b2Xi. 48 DERIVING LINEAR REGRESSION COEFFICIENTS True model Fitted model With this, we obtain the expression for the sum of the squares of the residuals. 49 DERIVING LINEAR REGRESSION COEFFICIENTS True model Fitted model We differentiate with respect to b2. The OLS estimator is the value that makes this slope equal to zero (the first-order condition for a minimum). Note that we have differentiated properly between the general b2 and the specific b2OLS. 50 DERIVING LINEAR REGRESSION COEFFICIENTS True model Fitted model Hence, we obtain the OLS estimator of b2 for this model. 51 DERIVING LINEAR REGRESSION COEFFICIENTS True model Fitted model The second derivative is positive, confirming that we have found a minimum. 52 |
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# 0.01410
Author Message
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0.01410 [#permalink] 25 May 2018, 09:41
Expert's post
00:00
Question Stats:
52% (00:12) correct 47% (00:19) wrong based on 34 sessions
Quantity A Quantity B $$0.0 \overline{1410}$$ $$0. \overline{0141}$$
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA
_________________
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Joined: 26 May 2018
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Re: 0.01410 [#permalink] 26 May 2018, 12:52
1
KUDOS
C option
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Re: 0.01410 [#permalink] 17 Mar 2019, 19:47
1
KUDOS
so we can just ignore the last digit?
for A replaced last digit "0" with 41
for B continue with 41
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Re: 0.01410 [#permalink] 20 Mar 2019, 08:48
Expert's post
The best thing to do here is to just write out a few repetitions and see what happens:
Quantity A: 0.01410141014101410...
Quantity B: 0.01410141014101410...
And now we can see that they're both the same.
(original post edited to reflect the better formatting)
Last edited by MagooshStudentHelp on 21 Mar 2019, 15:45, edited 1 time in total.
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Re: 0.01410 [#permalink] 21 Mar 2019, 02:40
Expert's post
Sorry for the inconvenience.
The question comes from the time the board was in its infant stage. From then a lot of things have been improved.
I have formatted the question properly now. The answer is still C
Thank you for your support guys.
Regards
_________________
Re: 0.01410 [#permalink] 21 Mar 2019, 02:40
Display posts from previous: Sort by | 776 | 2,410 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-13 | latest | en | 0.789179 | It is currently 26 Mar 2019, 04:07 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # 0.01410 Author Message TAGS: Moderator Joined: 18 Apr 2015 Posts: 5917 Followers: 96 Kudos [?]: 1158 [0], given: 5488 0.01410 [#permalink] 25 May 2018, 09:41 Expert's post 00:00 Question Stats: 52% (00:12) correct 47% (00:19) wrong based on 34 sessions Quantity A Quantity B $$0.0 \overline{1410}$$ $$0. \overline{0141}$$ A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given. [Reveal] Spoiler: OA _________________ Intern Joined: 26 May 2018 Posts: 31 Followers: 0 Kudos [?]: 5 [1] , given: 2 Re: 0.01410 [#permalink] 26 May 2018, 12:52 1 KUDOS C option Intern Joined: 08 Dec 2018 Posts: 8 Followers: 0 Kudos [?]: 6 [1] , given: 2 Re: 0.01410 [#permalink] 17 Mar 2019, 19:47 1 KUDOS so we can just ignore the last digit? for A replaced last digit "0" with 41 for B continue with 41 Manager Joined: 04 Feb 2019 Posts: 168 Followers: 4 Kudos [?]: 58 [0], given: 0 Re: 0.01410 [#permalink] 20 Mar 2019, 08:48 Expert's post The best thing to do here is to just write out a few repetitions and see what happens: Quantity A: 0.01410141014101410... Quantity B: 0.01410141014101410... And now we can see that they're both the same. (original post edited to reflect the better formatting) Last edited by MagooshStudentHelp on 21 Mar 2019, 15:45, edited 1 time in total. Moderator Joined: 18 Apr 2015 Posts: 5917 Followers: 96 Kudos [?]: 1158 [0], given: 5488 Re: 0.01410 [#permalink] 21 Mar 2019, 02:40 Expert's post Sorry for the inconvenience. The question comes from the time the board was in its infant stage. From | then a lot of things have been improved. I have formatted the question properly now. The answer is still C Thank you for your support guys. Regards _________________ Re: 0.01410 [#permalink] 21 Mar 2019, 02:40 Display posts from previous: Sort by |
https://rebab.net/how-many-ml-are-in-a-gallon/ | 1,660,484,147,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572033.91/warc/CC-MAIN-20220814113403-20220814143403-00568.warc.gz | 428,303,265 | 4,500 | To transform US gallons right into milliliters, you must multiply by the conversion aspect of 3785.41.
You are watching: How many ml are in a gallon
For example, if you desire to recognize how numerous milliliters space in 1 united state gallon, multiply the number of gallons by 3785.41 to acquire the answer in milliliters.
1 united state gallon x 3785.41 (conversion factor) = 3785.41 milliliters.
The answer, which is 3785.41, tells you the there room that numerous milliliters in 1 united state gallon.
If you want to reverse the question and also figure the end how many US gallons are in a certain variety of milliliters, you would certainly divide the number of milliliters by 3785.41.
For example, if you have actually 3785.41 milliliters, you have the right to divide the by 3785.41 to acquire 1 united state gallon.
This way that over there is 1 united state gallon in 3785.41 milliliters.
If girlfriend don’t feel like doing the math, use our digital conversion calculator to transform different devices of measurements and quickly give you the answer.
You can likewise use the complying with table to convert US gallons into milliliters.
US gallons right into milliliters conversion table
GallonsMillilitersGallonsMilliliters
13785.411141639.5
27570.821245424.9
311356.21349210.4
415141.61452995.8
518927.11556781.2
622712.51660566.6
726497.91764352
830283.31868137.4
934068.71971922.8
1037854.12075708.2
US gallons to millimeter conversion table
You can additionally convert 1 us gallon into other units of dimensions such as liters, cups, ounces, pints, and also quarts.
1 united state gallon = 15.77 cups1 united state gallon = 3.78541 liters1 united state gallon = 128 liquid ounces1 united state gallon = 4 quarts1 united state gallon = 8 pints
US gallons measurement
A united state gallon is a unit of volume measure up in liquid type that is typically used because that measuring bigger containers the liquid. The us gallon is equal to 3.785 liters.
See more: What Are The Conical Sections Of The Renal Medulla Called? Chapter 18 Flashcards
Milliliters measurement
A milliliter is a unit that volume offered to measure up liquid generally in smaller containers. Milliliters are used in the global system of devices (SI) and also is same to 1/1000 the a liter. | 591 | 2,297 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2022-33 | latest | en | 0.833859 | To transform US gallons right into milliliters, you must multiply by the conversion aspect of 3785.41. You are watching: How many ml are in a gallon For example, if you desire to recognize how numerous milliliters space in 1 united state gallon, multiply the number of gallons by 3785.41 to acquire the answer in milliliters. 1 united state gallon x 3785.41 (conversion factor) = 3785.41 milliliters. The answer, which is 3785.41, tells you the there room that numerous milliliters in 1 united state gallon. If you want to reverse the question and also figure the end how many US gallons are in a certain variety of milliliters, you would certainly divide the number of milliliters by 3785.41. For example, if you have actually 3785.41 milliliters, you have the right to divide the by 3785.41 to acquire 1 united state gallon. This way that over there is 1 united state gallon in 3785.41 milliliters. If girlfriend don’t feel like doing the math, use our digital conversion calculator to transform different devices of measurements and quickly give you the answer. You can likewise use the complying with table to convert US gallons into milliliters. US gallons right into milliliters conversion table GallonsMillilitersGallonsMilliliters 13785.411141639.5 27570.821245424.9 311356.21349210.4 415141.61452995.8 518927.11556781.2 622712.51660566.6 726497.91764352 830283.31868137.4 934068.71971922.8 1037854.12075708.2 US gallons to millimeter conversion table You can additionally convert 1 us gallon into other units of dimensions such as liters, cups, ounces, pints, and also quarts. 1 united state gallon = 15.77 cups1 united state gallon = 3.78541 liters1 united state gallon = 128 liquid ounces1 united state gallon = 4 quarts1 united state gallon = 8 pints US gallons measurement A united state gallon is a unit of volume measure up in liquid type that is typically used because that measuring bigger containers the liquid. The us gallon is equal to 3.785 liters. See more: What Are The Conical Sections Of The Renal Medulla Called? Chapter 18 Flashcards Milliliters | measurement A milliliter is a unit that volume offered to measure up liquid generally in smaller containers. Milliliters are used in the global system of devices (SI) and also is same to 1/1000 the a liter. |
http://www.freekent.com/1-john-s-sales-last-week-were-251-less-than-three/ | 1,601,086,763,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400232211.54/warc/CC-MAIN-20200926004805-20200926034805-00537.warc.gz | 168,902,684 | 6,953 | Question
# 1) John’s sales last week were \$251 less than three times Nancy’s sales. Together they sold \$1123. Determine how
much each person sold last week.
2) A restaurant holds 50 seats. Two types of seats are available the Friday night jazz festival: stage and dining. The cost of a stage is \$15.00 and the cost of a dining is \$5.00. If all 50 seats are sold the restaurant would collect \$400.00. How many of each type of seat is available?
3) A polishing machine requires 1 hours to make a unit of Product A, and 4 hours to make a unit of Product B. The polishing machine operated for 200 hours producing a total of 80 units. How many hours were used to manufacture units of Product A?
4) Edgar and Janet divide a profit of \$15 700. If Janet is to receive \$4200 more than one-fifths of Edgar’s share, how much will Edgar receive?
Math | 210 | 851 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2020-40 | latest | en | 0.951039 | Question # 1) John’s sales last week were \$251 less than three times Nancy’s sales. Together they sold \$1123. Determine how much each person sold last week. 2) A restaurant holds 50 seats. Two types of seats are available the Friday night jazz festival: stage and dining. The cost of a stage is \$15.00 and the cost of a dining is \$5.00. If all 50 seats are sold the restaurant would collect \$400.00. How many of each type of seat is available? 3) A polishing machine requires 1 hours to make a unit of Product A, and 4 hours to make a unit of Product B. The polishing machine operated for 200 hours producing a total of 80 units. How many hours were used to manufacture units of Product A? 4) Edgar and Janet divide a profit of \$15 700. If Janet | is to receive \$4200 more than one-fifths of Edgar’s share, how much will Edgar receive? Math |
https://books.google.no/books?id=5CkEAAAAQAAJ&q=angle+ABC&dq=editions:UOM39015067252117&lr=&hl=no&output=html&source=gbs_word_cloud_r&cad=5 | 1,632,127,959,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00352.warc.gz | 194,070,890 | 10,092 | # The Elements of Euclid, the parts read in the University of Cambridge [book 1-6 and parts of book 11,12] with geometrical problems, by J.W. Colenso
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Populære avsnitt
Side 42 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Side 4 - Let it be granted that a straight line may be drawn from any one point to any other point.
Side 33 - F, which is the common vertex of the triangles: that is », together with four right angles. Therefore all the angles of the figure, together with four right angles are equal to twice as many right angles as the figure has sides.
Side 62 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Side 22 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.
Side 58 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Side 146 - ... may be demonstrated from what has been said of the pentagon : and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon.
Side 194 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to the circumference are equal to one another : 16.
Side 69 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part. | 533 | 2,294 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.911593 | # The Elements of Euclid, the parts read in the University of Cambridge [book 1-6 and parts of book 11,12] with geometrical problems, by J.W. Colenso ### Hva folk mener -Skriv en omtale Vi har ikke funnet noen omtaler på noen av de vanlige stedene. ### Populære avsnitt Side 42 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Side 4 - Let it be granted that a straight line may be drawn from any one point to any other point. Side 33 - F, which is the common vertex of the triangles: that is », together with four right angles. Therefore all the angles of the figure, together with four right angles are equal to twice as many right angles as the figure has sides. Side 62 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. Side 22 - If from the ends of a side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Side 58 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Side 146 - ... may be demonstrated from what has been said of the pentagon : and likewise a circle may be inscribed in a given equilateral and equiangular hexagon, and circumscribed about it, by a method like to that used for the pentagon. Side 194 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar. Side 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such, that all straight lines drawn from a certain point within the figure to | the circumference are equal to one another : 16. Side 69 - To divide a given straight line into two parts, so that the rectangle contained by the whole, and one of the parts, may be equal to the square of the other part. |
https://www.physicsforums.com/threads/internal-combustion-engine-calculations.730736/ | 1,519,369,644,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814493.90/warc/CC-MAIN-20180223055326-20180223075326-00758.warc.gz | 927,727,529 | 15,977 | # Internal combustion engine calculations.
1. Dec 31, 2013
### gaber2611
Hello everybody,
I am working on the whole vehicle calculations, engine, brakes, suspension and engine simulation, using that software, take a look:
http://www.speed-wiz.com/calculations/engine-simulation/engine-simulation-conditions.htm
That software, you input the values at the top, and it calculates and outputs the results shown in the bottom of the screen
my work is to get the formulas which are used to get those ouputs, and i am just stuck with the intake air density, of those inputs in the screen i sent.
well, here are the formulas i used for "outside air density" calculations, and it gave me 0.0771 Ibs/cubic.ft, see how i got it, to help you understanding what i am requiring, for the "intake air density".
3- outside air density:
By substituting in the formula: Density = P / (R * T)
Where p: is the outside air pressure (Ibs/inch^2),
R: is the Gas constant = 53.35 (ft-lb)/ (lbm•R), and T: is the outside air temperature in °R,
And to convert P from (Ibs/inch^2) to (Ibs/ ft^2), we divide by (0.083) ^2, so we get:
Air pressure P = 14.68 /.083^2 = 2130.933 (Ibs/ft^2), so we get:
Density = 2130.933/ (53.35*518.67) = 0.077I Ibs/cf.
i am sure we will use same formula, but how can i calculate or get the intake air pressure to substitute in that formula, as the R is known, and T: is the intake air temperature is given as 80 F in the inputs.
hope anyone familiar with internal combustion engines help me with that,
will appreciate your help a lot,
Thanks.
2. Dec 31, 2013
### SteamKing
Staff Emeritus
Let me save you some time on some of the conversions:
If you want to convert pressure in psi to psf, multiply by 144 ( = 12^2) rather than divide by 0.083^2 (= 1/12^2)
There are 144 in^2 in 1 ft^2.
I'm not sure I understand your question. Air density varies with temperature and altitude. I notice you have specified an altitude of 1000 feet. Are you trying to calculate the air pressure at that altitude?
3. Jan 1, 2014
### gaber2611
Thanks steamKing for the reply, and no worries about the conversions, as long it gives same result, or very close to the software results
simply, what i am trying to calculate is the intake air density using that screen inputs which i sent its link, so how would calculate it, and getting same results of that screen, which is 0.002472 slugs/cu.ft, or 0.079452 Ibs/cu.ft, or 0.000040 grams/cc, so any help?
i hope you understand now, and that i could let you understand what i need,
Thanks again
4. Jan 1, 2014
### rcgldr
Are you trying to calculate the ambient pressure or the intake manifold pressure which will be significantly less, depending on the intake setup and throttle position?
5. Jan 1, 2014
### gaber2611
the intake manifold pressure , what is i am trying to calculate rcgldr, how to calculate using the software inputs?
6. Jan 2, 2014
### Kozy
Do you develop those applications, or are you simply trying to work out what the underlying equations are?
7. Jan 3, 2014
### gaber2611
hello kozy, thanks for the reply,
well, simply i am trying to work out the equations used for calculations, to substitute in and get same results as the software outputs,
i gave an example, for the outside air density calculations, i got the formula, substituted in it and got the results of 0.0771 Ibs/cubic feet
what i amstuck with is the inside air density, which formula used, and how to get the output of 0.079452 Ibs/cubic feet, look again at the link i sent, to see that result there, in same line of the "intake air density"
so can you help with how to get that value?, or anyone else here in the forum can help with that?
thanks | 982 | 3,700 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2018-09 | longest | en | 0.883222 | # Internal combustion engine calculations. 1. Dec 31, 2013 ### gaber2611 Hello everybody, I am working on the whole vehicle calculations, engine, brakes, suspension and engine simulation, using that software, take a look: http://www.speed-wiz.com/calculations/engine-simulation/engine-simulation-conditions.htm That software, you input the values at the top, and it calculates and outputs the results shown in the bottom of the screen my work is to get the formulas which are used to get those ouputs, and i am just stuck with the intake air density, of those inputs in the screen i sent. well, here are the formulas i used for "outside air density" calculations, and it gave me 0.0771 Ibs/cubic.ft, see how i got it, to help you understanding what i am requiring, for the "intake air density". 3- outside air density: By substituting in the formula: Density = P / (R * T) Where p: is the outside air pressure (Ibs/inch^2), R: is the Gas constant = 53.35 (ft-lb)/ (lbm•R), and T: is the outside air temperature in °R, And to convert P from (Ibs/inch^2) to (Ibs/ ft^2), we divide by (0.083) ^2, so we get: Air pressure P = 14.68 /.083^2 = 2130.933 (Ibs/ft^2), so we get: Density = 2130.933/ (53.35*518.67) = 0.077I Ibs/cf. i am sure we will use same formula, but how can i calculate or get the intake air pressure to substitute in that formula, as the R is known, and T: is the intake air temperature is given as 80 F in the inputs. hope anyone familiar with internal combustion engines help me with that, will appreciate your help a lot, Thanks. 2. Dec 31, 2013 ### SteamKing Staff Emeritus Let me save you some time on some of the conversions: If you want to convert pressure in psi to psf, multiply by 144 ( = 12^2) rather than divide by 0.083^2 (= 1/12^2) There are 144 in^2 in 1 ft^2. I'm not sure I understand your question. Air density varies with temperature and altitude. I notice you have specified an altitude of 1000 feet. Are you trying to calculate the air pressure at that altitude? 3. Jan 1, 2014 ### gaber2611 Thanks steamKing for the reply, and no worries about the conversions, as long it gives same result, or very close to the software results simply, what i am trying to calculate is the intake air density using that screen inputs which i sent its link, so how would calculate it, and getting same results of that screen, which is 0.002472 slugs/cu.ft, or 0.079452 Ibs/cu.ft, or 0.000040 grams/cc, so any help? i hope you understand now, and that i could let you understand what i need, Thanks again 4. Jan 1, 2014 ### rcgldr Are you trying to calculate the ambient pressure or the intake manifold pressure which will be significantly less, depending on the intake setup and throttle position? 5. Jan 1, 2014 ### gaber2611 the intake manifold pressure , what is i am trying to calculate rcgldr, how to calculate using the software inputs? 6. Jan 2, 2014 ### Kozy Do you develop those applications, or are you simply trying to work out what the underlying equations are? 7. Jan 3, 2014 ### gaber2611 hello kozy, thanks for the reply, well, simply i am trying to work out the equations used for calculations, to substitute in and get same results as the software outputs, i gave an example, for the outside air density calculations, i got the formula, substituted in it and got the results of 0.0771 Ibs/cubic | feet what i amstuck with is the inside air density, which formula used, and how to get the output of 0.079452 Ibs/cubic feet, look again at the link i sent, to see that result there, in same line of the "intake air density" so can you help with how to get that value?, or anyone else here in the forum can help with that? thanks |
https://usq.pressbooks.pub/statisticsforresearchstudents/chapter/comparing-two-independent-conditions-the-mann-whitney-u-test/ | 1,726,335,668,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651580.73/warc/CC-MAIN-20240914161327-20240914191327-00025.warc.gz | 545,013,991 | 20,653 | # Section 9.3: Comparing Two Independent Conditions: The Mann– Whitney U Test
Learning Objectives
At the end of this section you should be able to answer the following questions:
• When examining differences between two independent groups, which nonparametric test can be used?
• When examining differences between two dependent groups, which nonparametric test can be used?
## The Mann-Whitney U Test for two Independent Samples
When examining differences between two groups, Mann-Whitney U Test is best. This test examines the differences in median scores, as well as the size of the differences. Example: Is there a difference in the median number of Facebook Friends for male and female internet users? If a researcher wanted to compare Two Related Conditions, the test to use would be the Wilcoxon Signed-Rank Test.
## Interpretation for the Mann-Whitney U Test
As can be seen in the blue, there is a statistically significant difference, note the p value. The chi-squared value, and degrees of freedom are also needed for reporting. The median ranks indicate that female internet users have more Facebook Friends than male users.
## Write-up
The results of the Mann-Whitney U Test indicate that female internet users reported having a statistically significantly higher number of Facebook Friends (Median = 191.06) than male users (Median = 159.46; U = 5.65, p = .017).
PowerPoint: Mann-Whitney
Please click on the slides below to see an example of interpretation for the Mann-Whitney U Test. | 329 | 1,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2024-38 | latest | en | 0.884339 | # Section 9.3: Comparing Two Independent Conditions: The Mann– Whitney U Test Learning Objectives At the end of this section you should be able to answer the following questions: • When examining differences between two independent groups, which nonparametric test can be used? • When examining differences between two dependent groups, which nonparametric test can be used? ## The Mann-Whitney U Test for two Independent Samples When examining differences between two groups, Mann-Whitney U Test is best. This test examines the differences in median scores, as well as the size of the differences. Example: Is there a difference in the median number of Facebook Friends for male and female internet users? If a researcher wanted to compare Two Related Conditions, the test to use would be the Wilcoxon Signed-Rank Test. ## Interpretation for the Mann-Whitney U Test As can be seen in the blue, there is a statistically significant difference, note the p value. The chi-squared value, and degrees of freedom are also needed for reporting. The median ranks indicate that female internet users have more Facebook Friends than male users. ## Write-up The results of the Mann-Whitney U Test indicate that female internet users reported having a statistically significantly higher number of Facebook Friends (Median = 191.06) than male users (Median = 159.46; U | = 5.65, p = .017). PowerPoint: Mann-Whitney Please click on the slides below to see an example of interpretation for the Mann-Whitney U Test. |
https://www.proofwiki.org/wiki/Composition_of_Three_Mappings_which_form_Identity_Mapping | 1,685,855,301,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649439.65/warc/CC-MAIN-20230604025306-20230604055306-00525.warc.gz | 1,027,747,757 | 11,770 | # Composition of Three Mappings which form Identity Mapping
## Theorem
Let $A$, $B$ and $C$ be non-empty sets.
Let $f: A \to B$, $g: B \to C$ and $h: C \to A$ be mappings.
Let the following hold:
$\ds h \circ g \circ f$ $=$ $\ds I_A$ $\ds f \circ h \circ g$ $=$ $\ds I_B$ $\ds g \circ f \circ h$ $=$ $\ds I_C$
where:
$g \circ f$ (and so on) denote composition of mappings
$I_A$ (and so on) denote the identity mappings.
Then each of $f$, $g$ and $h$ are bijections, and:
$\ds f^{-1}$ $=$ $\ds h \circ g$ $\ds g^{-1}$ $=$ $\ds f \circ h$ $\ds h^{-1}$ $=$ $\ds g \circ f$
where $f^{-1}$ (and so on) denote the inverse mappings.
## Proof
First note that from Composition of Mappings is Associative:
$\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$
and so on.
However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done.
We have that:
$\paren {h \circ g} \circ f = I_A$
It follows from Injection iff Left Inverse that $f$ is an injection.
We also have that:
$f \circ \paren {h \circ g} = I_B$
It follows from Surjection iff Right Inverse that $f$ is a surjection.
So $f$ is a bijection.
It follows from the corollary to Bijection iff Left and Right Inverse that $h \circ g$ is also a bijection.
Thus we have that $h \circ g$ is both a left inverse and a right inverse of $f$.
It follows by definition that $h \circ g$ is the inverse of $f$:
$f^{-1} = h \circ g$
The same argument, mutatis mutandis, can be used to show that:
$g$ and $h$ are bijections
the inverse of $g$ is $f \circ h$
the inverse of $h$ is $g \circ f$.
$\blacksquare$ | 531 | 1,696 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-23 | latest | en | 0.809558 | # Composition of Three Mappings which form Identity Mapping ## Theorem Let $A$, $B$ and $C$ be non-empty sets. Let $f: A \to B$, $g: B \to C$ and $h: C \to A$ be mappings. Let the following hold: $\ds h \circ g \circ f$ $=$ $\ds I_A$ $\ds f \circ h \circ g$ $=$ $\ds I_B$ $\ds g \circ f \circ h$ $=$ $\ds I_C$ where: $g \circ f$ (and so on) denote composition of mappings $I_A$ (and so on) denote the identity mappings. Then each of $f$, $g$ and $h$ are bijections, and: $\ds f^{-1}$ $=$ $\ds h \circ g$ $\ds g^{-1}$ $=$ $\ds f \circ h$ $\ds h^{-1}$ $=$ $\ds g \circ f$ where $f^{-1}$ (and so on) denote the inverse mappings. ## Proof First note that from Composition of Mappings is Associative: $\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$ and so on. However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done. We have that: $\paren {h \circ g} \circ f = I_A$ It follows from Injection iff Left Inverse that $f$ is an injection. We also have that: $f \circ \paren {h \circ g} = I_B$ It follows from Surjection iff Right Inverse that $f$ is a surjection. So $f$ is a bijection. It follows from the corollary to Bijection iff Left and Right Inverse that $h \circ g$ is also a bijection. Thus we have that $h \circ g$ is both a left inverse and a right inverse of $f$. It follows by definition that $h \circ g$ is the inverse of $f$: $f^{-1} = h \circ g$ | The same argument, mutatis mutandis, can be used to show that: $g$ and $h$ are bijections the inverse of $g$ is $f \circ h$ the inverse of $h$ is $g \circ f$. $\blacksquare$ |
http://www.puzzlevilla.com/puzzles/puzzle/182 | 1,596,527,426,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735867.23/warc/CC-MAIN-20200804073038-20200804103038-00136.warc.gz | 149,403,885 | 10,752 | # PUZZLEVILLA
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Holmes Puzzle: Members in Each Group
Difficulty Level
In order to solve a case, Holmes was forced to work in disguise of a travelling man and found himself living with a group of nomads. One day, while he spent time with them, he was able to note the following few things:
1. 10% of the group comprised of young girls
2. 22% were young boys
3. 14% were old men
4. 10% of the total group members were old women
5. Out of the remaining 22 nomads, 50% were men.
Can you tell the total number of members in the group and the number of males?
View Answer Comments(1)
Answer: There were a total of fifty nomads in the group out of which 29 were male.
Assume that the total number of nomads in the group is x
According to the question, 10% of x=young girls
10% of x= old women
22% of x= young boys
14% of x= old men
Therefore, 0.1x+0.1x+0.22x+0.14x=x-22
0.56x=x-22
x-0.56x=22
0.44x=22
x = 50
Guest Said:Posted On 2016-06-17
impossible
Submit | 313 | 1,005 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-34 | latest | en | 0.96738 | # PUZZLEVILLA Login /Signup or Remember me ### Register OR Holmes Puzzle: Members in Each Group Difficulty Level In order to solve a case, Holmes was forced to work in disguise of a travelling man and found himself living with a group of nomads. One day, while he spent time with them, he was able to note the following few things: 1. 10% of the group comprised of young girls 2. 22% were young boys 3. 14% were old men 4. 10% of the total group members were old women 5. Out of the remaining 22 nomads, 50% were men. Can you tell the total number of members in the group and the number of males? View Answer Comments(1) Answer: There were a total of fifty nomads in the group out of which 29 were male. Assume that the total number of nomads in the group is x According to the question, 10% of x=young girls 10% of x= old women 22% of x= young boys | 14% of x= old men Therefore, 0.1x+0.1x+0.22x+0.14x=x-22 0.56x=x-22 x-0.56x=22 0.44x=22 x = 50 Guest Said:Posted On 2016-06-17 impossible Submit |
http://caponigroconstruction.com/assassin-s-magc/importance-of-line-graph-d204ab | 1,680,207,352,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949387.98/warc/CC-MAIN-20230330194843-20230330224843-00577.warc.gz | 10,168,506 | 9,710 | matrix of the graph line-graph L(G). Pros Click Here to subscribe to the CoolStuff Newsletter and get notified when our next blog is released. The Importance of Statistics to the User. Statistics is important in geography because of the following reasons: 1. Grid, Line and Path Games Graphing. The picture at the top of this page shows an example of a bar graph, a line graph, a pie chart, and a table. 5. Line graphs are very common in IELTS academic task 1 writing. 272 THE PHYSICS TEACHER Vol. Process 3 calculates the importance of the vertices of the graph L(G) equivalent to the importance of the edges of G. There are many different kinds of graphs. When using a line graph, why is it inportant to only graph 1 - 3 series of data? Each of these line graphs shows a change in data over time. IELTS Line Graph Answer. In this lesson, we will look at a Model Answer for CO2 emissions for 4 European countries and an analysis. The line graph consists of a … Once we graph something, it only makes sense to count elements from the graph, as well as to compare and contrast them. The important use of line graph is to track the changes over the short and long period of time. Typical examples of the types of data that can be presented using line graphs are monthly rainfall and annual unemployment rates. Each division should be equally spaced. or behaviors we want to increase (requesting for information, learning to read, counting, etc. Abstract: Graphs have been used to report scientific data for centuries; however, creating effective graphs can prove challenging. Line charts, or line graphs, are powerful visual tools that illustrate trends in data over a period of time or a particular correlation. The line graph is one of the simplest graphs you can make in Excel. How to Make a Line Graph in Excel: Explained Step-by-Step. Line graphs show you how numbers have changed over time. People use graphs to compare amounts of things or other numbers. In this post, we’ll talk about how a line graph works, plus: how to build one that provides meaningful information and context, what kind of data can be shown on a line graph… It is important to know that all line graphs must have a title. Using Line Graphs: An Example. Statistic is an important part of business,mathematics,representation of surveys,etc.It is the study of representing the lemark form of information into conc view the full answer. 3. The line graph represents the most frequently used display for visual analysis and subsequent interpretation and communication of experimental findings. Featured Products The Importance of the Best Fit Line Introduction One of the most important skills that you will learn in physics class is drawing a best fit line. Expert Answer . Label the x axis (horizontal axis) with the independent variable. By plotting sales figures on a line graph (as shown in figure 3), you can see the main fluctuations during the course of a year. Number each axis division (line). What are the importance of using a line graph. The line graph illustrates the amount of three kinds of spreads (margarine, low fat and reduced spreads and butter) which were consumed over 26 years from 1981 to 2007. Fibre bundles are not Cartesian products, but appear to be up close. Another name for a line graph is a line chart. Despite its apparent simplicity, the applications of the straight-line graph are often underesti- It's important to organise your graph clearly, draw out the key trends and make comparisons. 37, May 1999 The Importance of Graphs in Undergraduate Physics graph gives velocity, or a graph of velocity versus time will give acceleration. The title of a line graph provides a general overview of what is being displayed. In the next of our lessons on line graphs we look at how to change the verb/adverb collocations that we have already learnt into adjective/noun collocations. Some of the most common kinds are circle graphs, bar graphs, and line graphs. Just make sure the periods are equally distributed on the x-axis. Line graphs (or line charts) are best when you want to show how the value of something changes over time. Uses of Line Graph. There are four types of lines graphs: 1) Simple: These are drawn to … Important Vocabulary for Charts and Graphs. line graph depresses the trend and decreases variability. For too long we as humans have taken to much work upon our shoulders, it's time to simplify our lives and to use the best tools for the job. But that doesn’t mean it’s not one of the best.. ). Given the importance of reporting data effectively, it is worth your time to learn how to do so. Continuing with the sites IELTS line graph examples, this is an example of a line graph comparing car theft. Hello - I got a request to modify our Atterberg Limits graph to include the U-line in addition to the A-line. Page Content Offline Michael August Tue, Apr 22 2014 1:24 PM. Nevertheless, line graphs can also be applied to indicate tendencies based on other continuous periodic values, such as speed, temperature, distance etc. Label the y-axis (vertical axis) with the dependent variable. Line graph also has its pros and cons. Graphs are also known as charts. Usually what happens is you take two related variables and chart how they relate to each other. The chart on the right shows another bar graph, a diagram (sketch or picture), and a flow chart. If you are asking about charts and graphics in astrology, the answer is nothing, there isn’t any. A line graph in Microsoft Excel will not allow you to graph more than 3 series of data. Graphs are useful because they can be easier to understand than numbers and words alone. The graph of a function is contained in a Cartesian product of sets. This article describes the components of graphs, discusses general considerations in preparing graphs and graph selection, and addresses the most common problems in graphs. These behaviors can include behaviors we want to decrease (aggression, screaming, tantrums, pinching, self-injurious behavior, etc.) An X–Y plane is a cartesian product of two lines, called X and Y, while a cylinder is a cartesian product of a line and a circle, whose height, radius, and angle assign precise locations of the points. We take data on many different behaviors. Line Graph What is a Line Graph? Despite this common framework, the distinct features of bar and line graphs result in significant differences in their interpretation. Let's define the various parts of a line graph. Cartesian graphs have numbers on both axes, which therefore allow you to show how changes in one thing affect another. 3 Line Graphs 3.1 Line Graphs of an Undirected Graph Given an undirected graph G= (V;E), the line graph of is an undirected graph G = (E;F), in which there is an edge f2Fthat connects the nodes e;e02Eif and only if there is a node v2Vsuch that both eand 0incide on vin G. Line graphs have particular characteristics. Previous question Next question Get more help from Chegg. If you can use these adverbs effectively then you will improve you Task Achievement band and Lexical Resource band.. As well as frequency tables, data can be displayed in a a variety of graphs and charts too. Academic Writing task 1: C02 Emissions line graph. The lines are too thick to use more than 3 series. One of the reasons that I like graphing so much is that it gives my class a real-life, relevant reason to use math concepts. ABC Enterprises' sales vary throughout the year. You can find this book on Amazon. A line graph is useful for displaying data or information that changes continuously over time. The range will set the scale. Some examples of graphs used in ABA includes line graphs, bar graphs, and pie charts. The graph below will be used to help us define the parts of a line graph. IELTS Line Graph Examples. Written by co-founder Kasper Langmann, Microsoft Office Specialist.. Graphs are used in everyday life, from the local newspaper to the magazine stand. It will not show changes over time of you use more than 3 series of data. That has been done well in this answer. Line graphs can also be used to compare changes over the same period of time for more than one group. The most effective visuals are often the simplest—and line charts (another name for the same graph) are some of the easiest to understand. For example, a finance department may plot the change in the amount of cash the company has on hand over time. It is one of those skills that you simply cannot do without. Here, sales drop off in June and July, and again towards the end of the year. Like all graphs , bar graphs give a visual representation of numerical data. Determine the range of your data that must fit on each axis. Usually the line graph for data with the highest values is drawn first. 2. Simplifying your life is the way of the future. 584 Educ Psychol Rev (2017) 29:583–598. How to make a line graph 1. Atterberg Limits Graph - adding the U-Line to the graph. The separate lines that are made by connecting the points for each city. Today (finally) I’m going to talk about graphing. The input data (Data2) is the adjacency matrix A(G) of the graph G. Process 1 and 2 have been implemented using the programming language C/C++. 4. This line graph comes from Cambridge IELTS 11 academic. Table 1 Quality features of a lin e graph and measurement Essential structure Function Measured Vertical axis labeled with quantitative measure; horizontal axis labeled with time unit Line graphs are usually used to show time series data - that is how one or more variables vary over a continuous period of time. For example, one axis of the graph might represent a variable value, while the other axis often displays a timeline. Units are measured in grams. Bar Graph Examples Line Graphs. More than 3 series of data causes too many lines on the graph, which makes ti confusing to read.•• Graphing is one of those tools that you just cannot be without. It enables the geographers to handle large sets of data and summarize them in … Pros and Cons of Using Line Graphs As is known to all, every coin has two sides. Line graphs. Two of the most common are Line Graphs and Bar Graphs, which we will see examples of in this page. The points on the double line graph show the average monthly rainfall in the two cities (city 1, city 2). Line Graph: A line graph is a graph that measures change over time by plotting individual data points connected by straight lines. A line graph, also known as a line chart, is a type of chart used to visualize the value of something over time. They are used when you have data that are connected, and to show trends, for example, average night-time temperature in each month of the year. 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Are the importance of using line graphs: an example is nothing, there isn ’ t mean ’. May plot the change in the amount of cash the company has on hand over time of use! Thing affect another in Microsoft Excel will not show changes over time you. To include the U-Line to the graph, as well as to compare of... U-Line to the magazine stand you take two related variables and chart how they relate to each other points! ( sketch or picture ), and pie charts of something changes over time of you use than. The importance of reporting data effectively, it only makes sense to count elements from the newspaper! Another bar graph, a finance department may plot the change in the amount of cash the has.: 1 have a title to each other plot the change in the amount of cash company..., draw out the key trends and make comparisons, why is inportant! Excel: Explained Step-by-Step effectively, it is worth your time to learn how to make a line graph a... Aggression, screaming, tantrums, pinching, self-injurious behavior, etc. some of the types of data importance of line graph. Common in IELTS academic task 1: C02 Emissions line graph that simply. ( vertical axis ) with the sites IELTS line graph comparing car theft things! Do so ’ m going to talk about graphing graphs can prove challenging end the... The two cities ( city 1, city 2 ) sets of.. For example, one axis of the following reasons: 1 Get more help from Chegg graphics astrology. Axis ) with the independent variable t mean it ’ s not one of those skills you! Question Next question Get more help from Chegg or line charts ) best. Some examples of graphs and charts too are used in everyday life, the. Is you take two related variables and chart how they relate to other! The importance of reporting data effectively, it is important to know that all line graphs is. Read.•• IELTS line graph in Microsoft Excel will not allow you to graph more than 3 series data... | 4,122 | 19,376 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2023-14 | latest | en | 0.930564 | matrix of the graph line-graph L(G). Pros Click Here to subscribe to the CoolStuff Newsletter and get notified when our next blog is released. The Importance of Statistics to the User. Statistics is important in geography because of the following reasons: 1. Grid, Line and Path Games Graphing. The picture at the top of this page shows an example of a bar graph, a line graph, a pie chart, and a table. 5. Line graphs are very common in IELTS academic task 1 writing. 272 THE PHYSICS TEACHER Vol. Process 3 calculates the importance of the vertices of the graph L(G) equivalent to the importance of the edges of G. There are many different kinds of graphs. When using a line graph, why is it inportant to only graph 1 - 3 series of data? Each of these line graphs shows a change in data over time. IELTS Line Graph Answer. In this lesson, we will look at a Model Answer for CO2 emissions for 4 European countries and an analysis. The line graph consists of a … Once we graph something, it only makes sense to count elements from the graph, as well as to compare and contrast them. The important use of line graph is to track the changes over the short and long period of time. Typical examples of the types of data that can be presented using line graphs are monthly rainfall and annual unemployment rates. Each division should be equally spaced. or behaviors we want to increase (requesting for information, learning to read, counting, etc. Abstract: Graphs have been used to report scientific data for centuries; however, creating effective graphs can prove challenging. Line charts, or line graphs, are powerful visual tools that illustrate trends in data over a period of time or a particular correlation. The line graph is one of the simplest graphs you can make in Excel. How to Make a Line Graph in Excel: Explained Step-by-Step. Line graphs show you how numbers have changed over time. People use graphs to compare amounts of things or other numbers. In this post, we’ll talk about how a line graph works, plus: how to build one that provides meaningful information and context, what kind of data can be shown on a line graph… It is important to know that all line graphs must have a title. Using Line Graphs: An Example. Statistic is an important part of business,mathematics,representation of surveys,etc.It is the study of representing the lemark form of information into conc view the full answer. 3. The line graph represents the most frequently used display for visual analysis and subsequent interpretation and communication of experimental findings. Featured Products The Importance of the Best Fit Line Introduction One of the most important skills that you will learn in physics class is drawing a best fit line. Expert Answer . Label the x axis (horizontal axis) with the independent variable. By plotting sales figures on a line graph (as shown in figure 3), you can see the main fluctuations during the course of a year. Number each axis division (line). What are the importance of using a line graph. The line graph illustrates the amount of three kinds of spreads (margarine, low fat and reduced spreads and butter) which were consumed over 26 years from 1981 to 2007. Fibre bundles are not Cartesian products, but appear to be up close. Another name for a line graph is a line chart. Despite its apparent simplicity, the applications of the straight-line graph are often underesti- It's important to organise your graph clearly, draw out the key trends and make comparisons. 37, May 1999 The Importance of Graphs in Undergraduate Physics graph gives velocity, or a graph of velocity versus time will give acceleration. The title of a line graph provides a general overview of what is being displayed. In the next of our lessons on line graphs we look at how to change the verb/adverb collocations that we have already learnt into adjective/noun collocations. Some of the most common kinds are circle graphs, bar graphs, and line graphs. Just make sure the periods are equally distributed on the x-axis. Line graphs (or line charts) are best when you want to show how the value of something changes over time. Uses of Line Graph. There are four types of lines graphs: 1) Simple: These are drawn to … Important Vocabulary for Charts and Graphs. line graph depresses the trend and decreases variability. For too long we as humans have taken to much work upon our shoulders, it's time to simplify our lives and to use the best tools for the job. But that doesn’t mean it’s not one of the best.. ). Given the importance of reporting data effectively, it is worth your time to learn how to do so. Continuing with the sites IELTS line graph examples, this is an example of a line graph comparing car theft. Hello - I got a request to modify our Atterberg Limits graph to include the U-line in addition to the A-line. Page Content Offline Michael August Tue, Apr 22 2014 1:24 PM. Nevertheless, line graphs can also be applied to indicate tendencies based on other continuous periodic values, such as speed, temperature, distance etc. Label the y-axis (vertical axis) with the dependent variable. Line graph also has its pros and cons. Graphs are also known as charts. Usually what happens is you take two related variables and chart how they relate to each other. The chart on the right shows another bar graph, a diagram (sketch or picture), and a flow chart. If you are asking about charts and graphics in astrology, the answer is nothing, there isn’t any. A line graph in Microsoft Excel will not allow you to graph more than 3 series of data. Graphs are useful because they can be easier to understand than numbers and words alone. The graph of a function is contained in a Cartesian product of sets. This article describes the components of graphs, discusses general considerations in preparing graphs and graph selection, and addresses the most common problems in graphs. These behaviors can include behaviors we want to decrease (aggression, screaming, tantrums, pinching, self-injurious behavior, etc.) An X–Y plane is a cartesian product of two lines, called X and Y, while a cylinder is a cartesian product of a line and a circle, whose height, radius, and angle assign precise locations of the points. We take data on many different behaviors. Line Graph What is a Line Graph? Despite this common framework, the distinct features of bar and line graphs result in significant differences in their interpretation. Let's define the various parts of a line graph. Cartesian graphs have numbers on both axes, which therefore allow you to show how changes in one thing affect another. 3 Line Graphs 3.1 Line Graphs of an Undirected Graph Given an undirected graph G= (V;E), the line graph of is an undirected graph G = (E;F), in which there is an edge f2Fthat connects the nodes e;e02Eif and only if there is a node v2Vsuch that both eand 0incide on vin G. Line graphs have particular characteristics. Previous question Next question Get more help from Chegg. If you can use these adverbs effectively then you will improve you Task Achievement band and Lexical Resource band.. As well as frequency tables, data can be displayed in a a variety of graphs and charts too. Academic Writing task 1: C02 Emissions line graph. The lines are too thick to use more than 3 series. One of the reasons that I like graphing so much is that it gives my class a real-life, relevant reason to use math concepts. ABC Enterprises' sales vary throughout the year. You can find this book on Amazon. A line graph is useful for displaying data or information that changes continuously over time. The range will set the scale. Some examples of graphs used in ABA includes line graphs, bar graphs, and pie charts. The graph below will be used to help us define the parts of a line graph. IELTS Line Graph Examples. Written by co-founder Kasper Langmann, Microsoft Office Specialist.. Graphs are used in everyday life, from the local newspaper to the magazine stand. It will not show changes over time of you use more than 3 series of data. That has been done well in this answer. Line graphs can also be used to compare changes over the same period of time for more than one group. The most effective visuals are often the simplest—and line charts (another name for the same graph) are some of the easiest to understand. For example, a finance department may plot the change in the amount of cash the company has on hand over time. It is one of those skills that you simply cannot do without. Here, sales drop off in June and July, and again towards the end of the year. Like all graphs , bar graphs give a visual representation of numerical data. Determine the range of your data that must fit on each axis. Usually the line graph for data with the highest values is drawn first. 2. Simplifying your life is the way of the future. 584 Educ Psychol Rev (2017) 29:583–598. How to make a line graph 1. Atterberg Limits Graph - adding the U-Line to the graph. The separate lines that are made by connecting the points for each city. Today (finally) I’m going to talk about graphing. The input data (Data2) is the adjacency matrix A(G) of the graph G. Process 1 and 2 have been implemented using the programming language C/C++. 4. This line graph comes from Cambridge IELTS 11 academic. Table 1 Quality features of a lin e graph and measurement Essential structure Function Measured Vertical axis labeled with quantitative measure; horizontal axis labeled with time unit Line graphs are usually used to show time series data - that is how one or more variables vary over a continuous period of time. For example, one axis of the graph might represent a variable value, while the other axis often displays a timeline. Units are measured in grams. Bar Graph Examples Line Graphs. More than 3 series of data causes too many lines on the graph, which makes ti confusing to read.•• Graphing is one of those tools that you just cannot be without. It enables the geographers to handle large sets of data and summarize them in … Pros and Cons of Using Line Graphs As is known to all, every coin has two sides. Line graphs. Two of the most common are Line Graphs and Bar Graphs, which we will see examples of in this page. The points on the double line graph show the average monthly rainfall in the two cities (city 1, city 2). Line Graph: A line graph is a graph that measures change over time by plotting individual data points connected by straight lines. A line graph, also known as a line chart, is a type of chart used to visualize the value of something over time. They are used when you have data that are connected, and to show trends, for example, average night-time temperature in each month of the year. Chart on the right shows another bar graph, why is it inportant to graph. Atterberg Limits graph - adding the U-Line to the A-line in astrology, the distinct features of bar line! More help from Chegg creating effective graphs can also be used to help define! X axis ( horizontal axis ) with the independent variable what are the importance of using line. Visual representation of numerical data is nothing, there isn ’ t.. Graphs must have a title time for more than one group variable value while. 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Author: Jim Bumgardner | Website: http://www.krazydad.com/bestiary/
Here's a basic script that can make a movieclip point at another object.
```MovieClip.prototype.pointAt = function(x,y)
{
var dx = x - this._x; // distance to other object on x-axis
var dy = y - this._y; // distance on y-axis
var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem)
var angle;
// figure out angle in radians (0- 2*PI)
if (dy < 0)
angle = Math.PI*2-Math.acos(dx/dist);
else
angle = Math.acos(dx/dist);
// convert to rotation value (angle in degrees or 0-360)
this._rotation = angle*180/Math.PI;
}
```
#### Math Explanation
The function that does all the work here is Math.acos(). The Math.acos() function is the compliment of the Math.cos() function. On another tutorial here, you may have seen how to use Math.cos() and Math.sin() to draw circle shapes using polar coodinates.
Math.cos() takes an angle (expressed in radians, or going from 0 to 2*PI instead of 0 to 360) and converts it to a position on a circle.
Given an angle (expressed in degrees) and a radius, a point on a circle is given by:
```x = Math.cos(angle*Math.PI/180)*radius;
```
The Math.acos() function I'm using does the reverse, it converts the value returned by Math.cos() back into an angle in radians. So by reversing the math of the above equations for using polar coordinates, we can determine the original angle in radians. Then, to convert this angle to a rotation value (which is in degrees) we scale it by the ratio betwen degres and radians: 360 / (2*PI) which is the same as 180 / PI.
#### How to use the script
Here's a script, which works with the above script to make a movieclip point in the direction of the mouse.
```mc.pointAtMouse = function()
{
this.pointAt(this._parent._xmouse, this._parent._ymouse);
}
```
Let's say your have a movieclip that you want to point at the mouse, you can do the following to make this happpen:
```mc.onEnterFrame = mc.pointAtMouse;
```
OR you can call this.pointAtMouse() from within a more complex onEnterFrame function.
Finally, this script assumes that your object points to 3:00 (to the right) when it is in it's 'normal' (rotation=0) state. If it doesn't, you'll need to add an offset to the rotation value. Here's a version of the function which accepts an offset. If your object normally points at 12:00 (upwards) then use -90 to correct it.
```MovieClip.prototype.pointAt = function(x,y,offset)
{
var dx = x - this._x; // distance to other object on x-axis
var dy = y - this._y; // distance on y-axis
var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem)
var angle;
// figure out angle in radians (0- 2*PI)
if (dy < 0)
angle = Math.PI*2-Math.acos(dx/dist);
else
angle = Math.acos(dx/dist);
// convert to rotation value (angle in degrees or 0-360)
this._rotation = offset + angle*180/Math.PI;
}
```
Example usage for object which normally points upwards (12:00):
```this.pointAt(this._parent._xmouse, this._parent._ymouse, -90);
```
- jim
» Level Intermediate Added: 2004-03-15 Rating: 7.17 Votes: 24 1 2 3 4 5 6 7 8 9 10 (10 being the highest) » Author Professional C/C++ programmer and flash hobbyist. » Download Download the files used in this tutorial. Download (0 kb) » Forums More help? Search our boards for quick answers!
• There are no comments yet. Be the first to comment!
• You must have javascript enabled in order to post comments.
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» Title: RaiseTheBlocks
» Description: Raise all the blocks to win the game
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» Author: TomCat Carty
» Title: The Wood
» Description: Just a little game ending or it can maybe be looped. Recorders with music box and percussion to give the feel of well, I don't know, the woods? Free to use, just credit me. thank you
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» Author: Fábio FAFERS
» Description: I created this font for free use. Everyone can apply it in personal or business texts. Its free, but I want to be communicated in case of business use. Donations are accepted to keep the project of free fonts alive! Thank you all
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Audio Player
» Author: Davisigner
» Description: Hmm... what to say about this one? It's reminiscent of the closing notes of the opening music from the Three Stooges done in a church organ style with a closing cymbal crash. I'll give this one away gratis, but feel free to check out my free loops and potential upcoming license-mandated ones over in the respective part of Flashkit. | 1,229 | 4,724 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2019-13 | longest | en | 0.72799 | » Home » Movies » Tutorials » Submissions » Sound FX » Board » Links » Reviews » Feedback » Gallery » Fonts » The Lounge » Sound Loops First time here? Newsletter Search tutorials Author: Jim Bumgardner | Website: http://www.krazydad.com/bestiary/ Here's a basic script that can make a movieclip point at another object. ```MovieClip.prototype.pointAt = function(x,y) { var dx = x - this._x; // distance to other object on x-axis var dy = y - this._y; // distance on y-axis var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem) var angle; // figure out angle in radians (0- 2*PI) if (dy < 0) angle = Math.PI*2-Math.acos(dx/dist); else angle = Math.acos(dx/dist); // convert to rotation value (angle in degrees or 0-360) this._rotation = angle*180/Math.PI; } ``` #### Math Explanation The function that does all the work here is Math.acos(). The Math.acos() function is the compliment of the Math.cos() function. On another tutorial here, you may have seen how to use Math.cos() and Math.sin() to draw circle shapes using polar coodinates. Math.cos() takes an angle (expressed in radians, or going from 0 to 2*PI instead of 0 to 360) and converts it to a position on a circle. Given an angle (expressed in degrees) and a radius, a point on a circle is given by: ```x = Math.cos(angle*Math.PI/180)*radius; ``` The Math.acos() function I'm using does the reverse, it converts the value returned by Math.cos() back into an angle in radians. So by reversing the math of the above equations for using polar coordinates, we can determine the original angle in radians. Then, to convert this angle to a rotation value (which is in degrees) we scale it by the ratio betwen degres and radians: 360 / (2*PI) which is the same as 180 / PI. #### How to use the script Here's a script, which works with the above script to make a movieclip point in the direction of the mouse. ```mc.pointAtMouse = function() { this.pointAt(this._parent._xmouse, this._parent._ymouse); } ``` Let's say your have a movieclip that you want to point at the mouse, you can do the following to make this happpen: ```mc.onEnterFrame = mc.pointAtMouse; ``` OR you can call this.pointAtMouse() from within a more complex onEnterFrame function. Finally, this script assumes that your object points to 3:00 (to the right) when it is in it's 'normal' (rotation=0) state. If it doesn't, you'll need to add an offset to the rotation value. Here's a version of the function which accepts an offset. If your object normally points at 12:00 (upwards) then use -90 to correct it. ```MovieClip.prototype.pointAt = function(x,y,offset) { var dx = x - this._x; // distance to other object on x-axis var dy = y - this._y; // distance on y-axis var dist = Math.sqrt(dx*dx + dy*dy); // true distance (using Pythagorean theorem) var angle; // figure out angle in radians (0- 2*PI) if (dy < 0) angle = Math.PI*2-Math.acos(dx/dist); else angle = Math.acos(dx/dist); // convert to rotation value (angle in degrees or 0-360) this._rotation = offset + angle*180/Math.PI; } ``` Example usage for object which normally points upwards (12:00): ```this.pointAt(this._parent._xmouse, this._parent._ymouse, -90); ``` - jim » Level Intermediate Added: 2004-03-15 Rating: 7.17 Votes: 24 1 2 3 4 5 6 7 8 9 10 (10 being the highest) » Author Professional C/C++ programmer and flash hobbyist. » Download Download the files used in this tutorial. Download (0 kb) » Forums More help? Search our boards for quick answers! • There are no comments yet. Be the first to comment! • You must have javascript enabled in order to post comments. Featured Flash FLA » Author: Inocreato » Title: RaiseTheBlocks » Description: Raise all the blocks to win the game Featured Sound Loops Audio Player » Author: TomCat Carty » Title: The Wood » Description: Just a little game ending or it can maybe be looped. Recorders with music box and percussion to give the feel of well, I don't know, the woods? Free to use, just credit me. thank you Latest Font » Author: Fábio FAFERS » Description: I created this font for free use. Everyone can apply it in personal or business texts. Its free, but I want to be communicated in case of business use. Donations are accepted to | keep the project of free fonts alive! Thank you all Featured Sound Fx Audio Player » Author: Davisigner » Description: Hmm... what to say about this one? It's reminiscent of the closing notes of the opening music from the Three Stooges done in a church organ style with a closing cymbal crash. I'll give this one away gratis, but feel free to check out my free loops and potential upcoming license-mandated ones over in the respective part of Flashkit. |
https://www.jiskha.com/questions/1793090/what-is-the-nth-term-rule-of-the-linear-sequence-below-27-25-23-21-19 | 1,603,207,250,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107872746.20/warc/CC-MAIN-20201020134010-20201020164010-00542.warc.gz | 786,280,729 | 5,029 | # maths
What is the nth term rule of the linear sequence below?
27,25,23,21,19
1. 👍 5
2. 👎 0
3. 👁 527
1. Looks like an arithmetic sequence with a = 27 and d = -2
Term(n) = a + (n-1)d
sub in the given values and simplify
tell me what you get.
1. 👍 0
2. 👎 5
👨🏫
Reiny
2. cant you just tell us what it is were and have no cle what youre waffling on about
1. 👍 7
2. 👎 0
3. reiny thats not answering the question
1. 👍 2
2. 👎 0
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The fifth term of an arithmetic sequence is 23 and 12th term is 72. And they say 1. Determine the first three terms of the sequences and the nth term. 2. What is the value of the 10th term. 3. Which term has a value of 268. | 845 | 2,983 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2020-45 | latest | en | 0.929854 | # maths What is the nth term rule of the linear sequence below? 27,25,23,21,19 1. 👍 5 2. 👎 0 3. 👁 527 1. Looks like an arithmetic sequence with a = 27 and d = -2 Term(n) = a + (n-1)d sub in the given values and simplify tell me what you get. 1. 👍 0 2. 👎 5 👨🏫 Reiny 2. cant you just tell us what it is were and have no cle what youre waffling on about 1. 👍 7 2. 👎 0 3. reiny thats not answering the question 1. 👍 2 2. 👎 0 ## Similar Questions 1. ### math what are the similarities and differences between an arithmetic sequence and a linear equation? ok i know that arithmetic sequence is a sequence of real numbers for which each term is the previous term plus a constant (called the 2. ### Algebra ASAP pls pls help out A geometric sequence begins 36, 9, 9/4, 9/16, 9/64, . . . . (a) Find the common ratio r for this sequence. r = (b) Find a formula for the nth term an of the sequence. an = (c) Find the tenth term of the sequence. 3. ### Maths The 5th,9th and 16th terms of a linear sequence A.P are consecutive terms of an exponential sequence.Find the common difference of the linear sequence in terms of the first term. 4. ### Calculus 2 - Series I am so confused on how to do series problems...especially these. How can you tell the pattern and determining the formula for them? Can someone please help? 26) Write the first five terms of the sequence {an} whose nth term is 1. ### Pre-Calc/Trig... The nth term of a geometric sequence is given by an = 1/4(4)^n-1. Write the first five terms of this sequence. 2. ### mathematics The fifth ninth and sixteenth terms of a linear sequence ( ap) are consecutive terms of an exponential sequence 1 find the common difference of the linear sequence in terms of the 1st term 2 show that the 21st 37th and 65th terms 3. ### Dr.tsegay The 3rd term of arthmetic sequence is 3/4th of 6th term.and their sum is 60.find a,d and nth term. 4. ### Urgent math how do I do this can any body show me how? An arithmetic sequence begins 4, 9, 14, 19, 24, . . . . (a) Find the common difference d for this sequence. d = (b) Find a formula for the nth term an of the sequence. an = (c) Find the 1. ### Math The first, third and ninth terms of a linear sequence are the first three terms of an exponential sequence. If the seventh term of the linear sequence is 14 calculate the twentieth term of the linear sequence and the sum of the 2. ### alegebra the first third and ninth term of a linear sequence are the first three terms of an exponential sequence.if the 7th term of the linear sequence is 14 calcuate the 20th term of the linear sequnce 3. ### math sandra wrote the sequence below 2,5,10,17,... which equation represents the | rule for finding the nth term of this equation 4. ### maths The fifth term of an arithmetic sequence is 23 and 12th term is 72. And they say 1. Determine the first three terms of the sequences and the nth term. 2. What is the value of the 10th term. 3. Which term has a value of 268. |
https://www.unitmeasurement.com/power-conversion/femtowatts-to-kilowatts.html | 1,632,836,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780060803.2/warc/CC-MAIN-20210928122846-20210928152846-00076.warc.gz | 1,028,938,016 | 5,002 | # Femtowatts to Kilowatts Conversion
## You are currently converting Power units from Femtowatts to Kilowatts
1 Femtowatts (fW)
=
0 Kilowatts (kW)
Enter the number of Femtowatts(fW) to convert into Kilowatts(kW).
Femtowatts(fW) Value:
Results in Kilowatts(kW):
1 (fW) = 0 (kW)
Do you want to convert Kilowatts to Femtowatts?
How to Convert Femtowatts to Kilowatts
To convert Femtowatts to Kilowatts, multiply the Power by the conversion ratio. One Femtowatts is equal to 0 Kilowatts, so use this simple formula to convert:
Femtowatts = Kilowatts × 0
For example, here's how to convert 5 Femtowatts to Kilowatts using the formula above.
5 fW = (5 × 0) = 0 kW
1 Femtowatt is equal to how many Kilowatt?
1 Femtowatt is equal to 0 Kilowatts: 1 fW = 0 kW
There are 0 Kilowatts in 1 Femtowatt. To convert from Femtowatts to Kilowatts, multiply your figure by 0 (or divide by 1.0E+18) .
1 Kilowatt is equal to how many Femtowatt?
1 Kilowatt is equal to 1.0E+18 Femtowatts: 1 kW = 1.0E+18 fW
There are 1.0E+18 Femtowatts in 1 Kilowatt. To convert from Kilowatts to Femtowatts, multiply your figure by 1.0E+18 (or divide by 0) .
### Converting Femtowatts and Kilowatts
FemtowattsKilowattsKilowattsFemtowatts
1 fW0 kW1 kW1.0E+18 fW
2 fW0 kW2 kW2.0E+18 fW
3 fW0 kW3 kW3.0E+18 fW
4 fW0 kW4 kW4.0E+18 fW
5 fW0 kW5 kW5.0E+18 fW
6 fW0 kW6 kW6.0E+18 fW
7 fW0 kW7 kW7.0E+18 fW
8 fW0 kW8 kW8.0E+18 fW
9 fW0 kW9 kW9.0E+18 fW
10 fW0 kW10 kW1.0E+19 fW
11 fW0 kW11 kW1.1E+19 fW
12 fW0 kW12 kW1.2E+19 fW
13 fW0 kW13 kW1.3E+19 fW
14 fW0 kW14 kW1.4E+19 fW
15 fW0 kW15 kW1.5E+19 fW
16 fW0 kW16 kW1.6E+19 fW
17 fW0 kW17 kW1.7E+19 fW
18 fW0 kW18 kW1.8E+19 fW
19 fW0 kW19 kW1.9E+19 fW
20 fW0 kW20 kW2.0E+19 fW | 746 | 1,702 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2021-39 | latest | en | 0.659759 | # Femtowatts to Kilowatts Conversion ## You are currently converting Power units from Femtowatts to Kilowatts 1 Femtowatts (fW) = 0 Kilowatts (kW) Enter the number of Femtowatts(fW) to convert into Kilowatts(kW). Femtowatts(fW) Value: Results in Kilowatts(kW): 1 (fW) = 0 (kW) Do you want to convert Kilowatts to Femtowatts? How to Convert Femtowatts to Kilowatts To convert Femtowatts to Kilowatts, multiply the Power by the conversion ratio. One Femtowatts is equal to 0 Kilowatts, so use this simple formula to convert: Femtowatts = Kilowatts × 0 For example, here's how to convert 5 Femtowatts to Kilowatts using the formula above. 5 fW = (5 × 0) = 0 kW 1 Femtowatt is equal to how many Kilowatt? 1 Femtowatt is equal to 0 Kilowatts: 1 fW = 0 kW There are 0 Kilowatts in 1 Femtowatt. To convert from Femtowatts to Kilowatts, multiply your figure by 0 (or divide by 1.0E+18) . 1 Kilowatt is equal to how many Femtowatt? 1 Kilowatt is equal to 1.0E+18 Femtowatts: 1 kW = 1.0E+18 fW There are 1.0E+18 Femtowatts in 1 Kilowatt. To convert from Kilowatts to Femtowatts, multiply your figure by 1.0E+18 (or divide by 0) . ### Converting Femtowatts and Kilowatts FemtowattsKilowattsKilowattsFemtowatts 1 fW0 kW1 kW1.0E+18 fW 2 fW0 kW2 kW2.0E+18 fW 3 fW0 kW3 kW3.0E+18 fW 4 fW0 kW4 kW4.0E+18 fW 5 fW0 kW5 kW5.0E+18 fW 6 fW0 kW6 kW6.0E+18 fW 7 fW0 kW7 kW7.0E+18 fW 8 fW0 kW8 kW8.0E+18 fW 9 fW0 kW9 kW9.0E+18 fW 10 fW0 kW10 kW1.0E+19 fW 11 fW0 kW11 kW1.1E+19 fW 12 fW0 kW12 kW1.2E+19 fW 13 fW0 kW13 kW1.3E+19 fW 14 fW0 kW14 kW1.4E+19 | fW 15 fW0 kW15 kW1.5E+19 fW 16 fW0 kW16 kW1.6E+19 fW 17 fW0 kW17 kW1.7E+19 fW 18 fW0 kW18 kW1.8E+19 fW 19 fW0 kW19 kW1.9E+19 fW 20 fW0 kW20 kW2.0E+19 fW |
https://plainmath.org/algebra-i/104214-find-the-average-value-of-the | 1,720,814,304,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514452.62/warc/CC-MAIN-20240712192937-20240712222937-00232.warc.gz | 376,398,017 | 21,752 | Makayla English
2023-03-26
Find the average value of the function on the given interval?
anajusthings5mrf
Average performance of a function $=\frac{{\int }_{a}^{b}f\left(x\right)dx}{b-a}$
Here $f\left(x\right)=4x-{x}^{2}$ and $b=4$ , $a=0$
So,
${f\left(x\right)}_{avg}=\frac{{\int }_{0}^{4}\left(4x-{x}^{2}\right)dx}{4-0}$
${f\left(x\right)}_{avg}=\frac{{\left[\frac{4{x}^{2}}{2}-\frac{{x}^{3}}{3}\right]}_{0}^{4}}{4}$
${f\left(x\right)}_{avg}=\frac{{\left[\frac{4{x}^{2}}{2}-\frac{{x}^{3}}{3}\right]}_{0}^{4}}{4}$
${f\left(x\right)}_{avg}=\frac{\left(2×{4}^{2}-\frac{{4}^{3}}{3}\right)-\left(2×{0}^{2}-\frac{{0}^{3}}{3}\right)}{4}$
${f\left(x\right)}_{avg}=\frac{\frac{32}{3}}{4}$
${f\left(x\right)}_{avg}=\frac{8}{3}$
${f\left(x\right)}_{avg}=2.66$
Do you have a similar question? | 370 | 787 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 28, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-30 | latest | en | 0.343052 | Makayla English 2023-03-26 Find the average value of the function on the given interval? anajusthings5mrf Average performance of a function $=\frac{{\int }_{a}^{b}f\left(x\right)dx}{b-a}$ Here $f\left(x\right)=4x-{x}^{2}$ and $b=4$ , $a=0$ So, ${f\left(x\right)}_{avg}=\frac{{\int }_{0}^{4}\left(4x-{x}^{2}\right)dx}{4-0}$ ${f\left(x\right)}_{avg}=\frac{{\left[\frac{4{x}^{2}}{2}-\frac{{x}^{3}}{3}\right]}_{0}^{4}}{4}$ ${f\left(x\right)}_{avg}=\frac{{\left[\frac{4{x}^{2}}{2}-\frac{{x}^{3}}{3}\right]}_{0}^{4}}{4}$ ${f\left(x\right)}_{avg}=\frac{\left(2×{4}^{2}-\frac{{4}^{3}}{3}\right)-\left(2×{0}^{2}-\frac{{0}^{3}}{3}\right)}{4}$ ${f\left(x\right)}_{avg}=\frac{\frac{32}{3}}{4}$ ${f\left(x\right)}_{avg}=\frac{8}{3}$ ${f\left(x\right)}_{avg}=2.66$ Do | you have a similar question? |
http://www.ehow.com/how_5325828_estimate-cost-quartz-countertops.html | 1,480,740,921,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698540839.46/warc/CC-MAIN-20161202170900-00220-ip-10-31-129-80.ec2.internal.warc.gz | 442,826,512 | 17,548 | # How to Estimate the Cost of Quartz Countertops
Save
Durable quartz countertops offer a wider variety of colors than other stone alternatives, such as granite. The price for any material will vary according to availability, demand, price of labor and other economic factors, so estimate the cost of your new quartz countertop when you are ready to install it. If you wait too long, the cost may change.
### Things You'll Need
• Measuring tape
• Calculator
• Measure the length and width of the counter space in inches. If you are replacing an existing countertop, use its dimensions. If there is no existing countertop, measure the horizontal area that the new countertop will fill. Multiply the length by the width to find the total surface area in square inches. For example, if you are filling a space that is 24 inches by 84 inches, multiply 24 by 84 to get 2,016 square inches.
• Calculate any triangular surface area by multiplying the length of the triangle's base by its depth. The triangle's base is its longest side, and its depth is the distance from the base to its peak. After you have multiplied the base by the depth, divide that number by two to find the surface area in square inches. If you have a triangle that is 48 inches along the base and 24 inches from the base to the peak, multiply 48 by 24 to get 1,152, then divide by two for 576 square inches.
• Add the areas of all your individual rectangular and triangular surfaces to determine the total surface area in square inches of your new quartz countertops. Divide this number by 144 to calculate your total square footage. Adding the numbers above gives you a total of 2,592. Dividing this number by 144 yields a total of 18 square feet.
• Calculate your total estimated cost using a current price range. At the time of publication, the price of quartz ranged from \$50 to \$100 per square foot. Using the sample measurements above, you get an estimated total cost of \$900 to \$1,800.
## Tips & Warnings
• Several factors can influence your final costs, such as texture and thickness, edge styles and your contractor's price for labor. This is why prices range so widely. The use of a professional contractor is also almost always necessary, as quartz countertops are heavy and difficult to install without professional training and experience, so doing it yourself to avoid paying for labor likely is not an option.
• Use search engines to find online reviews and testimonials from previous customers, and consult your local Better Business Bureau office to determine the contractor's status or learn of any outstanding complaints against them. A more skilled, reliable contractor may be worth more to you than a less expensive worker with questionable business practices.
## References
• Photo Credit Hemera Technologies/AbleStock.com/Getty Images
Promoted By Zergnet
## Resources
### You May Also Like
• Granite vs. Quartz Countertops
Granite is a natural stone that is extracted from the earth, then cut into individual slabs for counters. Quartz is an engineered...
• Why Use Quartz Countertops?
Quartz is a material chosen for countertops mostly because of its strength, durability and low maintenance. Only a few other materials found...
• The Average Cost of Laminate Countertops
The cost of laminate countertops varies based on the installation method and the varying rates charged by different installers and contractors. In...
• Caesarstone vs. Silestone
Engineered stone uses a quartz mixture to replicate the features of granite or marble. While there are many different engineered stone companies,...
• Silestone vs. Zodiaq vs. Caesarstone vs. Granite
As of 2014, interior designer reported in the [National Kitchen and Bath Association annual survey](http://www.nkba.org/docs/default-source/market-research/2014-nkba-kitchen-and-bath-style-report.pdf?sfvrsn=2) that quartz -- Silestone, Zodiaq and Caesarstone...
• A Cost Comparison of Countertops
In the past, most kitchens and bathrooms sported laminate countertops. You might see the occasional ceramic tile or butcher block, but not...
• The Pricing for Corian vs. Granite
Corian and granite are two of the many types of countertops available to consumers. Brand name "Corian" countertops are more commonly referred...
• Quartz Vs. Corian Countertops
Quartz and Corian countertops are both attractive and affordable options. They do, however, have differences that you should know about before choosing.
• How to Estimate the Cost of a Corian Countertop
Corian is the trademarked brand name of a synthetic material manufactured by DuPont. As a nonporous, low-maintenance material, Corian is designed to...
• How to Estimate the Cost of a Silestone Countertop
As a naturally nonporous, sturdy quartz, Silestone doesn't scratch, stain or harbor bacteria. Its depth of color and easy maintenance make it...
• Granite Vs. Quartz
Granite and quartz countertops share so many traits that choosing between the two comes down to atomic differences. Granite is natural stone;...
• Are Quartz Countertops More Expensive Than Granite?
Quartz and granite countertops provide shine, durability and aesthetic appeal. Unless you have a preference for one of the materials, choosing one...
• Concrete Vs. Quartz Countertop
Quartz countertops are manufactured using approximately 95 percent stone and 5 percent filler or resin. This mixture results in a strong material...
## Related Searches
Check It Out
### 22 DIY Ways to Update Your Home on a Small Budget
M
Is DIY in your DNA? Become part of our maker community. | 1,175 | 5,590 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2016-50 | longest | en | 0.908178 | # How to Estimate the Cost of Quartz Countertops Save Durable quartz countertops offer a wider variety of colors than other stone alternatives, such as granite. The price for any material will vary according to availability, demand, price of labor and other economic factors, so estimate the cost of your new quartz countertop when you are ready to install it. If you wait too long, the cost may change. ### Things You'll Need • Measuring tape • Calculator • Measure the length and width of the counter space in inches. If you are replacing an existing countertop, use its dimensions. If there is no existing countertop, measure the horizontal area that the new countertop will fill. Multiply the length by the width to find the total surface area in square inches. For example, if you are filling a space that is 24 inches by 84 inches, multiply 24 by 84 to get 2,016 square inches. • Calculate any triangular surface area by multiplying the length of the triangle's base by its depth. The triangle's base is its longest side, and its depth is the distance from the base to its peak. After you have multiplied the base by the depth, divide that number by two to find the surface area in square inches. If you have a triangle that is 48 inches along the base and 24 inches from the base to the peak, multiply 48 by 24 to get 1,152, then divide by two for 576 square inches. • Add the areas of all your individual rectangular and triangular surfaces to determine the total surface area in square inches of your new quartz countertops. Divide this number by 144 to calculate your total square footage. Adding the numbers above gives you a total of 2,592. Dividing this number by 144 yields a total of 18 square feet. • Calculate your total estimated cost using a current price range. At the time of publication, the price of quartz ranged from \$50 to \$100 per square foot. Using the sample measurements above, you get an estimated total cost of \$900 to \$1,800. ## Tips & Warnings • Several factors can influence your final costs, such as texture and thickness, edge styles and your contractor's price for labor. This is why prices range so widely. The use of a professional contractor is also almost always necessary, as quartz countertops are heavy and difficult to install without professional training and experience, so doing it yourself to avoid paying for labor likely is not an option. • Use search engines to find online reviews and testimonials from previous customers, and consult your local Better Business Bureau office to determine the contractor's status or learn of any outstanding complaints against them. A more skilled, reliable contractor may be worth more to you than a less expensive worker with questionable business practices. ## References • Photo Credit Hemera Technologies/AbleStock.com/Getty Images Promoted By Zergnet ## Resources ### You May Also Like • Granite vs. Quartz Countertops Granite is a natural stone that is extracted from the earth, then cut into individual slabs for counters. Quartz is an engineered... • Why Use Quartz Countertops? Quartz is a material chosen for countertops mostly because of its strength, durability and low maintenance. Only a few other materials found... • The Average Cost of Laminate Countertops The cost of laminate countertops varies based on the installation method and the varying rates charged by different installers and contractors. In... • Caesarstone vs. Silestone Engineered stone uses a quartz mixture to replicate the features of granite or marble. While there are many different engineered stone companies,... • Silestone vs. Zodiaq vs. Caesarstone vs. Granite As of 2014, interior designer reported in the [National Kitchen and Bath Association annual survey](http://www.nkba.org/docs/default-source/market-research/2014-nkba-kitchen-and-bath-style-report.pdf?sfvrsn=2) that quartz -- Silestone, Zodiaq and Caesarstone... • A Cost Comparison of Countertops In the past, most kitchens and bathrooms sported laminate countertops. You might see the occasional ceramic tile or butcher block, but not... • The Pricing for Corian vs. Granite Corian and granite are two of the many types of countertops available to consumers. Brand name "Corian" countertops are more commonly referred... • Quartz Vs. Corian Countertops Quartz and Corian countertops are both attractive and affordable options. They do, however, have differences that you should know about before choosing. • How to Estimate the Cost of a Corian Countertop Corian is the trademarked brand name of a synthetic material manufactured by DuPont. As a nonporous, low-maintenance material, Corian is designed to... • How to Estimate the Cost of a Silestone Countertop As a naturally nonporous, sturdy quartz, Silestone doesn't scratch, stain or harbor bacteria. Its depth of color and easy maintenance make it... • Granite Vs. Quartz Granite and quartz countertops share so many traits that choosing between the two comes down to atomic differences. Granite | is natural stone;... • Are Quartz Countertops More Expensive Than Granite? Quartz and granite countertops provide shine, durability and aesthetic appeal. Unless you have a preference for one of the materials, choosing one... • Concrete Vs. Quartz Countertop Quartz countertops are manufactured using approximately 95 percent stone and 5 percent filler or resin. This mixture results in a strong material... ## Related Searches Check It Out ### 22 DIY Ways to Update Your Home on a Small Budget M Is DIY in your DNA? Become part of our maker community. |
https://yutsumura.com/tag/distance/ | 1,606,443,962,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141189038.24/warc/CC-MAIN-20201127015426-20201127045426-00666.warc.gz | 939,354,317 | 20,568 | # Tagged: distance
## Problem 689
For this problem, use the complex vectors
$\mathbf{w}_1 = \begin{bmatrix} 1 + i \\ 1 – i \\ 0 \end{bmatrix} , \, \mathbf{w}_2 = \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} , \, \mathbf{w}_3 = \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} .$
Suppose $\mathbf{w}_4$ is another complex vector which is orthogonal to both $\mathbf{w}_2$ and $\mathbf{w}_3$, and satisfies $\mathbf{w}_1 \cdot \mathbf{w}_4 = 2i$ and $\| \mathbf{w}_4 \| = 3$.
Calculate the following expressions:
(a) $\mathbf{w}_1 \cdot \mathbf{w}_2$.
(b) $\mathbf{w}_1 \cdot \mathbf{w}_3$.
(c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$.
(d) $\| \mathbf{w}_1 \| , \| \mathbf{w}_2 \|$, and $\| \mathbf{w}_3 \|$.
(e) $\| 3 \mathbf{w}_4 \|$.
(f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$?
## Problem 687
For this problem, use the real vectors
$\mathbf{v}_1 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} , \mathbf{v}_2 = \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} , \mathbf{v}_3 = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} .$ Suppose that $\mathbf{v}_4$ is another vector which is orthogonal to $\mathbf{v}_1$ and $\mathbf{v}_3$, and satisfying
$\mathbf{v}_2 \cdot \mathbf{v}_4 = -3 .$
Calculate the following expressions:
(a) $\mathbf{v}_1 \cdot \mathbf{v}_2$.
(b) $\mathbf{v}_3 \cdot \mathbf{v}_4$.
(c) $( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4$.
(d) $\| \mathbf{v}_1 \| , \, \| \mathbf{v}_2 \| , \, \| \mathbf{v}_3 \|$.
(e) What is the distance between $\mathbf{v}_1$ and $\mathbf{v}_2$?
## Problem 254
Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are
$\|\mathbf{a}\|=\|\mathbf{b}\|=1$ and the inner product
$\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$
Then determine the length $\|\mathbf{a}-\mathbf{b}\|$.
(Note that this length is the distance between $\mathbf{a}$ and $\mathbf{b}$.) | 801 | 1,937 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2020-50 | longest | en | 0.406505 | # Tagged: distance ## Problem 689 For this problem, use the complex vectors $\mathbf{w}_1 = \begin{bmatrix} 1 + i \\ 1 – i \\ 0 \end{bmatrix} , \, \mathbf{w}_2 = \begin{bmatrix} -i \\ 0 \\ 2 – i \end{bmatrix} , \, \mathbf{w}_3 = \begin{bmatrix} 2+i \\ 1 – 3i \\ 2i \end{bmatrix} .$ Suppose $\mathbf{w}_4$ is another complex vector which is orthogonal to both $\mathbf{w}_2$ and $\mathbf{w}_3$, and satisfies $\mathbf{w}_1 \cdot \mathbf{w}_4 = 2i$ and $\| \mathbf{w}_4 \| = 3$. Calculate the following expressions: (a) $\mathbf{w}_1 \cdot \mathbf{w}_2$. (b) $\mathbf{w}_1 \cdot \mathbf{w}_3$. (c) $((2+i)\mathbf{w}_1 – (1+i)\mathbf{w}_2 ) \cdot \mathbf{w}_4$. (d) $\| \mathbf{w}_1 \| , \| \mathbf{w}_2 \|$, and $\| \mathbf{w}_3 \|$. (e) $\| 3 \mathbf{w}_4 \|$. (f) What is the distance between $\mathbf{w}_2$ and $\mathbf{w}_3$? ## Problem 687 For this problem, use the real vectors $\mathbf{v}_1 = \begin{bmatrix} -1 \\ 0 \\ 2 \end{bmatrix} , \mathbf{v}_2 = \begin{bmatrix} 0 \\ 2 \\ -3 \end{bmatrix} , \mathbf{v}_3 = \begin{bmatrix} 2 \\ 2 \\ 3 \end{bmatrix} .$ Suppose that $\mathbf{v}_4$ is another vector which is orthogonal to $\mathbf{v}_1$ and $\mathbf{v}_3$, and satisfying $\mathbf{v}_2 \cdot \mathbf{v}_4 = -3 .$ Calculate the following expressions: (a) $\mathbf{v}_1 \cdot \mathbf{v}_2$. (b) $\mathbf{v}_3 \cdot \mathbf{v}_4$. (c) $( 2 \mathbf{v}_1 + 3 \mathbf{v}_2 – \mathbf{v}_3 ) \cdot \mathbf{v}_4$. (d) $\| \mathbf{v}_1 \| , \, \| \mathbf{v}_2 \| , \, \| \mathbf{v}_3 \|$. (e) What is the distance between $\mathbf{v}_1$ and $\mathbf{v}_2$? ## Problem 254 Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ | such that their length are $\|\mathbf{a}\|=\|\mathbf{b}\|=1$ and the inner product $\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.$ Then determine the length $\|\mathbf{a}-\mathbf{b}\|$. (Note that this length is the distance between $\mathbf{a}$ and $\mathbf{b}$.) |
https://convertoctopus.com/28-3-days-to-months | 1,643,017,642,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304528.78/warc/CC-MAIN-20220124094120-20220124124120-00096.warc.gz | 239,059,399 | 8,183 | ## Conversion formula
The conversion factor from days to months is 0.032854884083862, which means that 1 day is equal to 0.032854884083862 months:
1 d = 0.032854884083862 mo
To convert 28.3 days into months we have to multiply 28.3 by the conversion factor in order to get the time amount from days to months. We can also form a simple proportion to calculate the result:
1 d → 0.032854884083862 mo
28.3 d → T(mo)
Solve the above proportion to obtain the time T in months:
T(mo) = 28.3 d × 0.032854884083862 mo
T(mo) = 0.9297932195733 mo
The final result is:
28.3 d → 0.9297932195733 mo
We conclude that 28.3 days is equivalent to 0.9297932195733 months:
28.3 days = 0.9297932195733 months
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 month is equal to 1.07550795053 × 28.3 days.
Another way is saying that 28.3 days is equal to 1 ÷ 1.07550795053 months.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-eight point three days is approximately zero point nine three months:
28.3 d ≅ 0.93 mo
An alternative is also that one month is approximately one point zero seven six times twenty-eight point three days.
## Conversion table
### days to months chart
For quick reference purposes, below is the conversion table you can use to convert from days to months
days (d) months (mo)
29.3 days 0.963 months
30.3 days 0.996 months
31.3 days 1.028 months
32.3 days 1.061 months
33.3 days 1.094 months
34.3 days 1.127 months
35.3 days 1.16 months
36.3 days 1.193 months
37.3 days 1.225 months
38.3 days 1.258 months | 500 | 1,685 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2022-05 | latest | en | 0.820217 | ## Conversion formula The conversion factor from days to months is 0.032854884083862, which means that 1 day is equal to 0.032854884083862 months: 1 d = 0.032854884083862 mo To convert 28.3 days into months we have to multiply 28.3 by the conversion factor in order to get the time amount from days to months. We can also form a simple proportion to calculate the result: 1 d → 0.032854884083862 mo 28.3 d → T(mo) Solve the above proportion to obtain the time T in months: T(mo) = 28.3 d × 0.032854884083862 mo T(mo) = 0.9297932195733 mo The final result is: 28.3 d → 0.9297932195733 mo We conclude that 28.3 days is equivalent to 0.9297932195733 months: 28.3 days = 0.9297932195733 months ## Alternative conversion We can also convert by utilizing the inverse value of the conversion factor. In this case 1 month is equal to 1.07550795053 × 28.3 days. Another way is saying that 28.3 days is equal to 1 ÷ 1.07550795053 months. ## Approximate result For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-eight point three days is approximately zero point nine three months: 28.3 d ≅ 0.93 mo An alternative is also that one month is approximately one point zero seven six times twenty-eight point three days. ## Conversion table ### days to months chart For quick reference purposes, below is the conversion table you can use to convert from days to months days (d) months (mo) 29.3 days 0.963 months 30.3 days 0.996 months 31.3 days 1.028 | months 32.3 days 1.061 months 33.3 days 1.094 months 34.3 days 1.127 months 35.3 days 1.16 months 36.3 days 1.193 months 37.3 days 1.225 months 38.3 days 1.258 months |
https://es.scribd.com/doc/177153413/Algebraic-Expressions-Level | 1,560,862,605,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998724.57/warc/CC-MAIN-20190618123355-20190618145355-00278.warc.gz | 435,137,125 | 56,948 | Está en la página 1de 18
# SUBJECT : COURSE : TITLE : LEVEL : SET : START :
## MATHEMATICS FORM 1 TOPIC 7 LEVEL 1.0 31/01/2012 10:53:32 AM
1327978412 31/01/2012 10:53
1.
TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan.
Which of the following is not an unknown? Manakah di antara berikut bukan anu?
14338 14338 0 0
A. x. B. y. C. z. D. 3. 2. Which of the following is not an algebraic term with an an unknown? Manakah di antara berikut bukan sebutan algebra dengan satu anu ?
14339 0 0 14339
A. B. C.
D. 3. Which of the following is not an algebraic term ? Manakah di antara berikut bukan sebutan algebra?
0 0 14340 14340
A. B. C.
D.
4. Which of the following pairs are like terms? Manakah di antara pasangan berikut sebutan serupa?
14341 14341 0 0
A. B. C. D. 5. Find the difference between 5g and -2g. Cari perbezaan di antara 5g dan -2g.
0
14342
14342
## A. 3g. B. 7g. C. -3g. D. -7g. 6. x + y -(-2 y) = A. x + y. B. x - y.
14343 14343 0 0
C. x + 3y. D. x - 3y. 7. 81 m + 9 m = A. 9 m. B. 90 m. C. 90 + m. D. 2 + 9 m. 8. -7 x - 5 x = A. 12x. B. 2x. C. -12x. D. -2x. 9. 12p - ( - 8p) = A. 20p. B. 4p. C. 20 + p. D. - 2x. 10. Which of the following has 2 as its coefficient? Manakah di antara berikut mempunyai 2 sebagai pekalinya?
0 0 14347 14347 14346 14346 0 0 14345 14345 0 0 14344 14344 0 0
A. B. C. D.
## Sebutan serupa dengan -15p adalah ...
14348 14348 0 0
A. B. C.
D.
12. Which of the following is an expression? Manakah di antara berikut suatu kenyataan?
0
14349
14349
A. ab. B. 3c. C. 4ab. D. 4+a+3c. 13. 3 - 5x + 2x How many terms are there in the above expression? Berapa sebutankah yang terdapat di dalam kenyataan di atas?
14350 14350 0 0
A. 1. B. 2. C. 3. D. 4. 14. The coefficient of g in the expression -2g + 5g is... Pekali bagi g dalam kenyataan -2g + 5g adalah ...
14351 14351 0 0
14352 14352 0 0
14353 14353 0 0
## C. 8pq. D. 8 + p + q. 17. Simplify 3 - 2a + 9a - 1 = Permudahkan 3 - 2a + 9a - 1=
14354 14354 0 0
A. 7a + 4. B. 7a + 2. C. -7a + 2. D. -7a - 2. 18. 4r + 6 - 2r = A. 4r + 8. B. 10r - 2. C. 8r. D. 2r + 6. 19. Which of the following expression has the most number of terms? Manakah di antara kenyataan berikut mempunyai paling banyak sebutan?
14356 14355 14355 0 0
14356
A. 3m - 4. B. 5p + 6 - 8p. C. 7s - 7t + s. D. 2x - y + 3x + 5y. 20. Which of the following is the like term of 5d? Manakah di antara berikut sebutan bagi 5d?
14357
14357
A. I only / I sahaja. B. II only / II sahaja. C. I and II only / I dan II sahaja. D. II and III only / II dan III sahaja.
21
Submit
1. TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan.
P bought x ball pens and y note books for RMz. Which of the following is not an unknown? P telah membeli x pena mata ball dan y buku nota berharga RMz. Manakah di antara berikut bukan suatu anu?
14358 14358 0 0
A. P.
B. X. C. Y. D. Z. 2. An algebraic term with unknown g has a coefficient of 9. The algebraic term is... Suatu sebutan algebra dengan anu g mempunyai pekali 9. Sebutan algebra tersebut adalah ...
14359
14359
A.
B.
C.
D.
14360
## Pasangan manakah di antara berikut mengandungi sebutan serupa?
14360 0 0
A.
B.
C. 4. 5m - 8 - 2 ( m - 3 ) =
0 0 14370 14370
D.
A. 3m - 14.
B. 3m - 11. C. 3m - 5. D. 3m - 2. 5.
14371
14371
A.
B.
C.
D.
## 6. Which of the following is true?
14372 14372 0 0
Manakah di antara berikut benar? A. 2a + 3a = 6a. B. 4b + ( -2b ) = -2b. C. ( -8c ) - ( -5c ) = -3c. D. ( -5d ) + ( -4d ) = 9d. 7. 8 - f - ( f - 8 ) = B. 16 - 2f. C. 2f.
14373
14373
A. 16 + 2f.
D. -2f. 8. Which of the following algebraic terms are like terms for 9w ? Manakah di antara sebutan algebra berikut sebutan serupa bagi 9w?
14374 0 0
14374
A. C. D.
B.
## 9. Simplify 2y + 8y - ( -3y + 5 ). Permudahkan 2y + 8y - ( -3y + 5 ).
14375 14375 0 0
A. 3y - 5.
B. 10y + 5. C. 13y - 5. D. 13y + 5. 10. -12e - 8 + 6e + ( - 3f ) = A. 18e - 8 - 3f. B. -18e - 8 + 3f. C. -6e - 8 - 3f. D. -6e - 8 + 3f.
14376 14376 0 0
11. All of the following algebraic expressions can be simplified, except... Semua kenyataan algebra berikut boleh dipermudahkan, kecuali...
14377 14377 0 0
A. 2 + 5y - 5.
B. 4x - 10x + 9. C. 3x - y + 4z. D. 5p + 1 -2p - 3. 12. Find the perimeter of an equilateral triangle of side 3x cm. Cari perimeter suatu segitiga sama sisi dengan 3x cm.
14378 14378 0 0
A. 3x cm.
B. 6x cm. C. 9x cm. D. 12x cm. 13. All of the following expressions have the same number of algebraic terms except... Semua kenyataan berikut mempunyai bilangan sebutan algebra yang sama kecuali ...
14407
14407
B.
C.
14379 14379 0
D.
0
A.
B.
C.
D.
14380 14380 0
0
## D. -4t. 16. 3e subtracted from 9e is...
3e ditolak daripada 9e adalah... A. 3e - 9e. B. 9e + 3e. C. 9e - 3e. D. -9e - 3e. 17. 3y + (-2) - y + 6 = 12. B. 3y - 6. C. 2y + 4. D. 2y - 4. 18. A. B. Simplify -4 ( -3 + m ) + 7m =
14382
14381
14381
14382
A. 6y +
14383
14383
C.
D.
19.
## Permudahkan -4 ( -3 + m ) + 7m = A. 12 + 8m. B. 12 + 3m. C. -12 - 3m.
14384
14384
D. -12 + 8m. 20. The result, when a number x is multiplied by 2 and then 5 added to it, is... Hasil, apabila suatu nombor x di darab dengan 2 dan kemudian 5 ditambah kepadanya, adalah ...
14385 14385 0 0
A. 7.
B. x + 10. C. 2x + 5. D. 2x + 10.
## MATHEMATICS FORM 1 TOPIC 7 LEVEL 3.0 31/01/2012 10:55:50 AM
1327978550 31/01/2012 10:55
1.
TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan.
-8p - 5p - (-7) =
14386 14386 0 0
A. B. C. D.
## 2. Simplify 4s + 3 - ( -5s + 8 ). Permudahkan 4s + 3 - ( -5s + 8 ).
0 14387 14387 0
A. B.
9s + 11. 9s - 5.
C. -s + 11.
D.
-s - 5.
3. Find the perimeter of a square of side ( x + 4 )cm. Cari perimeter suatu segiempat sama bersisi ( x + 4 )cm.
14388 14388 0 0
A. B. C. D.
## 2x + 8. 3x + 12. 4x + 16. 5x + 20.
4. Calculate the area of the triangle below, in cm. Kirakan luas segitiga di bawah, dalam cm.
14389
14389
A. B. C. D.
5.
14390
14390
A. B. C. D.
6. Ahmad has 8x marbles. Bahrin has ( x + 2 ) marbles more than Ahmad. How many marbles does Bahrin have? Ahmad ada 8x biji guli. Bahrin ada ( x + 2 ) biji guli lebih banyak daripada Ahmad. Berapa banyak gulikah yang Bahrin ada?
14391 14391 0 0
A. B. C. D.
9x + 2. 8x + 2. 7x + 2. 7x - 2.
7. Siti bought a pen for RMv and then sold it at RM4w. What profit did she make? Siti telah membeli sebatang pen dengan harga RMv dan kemudian menjualnya dengan harga RM4w. Berapakah keuntungan yang dia dapat ?
14392 14392 0 0
A.
RM ( v - 4w ).
B. C. D.
RM ( v + 4w ). RM ( 4w - v ). RM ( 4w + v ).
8. A rectangular field measures ( 2x - 1)m by ( 3x + 2 )m. The perimeter of the field in meter, is... Sebuah padang segiempat berukuran ( 2x - 1)m dengan ( 3x + 2 )m. Perimeter padang itu, dalam meter, adalah ...
14393
14393
A. B. C. D.
5x + 1. 8x + 3. 10x + 2. 10x + 6.
9.
14394
14394
A. B.
C.
D.
10. The diagram below shows a triangle. Calculate the perimeter of the triangle.
Gambar rajah di bawah menunjukkan suatu segitiga. Kirakan perimeter segitiga tersebut.
14395
14395
A. B. C. D.
14396 14396 0 0
11.
## 2 [ k - ( 2k + 3 ) ] = A. B. C. D. -2k + 6. -2k - 6. -2k + 3. -2k - 3.
12.
6p - 3 - q - 5p - 3q + 8 = A. B. C. D. p - 4q + 11. p - 4q + 5. p - 3q + 11. p - 3q + 5.
14397
14397
13. Aini weights (4x - 2 )kg while her sister weights ( 3x + 6)kg. What is their total weight?
Berat Aini (4x - 2 )kg manakala berat adik perempuannya ialah ( 3x + 6)kg. Berapakah jumlah berat mereka? A. ( 5x + 3)kg. B. C. D. ( 6x + 9 )kg. ( 7x + 4 )kg. ( 12x - 8 )kg.
14398 14398 0 0
14. Rina has 2v sweets. Siti has w sweets less than Rina. While Tina has 8 sweets. How many sweets do they have altogether? Rina ada 2v biji gula-gula. Siti ada w biji gula-gula kurang daripada Rina. Manakala Tina ada 6 biji gula-gula. Berapa biji gula-gulakah yang mereka ada kesemuanya?
14399 14399 0 0
A. B. C. D.
2v - w + 8. 2v + 8. 4v - w + 8. 8 - 4v - w.
15. A watermelon costs RM2u and an apple costs 60sen. Find the total cost of the fruits in sen. Sebiji tembikai berharga RM2u dan sebiji epal berharga 60sen. Cari jumlah harga buahbuahan tersebut, dalam sen. A. 2u + 60. B. C. D. 2u - 60. 20u + 60. 200u + 60.
14400 14400 0 0
16. A box contains 5h apples. 10 of the apples have been eated. How many apples are there now? Sebuah kotak mengandungi 5h biji epal, 10 biji epal telah dimakan. Berapakah baki epal itu
sekarang? A. 5h - 10. B. C. D.
14401
14401
## 5h + 10. 10 - 5h. 10 + 5h.
14402 14402 0 0
17.
2( 3p - q ) + 3( p + q ) = A. B. C. D. 9p. 9p - q. 9p + q. 9p + 3q.
18. There are 55k students in a school. 39 + 12k of them are girls. How many boys are there in the school? Di sebuah sekolah terdapat 55k orang pelajar. 39 + 12k daripada mereka adalah perempuan. Berapa orang lelakikah dalam sekolah tersebut?
0 0
14403 14403
A. B. C. D.
## 11 + 12k. 43k + 11. 43k + 39. 43k - 39.
19. The table below shows the number of pens in a box. If the total number of pens in the box is U, express U in terms of w. Jadual dibawah menunjukkan bilangan pen dalam sebuah kotak. Jika jumlah bilangan pen di dalam kotak itu adalah U, nyatakan U dalam sebutan w.
14404
14404
A. B.
C.
D.
20.
14405
14405
A. B. C. D.
3w - 2 2w - 7 5w + 2 3w - 10 | 4,267 | 9,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-26 | latest | en | 0.264883 | Está en la página 1de 18 # SUBJECT : COURSE : TITLE : LEVEL : SET : START : ## MATHEMATICS FORM 1 TOPIC 7 LEVEL 1.0 31/01/2012 10:53:32 AM 1327978412 31/01/2012 10:53 1. TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan. Which of the following is not an unknown? Manakah di antara berikut bukan anu? 14338 14338 0 0 A. x. B. y. C. z. D. 3. 2. Which of the following is not an algebraic term with an an unknown? Manakah di antara berikut bukan sebutan algebra dengan satu anu ? 14339 0 0 14339 A. B. C. D. 3. Which of the following is not an algebraic term ? Manakah di antara berikut bukan sebutan algebra? 0 0 14340 14340 A. B. C. D. 4. Which of the following pairs are like terms? Manakah di antara pasangan berikut sebutan serupa? 14341 14341 0 0 A. B. C. D. 5. Find the difference between 5g and -2g. Cari perbezaan di antara 5g dan -2g. 0 14342 14342 ## A. 3g. B. 7g. C. -3g. D. -7g. 6. x + y -(-2 y) = A. x + y. B. x - y. 14343 14343 0 0 C. x + 3y. D. x - 3y. 7. 81 m + 9 m = A. 9 m. B. 90 m. C. 90 + m. D. 2 + 9 m. 8. -7 x - 5 x = A. 12x. B. 2x. C. -12x. D. -2x. 9. 12p - ( - 8p) = A. 20p. B. 4p. C. 20 + p. D. - 2x. 10. Which of the following has 2 as its coefficient? Manakah di antara berikut mempunyai 2 sebagai pekalinya? 0 0 14347 14347 14346 14346 0 0 14345 14345 0 0 14344 14344 0 0 A. B. C. D. ## Sebutan serupa dengan -15p adalah ... 14348 14348 0 0 A. B. C. D. 12. Which of the following is an expression? Manakah di antara berikut suatu kenyataan? 0 14349 14349 A. ab. B. 3c. C. 4ab. D. 4+a+3c. 13. 3 - 5x + 2x How many terms are there in the above expression? Berapa sebutankah yang terdapat di dalam kenyataan di atas? 14350 14350 0 0 A. 1. B. 2. C. 3. D. 4. 14. The coefficient of g in the expression -2g + 5g is... Pekali bagi g dalam kenyataan -2g + 5g adalah ... 14351 14351 0 0 14352 14352 0 0 14353 14353 0 0 ## C. 8pq. D. 8 + p + q. 17. Simplify 3 - 2a + 9a - 1 = Permudahkan 3 - 2a + 9a - 1= 14354 14354 0 0 A. 7a + 4. B. 7a + 2. C. -7a + 2. D. -7a - 2. 18. 4r + 6 - 2r = A. 4r + 8. B. 10r - 2. C. 8r. D. 2r + 6. 19. Which of the following expression has the most number of terms? Manakah di antara kenyataan berikut mempunyai paling banyak sebutan? 14356 14355 14355 0 0 14356 A. 3m - 4. B. 5p + 6 - 8p. C. 7s - 7t + s. D. 2x - y + 3x + 5y. 20. Which of the following is the like term of 5d? Manakah di antara berikut sebutan bagi 5d? 14357 14357 A. I only / I sahaja. B. II only / II sahaja. C. I and II only / I dan II sahaja. D. II and III only / II dan III sahaja. 21 Submit 1. TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan. P bought x ball pens and y note books for RMz. Which of the following is not an unknown? P telah membeli x pena mata ball dan y buku nota berharga RMz. Manakah di antara berikut bukan suatu anu? 14358 14358 0 0 A. P. B. X. C. Y. D. Z. 2. An algebraic term with unknown g has a coefficient of 9. The algebraic term is... Suatu sebutan algebra dengan anu g mempunyai pekali 9. Sebutan algebra tersebut adalah ... 14359 14359 A. B. C. D. 14360 ## Pasangan manakah di antara berikut mengandungi sebutan serupa? 14360 0 0 A. B. C. 4. 5m - 8 - 2 ( m - 3 ) = 0 0 14370 14370 D. A. 3m - 14. B. 3m - 11. C. 3m - 5. D. 3m - 2. 5. 14371 14371 A. B. C. D. ## 6. Which of the following is true? 14372 14372 0 0 Manakah di antara berikut benar? A. 2a + 3a = 6a. B. 4b + ( -2b ) = -2b. C. ( -8c ) - ( -5c ) = -3c. D. ( -5d ) + ( -4d ) = 9d. 7. 8 - f - ( f - 8 ) = B. 16 - 2f. C. 2f. 14373 14373 A. 16 + 2f. D. -2f. 8. Which of the following algebraic terms are like terms for 9w ? Manakah di antara sebutan algebra berikut sebutan serupa bagi 9w? 14374 0 0 14374 A. C. D. B. ## 9. Simplify 2y + 8y - ( -3y + 5 ). Permudahkan 2y + 8y - ( -3y + 5 ). 14375 14375 0 0 A. 3y - 5. B. 10y + 5. C. 13y - 5. D. 13y + 5. 10. -12e - 8 + 6e + ( - 3f ) = A. 18e - 8 - 3f. B. -18e - 8 + 3f. C. -6e - 8 - 3f. D. -6e - 8 + 3f. 14376 14376 0 0 11. All of the following algebraic expressions can be simplified, except... Semua kenyataan algebra berikut boleh dipermudahkan, kecuali... 14377 14377 0 0 A. 2 + 5y - 5. B. 4x - 10x + 9. C. 3x - y + 4z. D. 5p + 1 -2p - 3. 12. Find the perimeter of an equilateral triangle of side 3x cm. Cari perimeter suatu segitiga sama sisi dengan 3x cm. 14378 14378 0 0 A. 3x cm. B. 6x cm. C. 9x cm. D. 12x cm. 13. All of the following expressions have the same number of algebraic terms except... Semua kenyataan berikut mempunyai bilangan sebutan algebra yang sama kecuali ... 14407 14407 B. C. 14379 14379 0 D. 0 A. B. C. D. 14380 14380 0 0 ## D. -4t. 16. 3e subtracted from 9e is... 3e ditolak daripada 9e adalah... A. 3e - 9e. B. 9e + 3e. C. 9e - 3e. D. -9e - 3e. 17. 3y + (-2) - y + 6 = 12. B. 3y - 6. C. 2y + 4. D. 2y - 4. 18. A. B. Simplify -4 ( -3 + m ) + 7m = 14382 14381 14381 14382 A. 6y + 14383 14383 C. D. 19. ## Permudahkan -4 ( -3 + m ) + 7m = A. 12 + 8m. B. 12 + 3m. C. -12 - 3m. 14384 14384 D. -12 + 8m. 20. The result, when a number x is multiplied by 2 and then 5 added to it, is... Hasil, apabila suatu nombor x di darab dengan 2 dan kemudian 5 ditambah kepadanya, adalah ... 14385 14385 0 0 A. 7. B. x + 10. C. 2x + 5. D. 2x + 10. ## MATHEMATICS FORM 1 TOPIC 7 LEVEL 3.0 31/01/2012 10:55:50 AM 1327978550 31/01/2012 10:55 1. TOPIC : ALGEBRAIC EXPRESSIONS Instruction: This test consists of 20 objective questions marked A,B,C and D. Choose only ONE correct answer. Answer all questions. Arahan: Ujian ini mengandungi 20 soalan berbentuk objektif bertanda A,B,C dan D. Pilih hanya SATU jawapan yang betul. Jawab semua soalan. -8p - 5p - (-7) = 14386 14386 0 0 A. B. C. D. ## 2. Simplify 4s + 3 - ( -5s + 8 ). Permudahkan 4s + 3 - ( -5s + 8 ). 0 14387 14387 0 A. B. 9s + 11. 9s - 5. C. -s + 11. D. -s - 5. 3. Find the perimeter of a square of side ( x + 4 )cm. Cari perimeter suatu segiempat sama bersisi ( x + 4 )cm. 14388 14388 0 0 A. B. C. D. ## 2x + 8. 3x + 12. 4x + 16. 5x + 20. 4. Calculate the area of the triangle below, in cm. Kirakan luas segitiga di bawah, dalam cm. 14389 14389 A. B. C. D. 5. 14390 14390 A. B. C. D. 6. Ahmad has 8x marbles. Bahrin has ( x + 2 ) marbles more than Ahmad. How many marbles does Bahrin have? Ahmad ada 8x biji guli. Bahrin ada ( x + 2 ) biji guli lebih banyak daripada Ahmad. Berapa banyak gulikah yang Bahrin ada? 14391 14391 0 0 A. B. C. D. 9x + 2. 8x + 2. 7x + 2. 7x - 2. 7. Siti bought a pen for RMv and then sold it at RM4w. What profit did she make? Siti telah membeli sebatang pen dengan harga RMv dan kemudian menjualnya dengan harga RM4w. Berapakah keuntungan yang dia dapat ? 14392 14392 0 0 A. RM ( v - 4w ). B. C. D. RM ( v + 4w ). RM ( 4w - v ). RM ( 4w + v ). 8. A rectangular field measures ( 2x - 1)m by ( 3x + 2 )m. The perimeter of the field in meter, is... Sebuah padang segiempat berukuran ( 2x - 1)m dengan ( 3x + 2 )m. Perimeter padang itu, dalam meter, adalah ... 14393 14393 A. B. C. D. 5x + 1. 8x + 3. 10x + 2. 10x + 6. 9. 14394 14394 A. B. C. D. 10. The diagram below shows a triangle. Calculate the perimeter of the triangle. Gambar rajah di bawah menunjukkan suatu segitiga. Kirakan perimeter segitiga tersebut. 14395 14395 A. B. C. D. 14396 14396 0 0 11. ## 2 [ k - ( 2k + 3 ) ] = A. B. C. D. -2k + 6. -2k - 6. -2k + 3. -2k - 3. 12. 6p - 3 - q - 5p - 3q + 8 = A. B. C. D. p - 4q + 11. p - 4q + 5. p - 3q + 11. p - 3q + 5. 14397 14397 13. Aini weights (4x - 2 )kg while her sister weights ( 3x + 6)kg. What is their total weight? Berat Aini (4x - 2 )kg manakala berat adik perempuannya ialah ( 3x + 6)kg. Berapakah jumlah berat mereka? A. ( 5x + 3)kg. B. C. D. ( 6x + 9 )kg. ( 7x + 4 )kg. ( 12x - 8 )kg. 14398 14398 0 0 14. Rina has 2v sweets. Siti has w sweets less than Rina. While Tina has 8 sweets. How many sweets do they have altogether? Rina ada 2v biji gula-gula. Siti ada w biji gula-gula kurang daripada Rina. Manakala Tina ada 6 biji gula-gula. Berapa biji gula-gulakah yang mereka ada kesemuanya? 14399 14399 0 0 A. B. C. D. 2v - w + 8. 2v + 8. 4v - w + 8. 8 - 4v - w. 15. A watermelon costs RM2u and an apple costs 60sen. Find the total cost of the fruits in sen. Sebiji tembikai berharga RM2u dan sebiji epal berharga 60sen. Cari jumlah harga buahbuahan tersebut, dalam sen. A. 2u + 60. B. C. D. 2u - 60. 20u + 60. 200u + 60. 14400 14400 0 0 16. A box contains 5h | apples. 10 of the apples have been eated. How many apples are there now? Sebuah kotak mengandungi 5h biji epal, 10 biji epal telah dimakan. Berapakah baki epal itu sekarang? A. 5h - 10. B. C. D. 14401 14401 ## 5h + 10. 10 - 5h. 10 + 5h. 14402 14402 0 0 17. 2( 3p - q ) + 3( p + q ) = A. B. C. D. 9p. 9p - q. 9p + q. 9p + 3q. 18. There are 55k students in a school. 39 + 12k of them are girls. How many boys are there in the school? Di sebuah sekolah terdapat 55k orang pelajar. 39 + 12k daripada mereka adalah perempuan. Berapa orang lelakikah dalam sekolah tersebut? 0 0 14403 14403 A. B. C. D. ## 11 + 12k. 43k + 11. 43k + 39. 43k - 39. 19. The table below shows the number of pens in a box. If the total number of pens in the box is U, express U in terms of w. Jadual dibawah menunjukkan bilangan pen dalam sebuah kotak. Jika jumlah bilangan pen di dalam kotak itu adalah U, nyatakan U dalam sebutan w. 14404 14404 A. B. C. D. 20. 14405 14405 A. B. C. D. 3w - 2 2w - 7 5w + 2 3w - 10 |
http://oeis.org/A001043 | 1,603,552,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00423.warc.gz | 75,730,637 | 5,104 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A001043 Numbers that are the sum of 2 successive primes. (Formerly M3780 N0968) 146
5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, 78, 84, 90, 100, 112, 120, 128, 138, 144, 152, 162, 172, 186, 198, 204, 210, 216, 222, 240, 258, 268, 276, 288, 300, 308, 320, 330, 340, 352, 360, 372, 384, 390, 396, 410, 434, 450, 456, 462, 472, 480, 492, 508, 520 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Arithmetic derivative (see A003415) of prime(n)*prime(n+1). - Giorgio Balzarotti, May 26 2011 A008472(a(n)) = A191583(n). - Reinhard Zumkeller, Jun 28 2011 With the exception of the first term, all terms are even. a(n) is divisible by 4 if the difference between prime(n) and prime(n + 1) is not divisible by 4; e.g., prime(n) = 1 mod 4 and prime(n + 1) = 3 mod 4. In general, for a(n) to be divisible by some even number m > 2 requires that prime(n + 1) - prime(n) not be a multiple of m. - Alonso del Arte, Jan 30 2012 REFERENCES Archimedeans Problems Drive, Eureka, 26 (1963), 12. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe) Albert Frank & Philippe Jacqueroux, International Contest, 2001. Item 22 Richard K. Guy, Letters to N. J. A. Sloane, June-August 1968 FORMULA a(n) = prime(n) + prime(n + 1). a(n) = A000040(n) + A000040(n+1). a(n) = A116366(n, n - 1) for n > 1. - Reinhard Zumkeller, Feb 06 2006 EXAMPLE 2 + 3 = 5. 3 + 5 = 8. 5 + 7 = 12. 7 + 11 = 18. MAPLE Primes:= select(isprime, [2, seq(2*i+1, i=1..1000)]): n:= nops(Primes): Primes[1..n-1] + Primes[2..n]; # Robert Israel, Aug 29 2014 MATHEMATICA Table[Prime[n] + Prime[n + 1], {n, 55}] (* Ray Chandler, Feb 12 2005 *) Total/@Partition[Prime[Range[60]], 2, 1] (* Harvey P. Dale, Aug 23 2011 *) Abs[Differences[Table[(-1)^n Prime[n], {n, 60}]]] (* Alonso del Arte, Feb 03 2016 *) PROG (Sage) BB = primes_first_n(56) L = [] for i in range(55): L.append(BB[1 + i] + BB[i]) L # Zerinvary Lajos, May 14 2007 (MAGMA) [(NthPrime(n+1) + NthPrime(n)): n in [1..100]]; // Vincenzo Librandi, Apr 02 2011 (PARI) p=2; forprime(q=3, 1e3, print1(p+q", "); p=q) \\ Charles R Greathouse IV, Jun 10 2011 (PARI) is(n)=precprime((n-1)/2)+nextprime(n/2)==n&&n>2 \\ Charles R Greathouse IV, Jun 21 2012 (Haskell) a001043 n = a001043_list !! (n-1) a001043_list = zipWith (+) a000040_list \$ tail a000040_list -- Reinhard Zumkeller, Oct 19 2011 CROSSREFS Cf. A000040. Sequence in context: A325438 A314411 A069102 * A118775 A025001 A020749 Adjacent sequences: A001040 A001041 A001042 * A001044 A001045 A001046 KEYWORD nonn,nice,easy AUTHOR EXTENSIONS More terms from Larry Reeves (larryr(AT)acm.org), Mar 17 2000 STATUS approved
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Last modified October 24 10:53 EDT 2020. Contains 337975 sequences. (Running on oeis4.) | 1,249 | 3,372 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-45 | latest | en | 0.594092 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A001043 Numbers that are the sum of 2 successive primes. (Formerly M3780 N0968) 146 5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, 78, 84, 90, 100, 112, 120, 128, 138, 144, 152, 162, 172, 186, 198, 204, 210, 216, 222, 240, 258, 268, 276, 288, 300, 308, 320, 330, 340, 352, 360, 372, 384, 390, 396, 410, 434, 450, 456, 462, 472, 480, 492, 508, 520 (list; graph; refs; listen; history; text; internal format) OFFSET 1,1 COMMENTS Arithmetic derivative (see A003415) of prime(n)*prime(n+1). - Giorgio Balzarotti, May 26 2011 A008472(a(n)) = A191583(n). - Reinhard Zumkeller, Jun 28 2011 With the exception of the first term, all terms are even. a(n) is divisible by 4 if the difference between prime(n) and prime(n + 1) is not divisible by 4; e.g., prime(n) = 1 mod 4 and prime(n + 1) = 3 mod 4. In general, for a(n) to be divisible by some even number m > 2 requires that prime(n + 1) - prime(n) not be a multiple of m. - Alonso del Arte, Jan 30 2012 REFERENCES Archimedeans Problems Drive, Eureka, 26 (1963), 12. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe) Albert Frank & Philippe Jacqueroux, International Contest, 2001. Item 22 Richard K. Guy, Letters to N. J. A. Sloane, June-August 1968 FORMULA a(n) = prime(n) + prime(n + 1). a(n) = A000040(n) + A000040(n+1). a(n) = A116366(n, n - 1) for n > 1. - Reinhard Zumkeller, Feb 06 2006 EXAMPLE 2 + 3 = 5. 3 + 5 = 8. 5 + 7 = 12. 7 + 11 = 18. MAPLE Primes:= select(isprime, [2, seq(2*i+1, i=1..1000)]): n:= nops(Primes): Primes[1..n-1] + Primes[2..n]; # Robert Israel, Aug 29 2014 MATHEMATICA Table[Prime[n] + Prime[n + 1], {n, 55}] (* Ray Chandler, Feb 12 2005 *) Total/@Partition[Prime[Range[60]], 2, 1] (* Harvey P. Dale, Aug 23 2011 *) Abs[Differences[Table[(-1)^n Prime[n], {n, 60}]]] (* Alonso del Arte, Feb 03 2016 *) PROG (Sage) BB = primes_first_n(56) L = [] for i in range(55): L.append(BB[1 + i] + BB[i]) L # Zerinvary Lajos, May 14 2007 (MAGMA) [(NthPrime(n+1) + NthPrime(n)): n in [1..100]]; // Vincenzo Librandi, Apr 02 2011 (PARI) p=2; forprime(q=3, 1e3, print1(p+q", "); p=q) \\ Charles R Greathouse IV, Jun 10 2011 (PARI) is(n)=precprime((n-1)/2)+nextprime(n/2)==n&&n>2 \\ Charles R Greathouse IV, Jun 21 2012 (Haskell) a001043 n = a001043_list !! (n-1) a001043_list = zipWith (+) a000040_list \$ tail a000040_list -- Reinhard Zumkeller, Oct 19 2011 CROSSREFS Cf. A000040. Sequence in context: A325438 A314411 A069102 * A118775 A025001 A020749 Adjacent sequences: A001040 A001041 A001042 * A001044 A001045 A001046 KEYWORD nonn,nice,easy AUTHOR EXTENSIONS More terms from Larry Reeves (larryr(AT)acm.org), Mar 17 2000 STATUS approved Lookup | Welcome | | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent The OEIS Community | Maintained by The OEIS Foundation Inc. Last modified October 24 10:53 EDT 2020. Contains 337975 sequences. (Running on oeis4.) |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-5-exponents-and-polynomials-5-1-addition-and-subtraction-of-polynomials-problem-set-5-1-page-195/20 | 1,537,552,967,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267157351.3/warc/CC-MAIN-20180921170920-20180921191320-00321.warc.gz | 754,570,770 | 13,823 | # Chapter 5 - Exponents and Polynomials - 5.1 - Addition and Subtraction of Polynomials - Problem Set 5.1 - Page 195: 20
$-3a+3$
#### Work Step by Step
We write an expression to express this addition problem: $3a^2+4a-7+(-3a^2-7a+10)$ We use the commutative and associative properties of addition to group like terms: $(3a^2+(-3a^2))+(4a+(-7a))+(-7+10)$ We combine the like terms: $0-3a+3=-3a+3$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 177 | 559 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2018-39 | longest | en | 0.781707 | # Chapter 5 - Exponents and Polynomials - 5.1 - Addition and Subtraction of Polynomials - Problem Set 5.1 - Page 195: 20 $-3a+3$ #### Work Step by Step We write an expression to express this addition problem: $3a^2+4a-7+(-3a^2-7a+10)$ We use the commutative and associative properties of addition to group like terms: $(3a^2+(-3a^2))+(4a+(-7a))+(-7+10)$ We combine the like terms: $0-3a+3=-3a+3$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the | submission and either publish your submission or provide feedback. |
https://socratic.org/questions/how-do-you-simplify-sin-2-b-sec-2-b-sec-2-b | 1,713,402,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817184.35/warc/CC-MAIN-20240417235906-20240418025906-00434.warc.gz | 493,544,420 | 6,002 | # How do you simplify sin^2 B sec^2 B - sec^2 B?
Mar 15, 2018
-1
#### Explanation:
Given: ${\sin}^{2} B {\sec}^{2} B - {\sec}^{2} B$
Apply the reciprocal identity:
$\sec B = \frac{1}{\cos} B$
So: ${\sin}^{2} B \cdot \frac{1}{\cos} ^ 2 B - {\sec}^{2} B$
${\sin}^{2} \frac{B}{\cos} ^ 2 B - {\sec}^{2} B$
Apply the quotient identity:
$\sin \frac{B}{\cos} B = T a n B$
So: ${\tan}^{2} B - {\sec}^{2} B$
Now apply the Pythagorean identity:
$1 + {\tan}^{2} B = {\sec}^{2} B$
Manipulated to: $- 1 = {\tan}^{2} B - {\sec}^{2} B$
So: ${\tan}^{2} B - {\sec}^{2} B = - 1$ | 256 | 570 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-18 | latest | en | 0.342379 | # How do you simplify sin^2 B sec^2 B - sec^2 B? Mar 15, 2018 -1 #### Explanation: Given: ${\sin}^{2} B {\sec}^{2} B - {\sec}^{2} B$ Apply the reciprocal identity: $\sec B = \frac{1}{\cos} B$ So: ${\sin}^{2} B \cdot \frac{1}{\cos} ^ 2 B - {\sec}^{2} B$ ${\sin}^{2} \frac{B}{\cos} ^ 2 B - {\sec}^{2} B$ Apply the quotient identity: $\sin \frac{B}{\cos} B = T a n B$ So: ${\tan}^{2} B - {\sec}^{2} B$ Now apply the Pythagorean identity: $1 + {\tan}^{2} B = {\sec}^{2} B$ Manipulated to: $- 1 = {\tan}^{2} B - | {\sec}^{2} B$ So: ${\tan}^{2} B - {\sec}^{2} B = - 1$ |
http://forum.solar-electric.com/discussion/8322/how-much-power-does-inverter-draw | 1,506,274,729,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690112.3/warc/CC-MAIN-20170924171658-20170924191658-00211.warc.gz | 140,942,139 | 10,015 | # How much power does inverter draw?
Solar Expert Posts: 40
I am a bit confused on how much power things draw from my (very small) 12 volt system. I have a 100 Ah battery and a 50 watt solar panel. I use a 1500 watt inverter. I know the inverter draws .5 amp when it is on, but idle.
What I don't know is: how much power does the inverter use (in addition to what the lamp draws) when it is making a/c--say, for a 20 watt lamp?
also:
If I run a 120 watt a/c appliance for an hour, does it drain 1 amp hour from my battery (since it's 120 watt a/c), or does it drain 10 amp hours (since the inverter is using 12 volt power?
• Super Moderators Posts: 26,666 admin
Re: How much power does inverter draw?
I am a bit confused on how much power things draw from my (very small) 12 volt system. I have a 100 Ah battery and a 50 watt solar panel. I use a 1500 watt inverter. I know the inverter draws .5 amp when it is on, but idle.
• 12 volts * 0.5 amps = 6 watt draw
What I don't know is: how much power does the inverter use (in addition to what the lamp draws) when it is making a/c--say, for a 20 watt lamp?
• Power = 6 watts + (20 watt load / 0.85 inverter efficiency) = 29.6 watts total
If I run a 120 watt a/c appliance for an hour, does it drain 1 amp hour from my battery (since it's 120 watt a/c), or does it drain 10 amp hours (since the inverter is using 12 volt power?
• Power = Volts * Amps
• 120 Watts = 120 volts * 1 amp
• 120 watts = 12 volts * 10 amp [fixed typo--thank you greenthumb]
Using the above example:
• Power-inverter-input = 6 watts + (120 watts / 0.85) = 147 watts DC load
• I = P/V = 147 Watts / 12 volts = 12.25 amps
-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Solar Expert Posts: 40
Re: How much power does inverter draw?
Thank you so much for your clear, concise explanation!!:D:D
(You might want to correct a typo in your formula, which reads:
120 watts = 12 volts * 1 amp)
• Super Moderators Posts: 26,666 admin
Re: How much power does inverter draw?
Thank you... Typo Fixed.
By the way, the above are all approximations... Inverters are typically less efficient near 100% loading, current input requirements change vs battery bank voltage (10.5-15.5 volts), etc.
But it does show way a large inverter (high idling losses) with small loads is not a good thing for a small system.
-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Solar Expert Posts: 40
Re: How much power does inverter draw?
It's a very dynamic system . . . hard to pin down exact numbers, but approximations help a lot!!
But it does show why a large inverter (high idling losses) with small loads is not a good thing for a small system.
Yeah--it didn't take long to figure that out!!
• Solar Expert Posts: 1,280 ✭✭✭
Re: How much power does inverter draw?
There are basically two pieces to inverter overhead.
The first one you already mentioned, in your 0.5 amps, just for the electronics control and switching of the gate control of the output MOSFET's.
A pure sinewave inverter generally has higher idle consumption then a modified sinewave because the electronic control circuitry is a bit more complicated and the output MOSFET's transistors and being chopped at a much higher frequency then a modifed sinewave inverter. A sinewave inverter has an output filter to remove the high frequency pulse width modulation so there is a clean sinewave output. This filter draws some output current through the filter capacitor whether there is an output load on the inverter or not.
The larger, the higher the wattage rating the inverter, the more idle power it generally consumes. Overkill on inverter size can hurt you in low power use cases.
The second part is dynamic, being a function of the load on the inverter.
Most of this dynamic loss is in resistive losses in the series equivalent resistance of the output MOSFET transistors, wire resistance in transformer, and interconnects. These losses will be in direct proportion to the output load current.
There is also dynamic loss in transformer core material. The transformer core losses usually don't become signficant until you are operating above 80% of the inverter rated power level.
You might think of dynamic losses as voltage drop (and associated power loss) in a long extension cord. Up to the point where transformer core losses start to become significant, the long extension cord analogy is very accurate.
At the risk of giving you more info then you want to know, power factor comes into play on inverter dynamic losses.
Let's say you have a Kill-A-Watt meter hooked up you your refrig and it tells you your refrig draws 160 watts with a power factor of 0.8. The V*A of the refrig is 200 watts. The inverter dynamic losses will be as if you are running a 200 watt load. Do not confuse this as 200 watts is translated back to battery, only that the inverter dynamic loss will be higher then if it was running a 160 watt, PF=1 load.
A bi-directional inverter, like a Xantrex XW series, or Outback FX series, react a little different to poor PF then a non bi-directional inverter like a true sinewave Power-To-Go model. The non bi-directional inverter will have a bit higher dynamic losses due to poor power factor. They also can be damaged if power factor is too poor.
• Solar Expert Posts: 40
Re: How much power does inverter draw?
I think I get the gist of what you're saying, but you use a lot of terminology that's over my head . . . | 1,409 | 5,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2017-39 | latest | en | 0.886413 | # How much power does inverter draw? Solar Expert Posts: 40 I am a bit confused on how much power things draw from my (very small) 12 volt system. I have a 100 Ah battery and a 50 watt solar panel. I use a 1500 watt inverter. I know the inverter draws .5 amp when it is on, but idle. What I don't know is: how much power does the inverter use (in addition to what the lamp draws) when it is making a/c--say, for a 20 watt lamp? also: If I run a 120 watt a/c appliance for an hour, does it drain 1 amp hour from my battery (since it's 120 watt a/c), or does it drain 10 amp hours (since the inverter is using 12 volt power? • Super Moderators Posts: 26,666 admin Re: How much power does inverter draw? I am a bit confused on how much power things draw from my (very small) 12 volt system. I have a 100 Ah battery and a 50 watt solar panel. I use a 1500 watt inverter. I know the inverter draws .5 amp when it is on, but idle. • 12 volts * 0.5 amps = 6 watt draw What I don't know is: how much power does the inverter use (in addition to what the lamp draws) when it is making a/c--say, for a 20 watt lamp? • Power = 6 watts + (20 watt load / 0.85 inverter efficiency) = 29.6 watts total If I run a 120 watt a/c appliance for an hour, does it drain 1 amp hour from my battery (since it's 120 watt a/c), or does it drain 10 amp hours (since the inverter is using 12 volt power? • Power = Volts * Amps • 120 Watts = 120 volts * 1 amp • 120 watts = 12 volts * 10 amp [fixed typo--thank you greenthumb] Using the above example: • Power-inverter-input = 6 watts + (120 watts / 0.85) = 147 watts DC load • I = P/V = 147 Watts / 12 volts = 12.25 amps -Bill Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset • Solar Expert Posts: 40 Re: How much power does inverter draw? Thank you so much for your clear, concise explanation!!:D:D (You might want to correct a typo in your formula, which reads: 120 watts = 12 volts * 1 amp) • Super Moderators Posts: 26,666 admin Re: How much power does inverter draw? Thank you... Typo Fixed. By the way, the above are all approximations... Inverters are typically less efficient near 100% loading, current input requirements change vs battery bank voltage (10.5-15.5 volts), etc. But it does show way a large inverter (high idling losses) with small loads is not a good thing for a small system. -Bill Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset • Solar Expert Posts: 40 Re: How much power does inverter draw? It's a very dynamic system . . . hard to pin down exact numbers, but approximations help a lot!! But it does show why a large inverter (high idling losses) with small loads is not a good thing for a small system. Yeah--it didn't take long to figure that out!! • Solar Expert Posts: 1,280 ✭✭✭ Re: How much power does inverter draw? There are basically two pieces to inverter overhead. The first one you already mentioned, in your 0.5 amps, just for the electronics control and switching of the gate control of the output MOSFET's. A pure sinewave inverter generally has higher idle consumption then a modified sinewave because the electronic control circuitry is a bit more complicated and the output MOSFET's transistors and being chopped at a much higher frequency then a modifed sinewave inverter. A sinewave inverter has an output filter to remove the high frequency pulse width modulation so there is a clean sinewave output. This filter draws some output current through the filter capacitor whether there is an output load on the inverter or not. The larger, the higher the wattage rating the inverter, the more idle power it generally consumes. Overkill on inverter size can hurt you in low power use cases. The second part is dynamic, being a function of the load on the inverter. Most of this dynamic loss is in resistive losses in the series equivalent resistance of the output MOSFET transistors, wire resistance in transformer, and interconnects. These losses will be in direct proportion to the output load current. There is also dynamic loss in transformer core material. The transformer core losses usually don't become signficant until you are operating above 80% of the inverter rated power level. You might think of dynamic losses as voltage drop (and associated power loss) in a long extension cord. Up to the point where transformer core losses start to become significant, the long extension cord analogy is very accurate. At the risk of giving you more info then you want to know, power factor comes into play on inverter dynamic losses. Let's say you have a Kill-A-Watt meter hooked up you your refrig and it tells you your refrig draws 160 watts with a power factor of 0.8. The V*A of the refrig is 200 watts. The inverter dynamic losses will be as if you are running a 200 watt load. Do not confuse this as 200 watts is translated back to battery, only that the inverter dynamic loss will be higher then | if it was running a 160 watt, PF=1 load. A bi-directional inverter, like a Xantrex XW series, or Outback FX series, react a little different to poor PF then a non bi-directional inverter like a true sinewave Power-To-Go model. The non bi-directional inverter will have a bit higher dynamic losses due to poor power factor. They also can be damaged if power factor is too poor. • Solar Expert Posts: 40 Re: How much power does inverter draw? I think I get the gist of what you're saying, but you use a lot of terminology that's over my head . . . |
https://en.wikipedia.org/wiki/Rigid_body | 1,722,836,124,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640434051.17/warc/CC-MAIN-20240805052118-20240805082118-00526.warc.gz | 180,951,939 | 36,848 | # Rigid body
In physics, a rigid body, also known as a rigid object,[2] is a solid body in which deformation is zero or negligible. The distance between any two given points on a rigid body remains constant in time regardless of external forces or moments exerted on it. A rigid body is usually considered as a continuous distribution of mass.
In the study of special relativity, a perfectly rigid body does not exist; and objects can only be assumed to be rigid if they are not moving near the speed of light. In quantum mechanics, a rigid body is usually thought of as a collection of point masses. For instance, molecules (consisting of the point masses: electrons and nuclei) are often seen as rigid bodies (see classification of molecules as rigid rotors).
## Kinematics
### Linear and angular position
The position of a rigid body is the position of all the particles of which it is composed. To simplify the description of this position, we exploit the property that the body is rigid, namely that all its particles maintain the same distance relative to each other. If the body is rigid, it is sufficient to describe the position of at least three non-collinear particles. This makes it possible to reconstruct the position of all the other particles, provided that their time-invariant position relative to the three selected particles is known. However, typically a different, mathematically more convenient, but equivalent approach is used. The position of the whole body is represented by:
1. the linear position or position of the body, namely the position of one of the particles of the body, specifically chosen as a reference point (typically coinciding with the center of mass or centroid of the body), together with
2. the angular position (also known as orientation', or attitude) of the body.
Thus, the position of a rigid body has two components: linear and angular, respectively.[3] The same is true for other kinematic and kinetic quantities describing the motion of a rigid body, such as linear and angular velocity, acceleration, momentum, impulse, and kinetic energy.[4]
The linear position can be represented by a vector with its tail at an arbitrary reference point in space (the origin of a chosen coordinate system) and its tip at an arbitrary point of interest on the rigid body, typically coinciding with its center of mass or centroid. This reference point may define the origin of a coordinate system fixed to the body.
There are several ways to numerically describe the orientation of a rigid body, including a set of three Euler angles, a quaternion, or a direction cosine matrix (also referred to as a rotation matrix). All these methods actually define the orientation of a basis set (or coordinate system) which has a fixed orientation relative to the body (i.e. rotates together with the body), relative to another basis set (or coordinate system), from which the motion of the rigid body is observed. For instance, a basis set with fixed orientation relative to an airplane can be defined as a set of three orthogonal unit vectors b1, b2, b3, such that b1 is parallel to the chord line of the wing and directed forward, b2 is normal to the plane of symmetry and directed rightward, and b3 is given by the cross product ${\displaystyle b_{3}=b_{1}\times b_{2}}$.
In general, when a rigid body moves, both its position and orientation vary with time. In the kinematic sense, these changes are referred to as translation and rotation, respectively. Indeed, the position of a rigid body can be viewed as a hypothetic translation and rotation (roto-translation) of the body starting from a hypothetic reference position (not necessarily coinciding with a position actually taken by the body during its motion).
### Linear and angular velocity
Velocity (also called linear velocity) and angular velocity are measured with respect to a frame of reference.
The linear velocity of a rigid body is a vector quantity, equal to the time rate of change of its linear position. Thus, it is the velocity of a reference point fixed to the body. During purely translational motion (motion with no rotation), all points on a rigid body move with the same velocity. However, when motion involves rotation, the instantaneous velocity of any two points on the body will generally not be the same. Two points of a rotating body will have the same instantaneous velocity only if they happen to lie on an axis parallel to the instantaneous axis of rotation.
Angular velocity is a vector quantity that describes the angular speed at which the orientation of the rigid body is changing and the instantaneous axis about which it is rotating (the existence of this instantaneous axis is guaranteed by the Euler's rotation theorem). All points on a rigid body experience the same angular velocity at all times. During purely rotational motion, all points on the body change position except for those lying on the instantaneous axis of rotation. The relationship between orientation and angular velocity is not directly analogous to the relationship between position and velocity. Angular velocity is not the time rate of change of orientation, because there is no such concept as an orientation vector that can be differentiated to obtain the angular velocity.
## Kinematical equations
### Addition theorem for angular velocity
The angular velocity of a rigid body B in a reference frame N is equal to the sum of the angular velocity of a rigid body D in N and the angular velocity of B with respect to D:[5]
${\displaystyle {}^{\mathrm {N} }\!{\boldsymbol {\omega }}^{\mathrm {B} }={}^{\mathrm {N} }\!{\boldsymbol {\omega }}^{\mathrm {D} }+{}^{\mathrm {D} }\!{\boldsymbol {\omega }}^{\mathrm {B} }.}$
In this case, rigid bodies and reference frames are indistinguishable and completely interchangeable.
### Addition theorem for position
For any set of three points P, Q, and R, the position vector from P to R is the sum of the position vector from P to Q and the position vector from Q to R:
${\displaystyle \mathbf {r} ^{\mathrm {PR} }=\mathbf {r} ^{\mathrm {PQ} }+\mathbf {r} ^{\mathrm {QR} }.}$
The norm of a position vector is the spatial distance. Here the coordinates of all three vectors must be expressed in coordinate frames with the same orientation.
### Mathematical definition of velocity
The velocity of point P in reference frame N is defined as the time derivative in N of the position vector from O to P:[6]
${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {P} }={\frac {{}^{\mathrm {N} }\mathrm {d} }{\mathrm {d} t}}(\mathbf {r} ^{\mathrm {OP} })}$
where O is any arbitrary point fixed in reference frame N, and the N to the left of the d/dt operator indicates that the derivative is taken in reference frame N. The result is independent of the selection of O so long as O is fixed in N.
### Mathematical definition of acceleration
The acceleration of point P in reference frame N is defined as the time derivative in N of its velocity:[6]
${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {P} }={\frac {^{\mathrm {N} }\mathrm {d} }{\mathrm {d} t}}({}^{\mathrm {N} }\mathbf {v} ^{\mathrm {P} }).}$
### Velocity of two points fixed on a rigid body
For two points P and Q that are fixed on a rigid body B, where B has an angular velocity ${\displaystyle \scriptstyle {^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }}}$ in the reference frame N, the velocity of Q in N can be expressed as a function of the velocity of P in N:[7]
${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {Q} }={}^{\mathrm {N} }\!\mathbf {v} ^{\mathrm {P} }+{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }.}$
where ${\displaystyle \mathbf {r} ^{\mathrm {PQ} }}$ is the position vector from P to Q.,[7] with coordinates expressed in N (or a frame with the same orientation as N.) This relation can be derived from the temporal invariance of the norm distance between P and Q.
### Acceleration of two points fixed on a rigid body
By differentiating the equation for the Velocity of two points fixed on a rigid body in N with respect to time, the acceleration in reference frame N of a point Q fixed on a rigid body B can be expressed as
${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {Q} }={}^{\mathrm {N} }\mathbf {a} ^{\mathrm {P} }+{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \left({}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }\right)+{}^{\mathrm {N} }{\boldsymbol {\alpha }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }}$
where ${\displaystyle \scriptstyle {{}^{\mathrm {N} }\!{\boldsymbol {\alpha }}^{\mathrm {B} }}}$ is the angular acceleration of B in the reference frame N.[7]
### Angular velocity and acceleration of two points fixed on a rigid body
As mentioned above, all points on a rigid body B have the same angular velocity ${\displaystyle {}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }}$ in a fixed reference frame N, and thus the same angular acceleration ${\displaystyle {}^{\mathrm {N} }{\boldsymbol {\alpha }}^{\mathrm {B} }.}$
### Velocity of one point moving on a rigid body
If the point R is moving in the rigid body B while B moves in reference frame N, then the velocity of R in N is
${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {R} }={}^{\mathrm {N} }\mathbf {v} ^{\mathrm {Q} }+{}^{\mathrm {B} }\mathbf {v} ^{\mathrm {R} }}$
where Q is the point fixed in B that is instantaneously coincident with R at the instant of interest.[8] This relation is often combined with the relation for the Velocity of two points fixed on a rigid body.
### Acceleration of one point moving on a rigid body
The acceleration in reference frame N of the point R moving in body B while B is moving in frame N is given by
${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {R} }={}^{\mathrm {N} }\mathbf {a} ^{\mathrm {Q} }+{}^{\mathrm {B} }\mathbf {a} ^{\mathrm {R} }+2{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times {}^{\mathrm {B} }\mathbf {v} ^{\mathrm {R} }}$
where Q is the point fixed in B that instantaneously coincident with R at the instant of interest.[8] This equation is often combined with Acceleration of two points fixed on a rigid body.
### Other quantities
If C is the origin of a local coordinate system L, attached to the body, the spatial or twist acceleration of a rigid body is defined as the spatial acceleration of C (as opposed to material acceleration above): ${\displaystyle {\boldsymbol {\psi }}(t,\mathbf {r} _{0})=\mathbf {a} (t,\mathbf {r} _{0})-{\boldsymbol {\omega }}(t)\times \mathbf {v} (t,\mathbf {r} _{0})={\boldsymbol {\psi }}_{c}(t)+{\boldsymbol {\alpha }}(t)\times A(t)\mathbf {r} _{0}}$ where
• ${\displaystyle \mathbf {r} _{0}}$ represents the position of the point/particle with respect to the reference point of the body in terms of the local coordinate system L (the rigidity of the body means that this does not depend on time)
• ${\displaystyle A(t)\,}$ is the orientation matrix, an orthogonal matrix with determinant 1, representing the orientation (angular position) of the local coordinate system L, with respect to the arbitrary reference orientation of another coordinate system G. Think of this matrix as three orthogonal unit vectors, one in each column, which define the orientation of the axes of L with respect to G.
• ${\displaystyle {\boldsymbol {\omega }}(t)}$ represents the angular velocity of the rigid body
• ${\displaystyle \mathbf {v} (t,\mathbf {r} _{0})}$ represents the total velocity of the point/particle
• ${\displaystyle \mathbf {a} (t,\mathbf {r} _{0})}$ represents the total acceleration of the point/particle
• ${\displaystyle {\boldsymbol {\alpha }}(t)}$ represents the angular acceleration of the rigid body
• ${\displaystyle {\boldsymbol {\psi }}(t,\mathbf {r} _{0})}$ represents the spatial acceleration of the point/particle
• ${\displaystyle {\boldsymbol {\psi }}_{c}(t)}$ represents the spatial acceleration of the rigid body (i.e. the spatial acceleration of the origin of L).
In 2D, the angular velocity is a scalar, and matrix A(t) simply represents a rotation in the xy-plane by an angle which is the integral of the angular velocity over time.
Vehicles, walking people, etc., usually rotate according to changes in the direction of the velocity: they move forward with respect to their own orientation. Then, if the body follows a closed orbit in a plane, the angular velocity integrated over a time interval in which the orbit is completed once, is an integer times 360°. This integer is the winding number with respect to the origin of the velocity. Compare the amount of rotation associated with the vertices of a polygon.
## Kinetics
Any point that is rigidly connected to the body can be used as reference point (origin of coordinate system L) to describe the linear motion of the body (the linear position, velocity and acceleration vectors depend on the choice).
However, depending on the application, a convenient choice may be:
• the center of mass of the whole system, which generally has the simplest motion for a body moving freely in space;
• a point such that the translational motion is zero or simplified, e.g. on an axle or hinge, at the center of a ball and socket joint, etc.
When the center of mass is used as reference point:
• The (linear) momentum is independent of the rotational motion. At any time it is equal to the total mass of the rigid body times the translational velocity.
• The angular momentum with respect to the center of mass is the same as without translation: at any time it is equal to the inertia tensor times the angular velocity. When the angular velocity is expressed with respect to a coordinate system coinciding with the principal axes of the body, each component of the angular momentum is a product of a moment of inertia (a principal value of the inertia tensor) times the corresponding component of the angular velocity; the torque is the inertia tensor times the angular acceleration.
• Possible motions in the absence of external forces are translation with constant velocity, steady rotation about a fixed principal axis, and also torque-free precession.
• The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates).
• The total kinetic energy is simply the sum of translational and rotational energy.
## Geometry
Two rigid bodies are said to be different (not copies) if there is no proper rotation from one to the other. A rigid body is called chiral if its mirror image is different in that sense, i.e., if it has either no symmetry or its symmetry group contains only proper rotations. In the opposite case an object is called achiral: the mirror image is a copy, not a different object. Such an object may have a symmetry plane, but not necessarily: there may also be a plane of reflection with respect to which the image of the object is a rotated version. The latter applies for S2n, of which the case n = 1 is inversion symmetry.
For a (rigid) rectangular transparent sheet, inversion symmetry corresponds to having on one side an image without rotational symmetry and on the other side an image such that what shines through is the image at the top side, upside down. We can distinguish two cases:
• the sheet surface with the image is not symmetric - in this case the two sides are different, but the mirror image of the object is the same, after a rotation by 180° about the axis perpendicular to the mirror plane.
• the sheet surface with the image has a symmetry axis - in this case the two sides are the same, and the mirror image of the object is also the same, again after a rotation by 180° about the axis perpendicular to the mirror plane.
A sheet with a through and through image is achiral. We can distinguish again two cases:
• the sheet surface with the image has no symmetry axis - the two sides are different
• the sheet surface with the image has a symmetry axis - the two sides are the same
## Configuration space
The configuration space of a rigid body with one point fixed (i.e., a body with zero translational motion) is given by the underlying manifold of the rotation group SO(3). The configuration space of a nonfixed (with non-zero translational motion) rigid body is E+(3), the subgroup of direct isometries of the Euclidean group in three dimensions (combinations of translations and rotations).
## Notes
1. ^ Lorenzo Sciavicco, Bruno Siciliano (2000). "§2.4.2 Roll-pitch-yaw angles". Modelling and control of robot manipulators (2nd ed.). Springer. p. 32. ISBN 1-85233-221-2.
2. ^ Andy Ruina and Rudra Pratap (2015). Introduction to Statics and Dynamics. Oxford University Press. (link: [1])
3. ^ In general, the position of a point or particle is also known, in physics, as linear position, as opposed to the angular position of a line, or line segment (e.g., in circular motion, the "radius" joining the rotating point with the center of rotation), or basis set, or coordinate system.
4. ^ In kinematics, linear means "along a straight or curved line" (the path of the particle in space). In mathematics, however, linear has a different meaning. In both contexts, the word "linear" is related to the word "line". In mathematics, a line is often defined as a straight curve. For those who adopt this definition, a curve can be straight, and curved lines are not supposed to exist. In kinematics, the term line is used as a synonym of the term trajectory, or path (namely, it has the same non-restricted meaning as that given, in mathematics, to the word curve). In short, both straight and curved lines are supposed to exist. In kinematics and dynamics, the following words refer to the same non-restricted meaning of the term "line":
• "linear" (= along a straight or curved line),
• "rectilinear" (= along a straight line, from Latin rectus = straight, and linere = spread),
• "curvilinear" (=along a curved line, from Latin curvus = curved, and linere = spread).
In topology and meteorology, the term "line" has the same meaning; namely, a contour line is a curve.
5. ^ Kane, Thomas; Levinson, David (1996). "2-4 Auxiliary Reference Frames". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
6. ^ a b Kane, Thomas; Levinson, David (1996). "2-6 Velocity and Acceleration". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
7. ^ a b c Kane, Thomas; Levinson, David (1996). "2-7 Two Points Fixed on a Rigid Body". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
8. ^ a b Kane, Thomas; Levinson, David (1996). "2-8 One Point Moving on a Rigid Body". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc.
## References
• Roy Featherstone (1987). Robot Dynamics Algorithms. Springer. ISBN 0-89838-230-0. This reference effectively combines screw theory with rigid body dynamics for robotic applications. The author also chooses to use spatial accelerations extensively in place of material accelerations as they simplify the equations and allow for compact notation.
• JPL DARTS page has a section on spatial operator algebra (link: [2]) as well as an extensive list of references (link: [3]).
• Andy Ruina and Rudra Pratap (2015). Introduction to Statics and Dynamics. Oxford University Press. (link: [4]).
• Prof. Dr. Dennis M. Kochmann, Dynamics Lecture Notes, ETH Zurich. [5] | 4,614 | 19,544 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 24, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-33 | latest | en | 0.938636 | # Rigid body In physics, a rigid body, also known as a rigid object,[2] is a solid body in which deformation is zero or negligible. The distance between any two given points on a rigid body remains constant in time regardless of external forces or moments exerted on it. A rigid body is usually considered as a continuous distribution of mass. In the study of special relativity, a perfectly rigid body does not exist; and objects can only be assumed to be rigid if they are not moving near the speed of light. In quantum mechanics, a rigid body is usually thought of as a collection of point masses. For instance, molecules (consisting of the point masses: electrons and nuclei) are often seen as rigid bodies (see classification of molecules as rigid rotors). ## Kinematics ### Linear and angular position The position of a rigid body is the position of all the particles of which it is composed. To simplify the description of this position, we exploit the property that the body is rigid, namely that all its particles maintain the same distance relative to each other. If the body is rigid, it is sufficient to describe the position of at least three non-collinear particles. This makes it possible to reconstruct the position of all the other particles, provided that their time-invariant position relative to the three selected particles is known. However, typically a different, mathematically more convenient, but equivalent approach is used. The position of the whole body is represented by: 1. the linear position or position of the body, namely the position of one of the particles of the body, specifically chosen as a reference point (typically coinciding with the center of mass or centroid of the body), together with 2. the angular position (also known as orientation', or attitude) of the body. Thus, the position of a rigid body has two components: linear and angular, respectively.[3] The same is true for other kinematic and kinetic quantities describing the motion of a rigid body, such as linear and angular velocity, acceleration, momentum, impulse, and kinetic energy.[4] The linear position can be represented by a vector with its tail at an arbitrary reference point in space (the origin of a chosen coordinate system) and its tip at an arbitrary point of interest on the rigid body, typically coinciding with its center of mass or centroid. This reference point may define the origin of a coordinate system fixed to the body. There are several ways to numerically describe the orientation of a rigid body, including a set of three Euler angles, a quaternion, or a direction cosine matrix (also referred to as a rotation matrix). All these methods actually define the orientation of a basis set (or coordinate system) which has a fixed orientation relative to the body (i.e. rotates together with the body), relative to another basis set (or coordinate system), from which the motion of the rigid body is observed. For instance, a basis set with fixed orientation relative to an airplane can be defined as a set of three orthogonal unit vectors b1, b2, b3, such that b1 is parallel to the chord line of the wing and directed forward, b2 is normal to the plane of symmetry and directed rightward, and b3 is given by the cross product ${\displaystyle b_{3}=b_{1}\times b_{2}}$. In general, when a rigid body moves, both its position and orientation vary with time. In the kinematic sense, these changes are referred to as translation and rotation, respectively. Indeed, the position of a rigid body can be viewed as a hypothetic translation and rotation (roto-translation) of the body starting from a hypothetic reference position (not necessarily coinciding with a position actually taken by the body during its motion). ### Linear and angular velocity Velocity (also called linear velocity) and angular velocity are measured with respect to a frame of reference. The linear velocity of a rigid body is a vector quantity, equal to the time rate of change of its linear position. Thus, it is the velocity of a reference point fixed to the body. During purely translational motion (motion with no rotation), all points on a rigid body move with the same velocity. However, when motion involves rotation, the instantaneous velocity of any two points on the body will generally not be the same. Two points of a rotating body will have the same instantaneous velocity only if they happen to lie on an axis parallel to the instantaneous axis of rotation. Angular velocity is a vector quantity that describes the angular speed at which the orientation of the rigid body is changing and the instantaneous axis about which it is rotating (the existence of this instantaneous axis is guaranteed by the Euler's rotation theorem). All points on a rigid body experience the same angular velocity at all times. During purely rotational motion, all points on the body change position except for those lying on the instantaneous axis of rotation. The relationship between orientation and angular velocity is not directly analogous to the relationship between position and velocity. Angular velocity is not the time rate of change of orientation, because there is no such concept as an orientation vector that can be differentiated to obtain the angular velocity. ## Kinematical equations ### Addition theorem for angular velocity The angular velocity of a rigid body B in a reference frame N is equal to the sum of the angular velocity of a rigid body D in N and the angular velocity of B with respect to D:[5] ${\displaystyle {}^{\mathrm {N} }\!{\boldsymbol {\omega }}^{\mathrm {B} }={}^{\mathrm {N} }\!{\boldsymbol {\omega }}^{\mathrm {D} }+{}^{\mathrm {D} }\!{\boldsymbol {\omega }}^{\mathrm {B} }.}$ In this case, rigid bodies and reference frames are indistinguishable and completely interchangeable. ### Addition theorem for position For any set of three points P, Q, and R, the position vector from P to R is the sum of the position vector from P to Q and the position vector from Q to R: ${\displaystyle \mathbf {r} ^{\mathrm {PR} }=\mathbf {r} ^{\mathrm {PQ} }+\mathbf {r} ^{\mathrm {QR} }.}$ The norm of a position vector is the spatial distance. Here the coordinates of all three vectors must be expressed in coordinate frames with the same orientation. ### Mathematical definition of velocity The velocity of point P in reference frame N is defined as the time derivative in N of the position vector from O to P:[6] ${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {P} }={\frac {{}^{\mathrm {N} }\mathrm {d} }{\mathrm {d} t}}(\mathbf {r} ^{\mathrm {OP} })}$ where O is any arbitrary point fixed in reference frame N, and the N to the left of the d/dt operator indicates that the derivative is taken in reference frame N. The result is independent of the selection of O so long as O is fixed in N. ### Mathematical definition of acceleration The acceleration of point P in reference frame N is defined as the time derivative in N of its velocity:[6] ${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {P} }={\frac {^{\mathrm {N} }\mathrm {d} }{\mathrm {d} t}}({}^{\mathrm {N} }\mathbf {v} ^{\mathrm {P} }).}$ ### Velocity of two points fixed on a rigid body For two points P and Q that are fixed on a rigid body B, where B has an angular velocity ${\displaystyle \scriptstyle {^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }}}$ in the reference frame N, the velocity of Q in N can be expressed as a function of the velocity of P in N:[7] ${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {Q} }={}^{\mathrm {N} }\!\mathbf {v} ^{\mathrm {P} }+{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }.}$ where ${\displaystyle \mathbf {r} ^{\mathrm {PQ} }}$ is the position vector from P to Q.,[7] with coordinates expressed in N (or a frame with the same orientation as N.) This relation can be derived from the temporal invariance of the norm distance between P and Q. ### Acceleration of two points fixed on a rigid body By differentiating the equation for the Velocity of two points fixed on a rigid body in N with respect to time, the acceleration in reference frame N of a point Q fixed on a rigid body B can be expressed as ${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {Q} }={}^{\mathrm {N} }\mathbf {a} ^{\mathrm {P} }+{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \left({}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }\right)+{}^{\mathrm {N} }{\boldsymbol {\alpha }}^{\mathrm {B} }\times \mathbf {r} ^{\mathrm {PQ} }}$ where ${\displaystyle \scriptstyle {{}^{\mathrm {N} }\!{\boldsymbol {\alpha }}^{\mathrm {B} }}}$ is the angular acceleration of B in the reference frame N.[7] ### Angular velocity and acceleration of two points fixed on a rigid body As mentioned above, all points on a rigid body B have the same angular velocity ${\displaystyle {}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }}$ in a fixed reference frame N, and thus the same angular acceleration ${\displaystyle {}^{\mathrm {N} }{\boldsymbol {\alpha }}^{\mathrm {B} }.}$ ### Velocity of one point moving on a rigid body If the point R is moving in the rigid body B while B moves in reference frame N, then the velocity of R in N is ${\displaystyle {}^{\mathrm {N} }\mathbf {v} ^{\mathrm {R} }={}^{\mathrm {N} }\mathbf {v} ^{\mathrm {Q} }+{}^{\mathrm {B} }\mathbf {v} ^{\mathrm {R} }}$ where Q is the point fixed in B that is instantaneously coincident with R at the instant of interest.[8] This relation is often combined with the relation for the Velocity of two points fixed on a rigid body. ### Acceleration of one point moving on a rigid body The acceleration in reference frame N of the point R moving in body B while B is moving in frame N is given by ${\displaystyle {}^{\mathrm {N} }\mathbf {a} ^{\mathrm {R} }={}^{\mathrm {N} }\mathbf {a} ^{\mathrm {Q} }+{}^{\mathrm {B} }\mathbf {a} ^{\mathrm {R} }+2{}^{\mathrm {N} }{\boldsymbol {\omega }}^{\mathrm {B} }\times {}^{\mathrm {B} }\mathbf {v} ^{\mathrm {R} }}$ where Q is the point fixed in B that instantaneously coincident with R at the instant of interest.[8] This equation is often combined with Acceleration of two points fixed on a rigid body. ### Other quantities If C is the origin of a local coordinate system L, attached to the body, the spatial or twist acceleration of a rigid body is defined as the spatial acceleration of C (as opposed to material acceleration above): ${\displaystyle {\boldsymbol {\psi }}(t,\mathbf {r} _{0})=\mathbf {a} (t,\mathbf {r} _{0})-{\boldsymbol {\omega }}(t)\times \mathbf {v} (t,\mathbf {r} _{0})={\boldsymbol {\psi }}_{c}(t)+{\boldsymbol {\alpha }}(t)\times A(t)\mathbf {r} _{0}}$ where • ${\displaystyle \mathbf {r} _{0}}$ represents the position of the point/particle with respect to the reference point of the body in terms of the local coordinate system L (the rigidity of the body means that this does not depend on time) • ${\displaystyle A(t)\,}$ is the orientation matrix, an orthogonal matrix with determinant 1, representing the orientation (angular position) of the local coordinate system L, with respect to the arbitrary reference orientation of another coordinate system G. Think of this matrix as three orthogonal unit vectors, one in each column, which define the orientation of the axes of L with respect to G. • ${\displaystyle {\boldsymbol {\omega }}(t)}$ represents the angular velocity of the rigid body • ${\displaystyle \mathbf {v} (t,\mathbf {r} _{0})}$ represents the total velocity of the point/particle • ${\displaystyle \mathbf {a} (t,\mathbf {r} _{0})}$ represents the total acceleration of the point/particle • ${\displaystyle {\boldsymbol {\alpha }}(t)}$ represents the angular acceleration of the rigid body • ${\displaystyle {\boldsymbol {\psi }}(t,\mathbf {r} _{0})}$ represents the spatial acceleration of the point/particle • ${\displaystyle {\boldsymbol {\psi }}_{c}(t)}$ represents the spatial acceleration of the rigid body (i.e. the spatial acceleration of the origin of L). In 2D, the angular velocity is a scalar, and matrix A(t) simply represents a rotation in the xy-plane by an angle which is the integral of the angular velocity over time. Vehicles, walking people, etc., usually rotate according to changes in the direction of the velocity: they move forward with respect to their own orientation. Then, if the body follows a closed orbit in a plane, the angular velocity integrated over a time interval in which the orbit is completed once, is an integer times 360°. This integer is the winding number with respect to the origin of the velocity. Compare the amount of rotation associated with the vertices of a polygon. ## Kinetics Any point that is rigidly connected to the body can be used as reference point (origin of coordinate system L) to describe the linear motion of the body (the linear position, velocity and acceleration vectors depend on the choice). However, depending on the application, a convenient choice may be: • the center of mass of the whole system, which generally has the simplest motion for a body moving freely in space; • a point such that the translational motion is zero or simplified, e.g. on an axle or hinge, at the center of a ball and socket joint, etc. When the center of mass is used as reference point: • The (linear) momentum is independent of the rotational motion. At any time it is equal to the total mass of the rigid body times the translational velocity. • The angular momentum with respect to the center of mass is the same as without translation: at any time it is equal to the inertia tensor times the angular velocity. When the angular velocity is expressed with respect to a coordinate system coinciding with the principal axes of the body, each component of the angular momentum is a product of a moment of inertia (a principal value of the inertia tensor) times the corresponding component of the angular velocity; the torque is the inertia tensor times the angular acceleration. • Possible motions in the absence of external forces are translation with constant velocity, steady rotation about a fixed principal axis, and also torque-free precession. • The net external force on the rigid body is always equal to the total mass times the translational acceleration (i.e., Newton's second law holds for the translational motion, even when the net external torque is nonzero, and/or the body rotates). • The total kinetic energy is simply the sum of translational and rotational energy. ## Geometry Two rigid bodies are said to be different (not copies) if there is no proper rotation from one to the other. A rigid body is called chiral if its mirror image is different in that sense, i.e., if it has either no symmetry or its symmetry group contains only proper rotations. In the opposite case an object is called achiral: the mirror image is a copy, not a different object. Such an object may have a symmetry plane, but not necessarily: there may also be a plane of reflection with respect to which the image of the object is a rotated version. The latter applies for S2n, of which the case n = 1 is inversion symmetry. For a (rigid) rectangular transparent sheet, inversion symmetry corresponds to having on one side an image without rotational symmetry and on the other side an image such that what shines through is the image at the top side, upside down. We can distinguish two cases: • the sheet surface with the image is not symmetric - in this case the two sides are different, but the mirror image of the object is the same, after a rotation by 180° about the axis perpendicular to the mirror plane. • the sheet surface with the image has a symmetry axis - in this case the two sides are the same, and the mirror image of the object is also the same, again after a rotation by 180° about the axis perpendicular to the mirror plane. A sheet with a through and through image is achiral. We can distinguish again two cases: • the sheet surface with the image has no symmetry axis - the two sides are different • the sheet surface with the image has a symmetry axis - the two sides are the same ## Configuration space The configuration space of a rigid body with one point fixed (i.e., a body with zero translational motion) is given by the underlying manifold of the rotation group SO(3). The configuration space of a nonfixed (with non-zero translational motion) rigid body is E+(3), the subgroup of direct isometries of the Euclidean group in three dimensions (combinations of translations and rotations). ## Notes 1. ^ Lorenzo Sciavicco, Bruno Siciliano (2000). "§2.4.2 Roll-pitch-yaw angles". Modelling and control of robot manipulators (2nd ed.). Springer. p. 32. ISBN 1-85233-221-2. 2. ^ Andy Ruina and Rudra Pratap (2015). Introduction to Statics and Dynamics. Oxford University Press. (link: [1]) 3. ^ In general, the position of a point or particle is also known, in physics, as linear position, as opposed to the angular position of a line, or line segment (e.g., in circular motion, the "radius" joining the rotating point with the center of rotation), or basis set, or coordinate system. 4. ^ In kinematics, linear means "along a straight or curved line" (the path of the particle in space). In mathematics, however, linear has a different meaning. In both contexts, the word "linear" is related to the word "line". In mathematics, a line is often defined as a straight curve. For those who adopt this | definition, a curve can be straight, and curved lines are not supposed to exist. In kinematics, the term line is used as a synonym of the term trajectory, or path (namely, it has the same non-restricted meaning as that given, in mathematics, to the word curve). In short, both straight and curved lines are supposed to exist. In kinematics and dynamics, the following words refer to the same non-restricted meaning of the term "line": • "linear" (= along a straight or curved line), • "rectilinear" (= along a straight line, from Latin rectus = straight, and linere = spread), • "curvilinear" (=along a curved line, from Latin curvus = curved, and linere = spread). In topology and meteorology, the term "line" has the same meaning; namely, a contour line is a curve. 5. ^ Kane, Thomas; Levinson, David (1996). "2-4 Auxiliary Reference Frames". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc. 6. ^ a b Kane, Thomas; Levinson, David (1996). "2-6 Velocity and Acceleration". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc. 7. ^ a b c Kane, Thomas; Levinson, David (1996). "2-7 Two Points Fixed on a Rigid Body". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc. 8. ^ a b Kane, Thomas; Levinson, David (1996). "2-8 One Point Moving on a Rigid Body". Dynamics Online. Sunnyvale, California: OnLine Dynamics, Inc. ## References • Roy Featherstone (1987). Robot Dynamics Algorithms. Springer. ISBN 0-89838-230-0. This reference effectively combines screw theory with rigid body dynamics for robotic applications. The author also chooses to use spatial accelerations extensively in place of material accelerations as they simplify the equations and allow for compact notation. • JPL DARTS page has a section on spatial operator algebra (link: [2]) as well as an extensive list of references (link: [3]). • Andy Ruina and Rudra Pratap (2015). Introduction to Statics and Dynamics. Oxford University Press. (link: [4]). • Prof. Dr. Dennis M. Kochmann, Dynamics Lecture Notes, ETH Zurich. [5] |
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# Example: Lottery probability
What is the probability of winning a 4-number lottery? Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• arent there 4! ways we can write the winning numbers . i mean the order doesnt matter so 3,15,46,49 should be the same as 15,3,46,49 but sal says that theres only one way of getting the correct lottery numbers why is that?
• Achu and Naveen,
Sal could have solved this problem in two ways. He could have taken the number of possible permutations with a favorable outcome and divided that by the total possible number of permutations –or—he could have taken the number of possible combinations with a favorable outcome and divided that by the total number of possible combinations (which is what he did).
Let us do it both ways, using the permutations first.
As you mentioned, there a 4! ways of writing the four numbers. Another way to say this is that there are 4! different ways to order the four numbers –or-- there are 4! different permutations of the four numbers that give us the favorable outcome. This can be written as 4*3*2*1.
The total number of different permutations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!), which can be written as 60*59*58*57.
Our answer using permutations would be the number of favorable outcomes/the number of possible outcomes which would be (4*3*2*1)/(60*59*58*57). This simplifies to 1/487,635.
Using combinations, there is only one (1) combination of numbers that gives us that favorable outcome (that one way, achu).
The number of possible combinations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!*4!). This can be written as (60*59*58*57)/(4*3*2*1) . This number is 487,635.
Our answer using combinations would be the number of favorable outcomes/the number of possible outcomes which would be 1/487,635. This is the same answer we got using permutations.
Consider combinations and permutations to be different “units”. You would not say that after moving 3 inches you are halfway to traveling 6 meters, even though 3/6 = ½. You cannot take the number of possible favorable permutation outcomes and divide that by the number of total possible combination outcomes and get the correct probability. Likewise you cannot take the number of possible favorable combination outcomes and divide that by the number of total possible permutation outcomes and get the correct probability.
• Is there any reason why I could not solve the problem this way? Because I did and it turned out ok, but I don't always trust my own leaps of logic:
If I am hoping to draw 4 particular numbers randomly out of 60, then I can say that on my first draw there are 4 numbers that I could hope to get, out of a total of 60, so I begin with
4/60 as my chances of getting one of those numbers on that first draw.
On my second draw, there are now 3 numbers left that I want, out of a possible 59 total remaining, so I begin to multiply:
(4/60) * (3/59)
The last two draws follow logically: 2 acceptable numbers out of 58, and then 1 acceptable number out of 57. So the probability of drawing any particular set of 4 numbers out of 60, if we cannot draw any number twice, is:
(4/60) * (3/59) * (2/58) * (1/57)
I can just calculate this and get the same answer that Sal gets, but what do I make of the fact that this expression is equal to
4! / (60! / (60-4)!)
or
4! / P(60,4)
I am trying to draw some sort of larger logical conclusion out of this, but am having some trouble not confusing myself. It seems that I could say this represents the number of possible arrangements of any particular set of 4 numbers distributed over, or divided by, the total number of possible ways to draw all sets of 4 numbers out of 60.
This makes some sense to me, but I tend to come at these things backwards, using intuition and numbers first and then going back over the solution to see whether or not I could claim any larger guiding principles and/or formulas. Can anyone do a better job of explaining why my intuition was right?
• This sounds like a tautology but your intuition is right because it is right.
Let's look at the different ways of arriving to the answer.
Your way: look at the situation one draw at a time.
4/60 * 3/59 * 2/58 * 1/57, which is equal to 4*3*2*1 / (60*59*58*57)
Sal's way:
4! / (60 P 4)
which is the same as
4! / (60!/(60-4)!)
which is the same as
4! * (60-4)! / 60!
which is the same as
4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
which is the same as
4*3*2*1 / (60*59*58*57)
So your and Sal's ways are the same thing just expressed in different ways.
Here's one more way to look at it:
There are (60 C 4) ways of choosing four numbers (all different) from 60. Only one combination of them is what we have chosen. So the probability is
1/(60 C 4)
which is the same as
1/( 60!/(4!*(60-4)!))
which is the same as
4! * (60-4)! / 60!
which, as we already have seen, is the same as
4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1)
which is the same as
4*3*2*1 / (60*59*58*57)
• If S=1+2+4+8+16+32................ Can this be taken as S=1+2(1+2+4+8+16.......)???
If it can be taken,
S=1+2S
S=-1 Is this a valid answer for such a big question???
• Your reasoning only works when the sum S is a real number and does not continue on to infinity. For example, if S = 1 + 1/2 + 1/4 + ... + 1/(2^n) + ... and so on forever, then your logic says that S = 1 + 1/2(S), which gives the right answer of S = 2.
The difference between the two problems is that the sum S = 1 + 2 + 4 + ... continues to get larger forever and does not stop, while S = 1 + 1/2 + 1/4 + ... has a limit that it can never reach, no matter how long you count. By taking your approach, you can short-cut the infinite series and figure out that limit without having to take forever (literally!).
• just wanted to add my 2 cents. I'm having a hard time explaining it all though so would love feedback. For the lottery question, another way to think of it is as below.
explanation:
think of this top part of the probability (numerator) as 4p4 since you have 4 numbers to pick from and you want to pick 4 numbers, the number of ways you can pick 4 numbers from 4 numbers is 4*3*2*1. 4p4. This gives you the total number of non-unqiue ways to choose these numbers.
on the bottom, you know that you have 4 numbers to pick and 60 to pick from. your total number of ways to pick 4 numbers from 60 is 60p4.
I know that we dont care about order, but you see when you look for probability, the probability of the order not mattering is the same as the probability of the order mattering. because the numerator and denominator increase by factors of 4!. since when you think about it, the numerator is the number of ways total that you can combine 4 numbers divided by the denominator which is the total number of ways to pick 4 numbers of 60. Both top and bottom are including all the different permutations of 4 numbers. so really it's including those groupings of duplicated values. individually looking at each unique grouping's probability is the same as looking at the probability across all groupings.
Alternate way:
4c4/60c4 - removes the order. same answer.
QUESTION: what if we're drawing the numbers, what is the probability of drawing these 4 numbers. Does the math change / probability change?
Draw 1: 4 c 1
Draw 2: 3 c 1
Draw 3: 2 c 1
Draw 4: 1 c 1
fundamental counting principle lets us rep drawing 1,2,3,4 as 4c1*3c1*2c1*1c1
on the bottom you total ways of drawing per draw.
Draw 1: total ways = 60 c 1
Draw 2: 59 c 1
Draw 3: 58 c 1
Draw 4: 57 c 1
the probability of drawing is top / bottom.
QUESTION: should i use P or C? logically speaking which one makes more sense. I know they yield the same answer but which one fits this equation / changes the equation's logic.
Also, does this any of what i wrote up apply beyond jsut this question? Is this some unique solution methodology.
is it a rule that says you can't divide a combination by permutation?
e.g. 4c4/60p4. not allowed
because i see a lot of people wondering about 1/60*1/59*1/58*1/57 as a possible answer. My response is the top only shows 1 permutation of arranging the 4 numbers and the bottom represents the total number of ways of picking 4 numbers. so it's like 4c4/60p4 which incorrectly represents the solution. I think it's actually ok, its just not ok for this question.
because the top yes can be 4c4, it represents 1 but really depending on the person answering. they could be incorrect due to 2 different reasons. 1 not realizing that the order doesn't matter so their method of only using 1 / 60 etc only calculates the probability of 1 method of arranging those 60 and then also not realizing that the bottom looks at all possible permutation. OR they are looking at the top correctly as a single combination and not evaluating the bottom correclty and failing to use also combinations below.
Apologies for so many thoughts, probabilities is really interesting and difficult for me and writing up my thoughts helps me. I hope it helps someone else too.
• As long as you’re consistent, you will get the correct answer. It makes more sense to use the permutation method (for both top and bottom) if you think of the numbers as picked one at a time, but it makes more sense to use the combination method (for both top and bottom) if you think of all four numbers as being picked at once.
(1 vote)
• I was just wondering what the nCr and nPr buttons on the calculator do. My teacher explaned it, but i forgot what the do and how to use them. Please help!
(1 vote)
• nCr is used for Combinations, while nPr is used in permutations. 'N' represents the total number of items you have to choose from, and 'R' represents the number you choose. So if you had 36C10, that would mean you have 36 items and you can choose 10, regardless of order, since it is a Combination.
• Isn't 59 factorial (!), 60*59*58*57*56*all the way down to 0??
(1 vote)
• No, there's no 60 or 0 involved. It's 59 through 1.
• if in this lottery, picking a number and putting it back is allowed... so that means you can pick a number a multiple of times... what would the probability be then?
64! / ((64-4!) * 4! ) ?
• Well, you'd choose 4 numbers from 60 numbers (1 to 60) and repetition is allowed, the probability of winning would be 1/(60^4/4!) = 4!/60^4 = 1/540000 ≈ 0.000002
(The 4! is there because the order of the numbers still doesn't matter. If the winning row was 7,34,34,50 and you had 34,7,50,34 you'd still win)
• I think I may have a fundamental misunderstanding of combinations and / or permutations. I tried to solve this problem by doing the following (60! / (56! * 4!)) / (60^4) which is the combinations formula divided by (I thought) the total number of possible outcomes with 60 numbers in 4 slots. Why is that incorrect?
• 60^4 isn't the total number of possible groups of 4, because the order of the 4 numbers doesn't matter for combinations. 60^4 is the number of permutations, not combinations.
and 60!/(56!*4!) is the total number of possible combinations of 4 numbers, so it is the sample space, not the # of successes meaning it should be the denominator.
• what if you want to know the probability of a number winning excluding some number already played that will not be played again?
• That's a fun calculation. Say you have 7 different items in a bag. The probability of pulling a certain one out is 1/7. After doing so you now have 6 items. The next time you pull one out the probability will be 1/6. Makes sense?
This reasoning is used in many situations where multiple items need to be pulled out of the bag in quick succession.
(1 vote)
• In the same Sal's case - 4 number ticket, let's assume that if I pay a little bit more for the lottery ticket I'm able to pick 5 numbers instead of 4. Then I would have a higher probability of winning. But how do I calculate that probability with an "extra number"?
(1 vote)
• Let find the easy part, which is the total # of outcomes. This is just going to be the number of different ways in which the host of the lottery prize is going to choose 4 numbers out of 60. This is just C(60,4).
Now the # of ways in which our event can happen is just the number of ways in which those 4 numbers match exactly 4 of our 5 numbers. i.e. 4 out 5 numbers: C(5,4).
Thus the probability is:
5/487,635 = 1/97,527
And it makes sense. If you paid the number 1, 2, 3 ,4 and 5, you will win in these situations:
[1 - 2 - 3 - 4]
[1 - 2 - 3 - 5]
[1 - 2 - 4 -5]
[1 - 3 - 4 -5]
[2 - 3 - 4 -5] | 3,420 | 12,803 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2023-23 | longest | en | 0.926465 | If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. # Example: Lottery probability What is the probability of winning a 4-number lottery? Created by Sal Khan and Monterey Institute for Technology and Education. ## Want to join the conversation? • arent there 4! ways we can write the winning numbers . i mean the order doesnt matter so 3,15,46,49 should be the same as 15,3,46,49 but sal says that theres only one way of getting the correct lottery numbers why is that? • Achu and Naveen, Sal could have solved this problem in two ways. He could have taken the number of possible permutations with a favorable outcome and divided that by the total possible number of permutations –or—he could have taken the number of possible combinations with a favorable outcome and divided that by the total number of possible combinations (which is what he did). Let us do it both ways, using the permutations first. As you mentioned, there a 4! ways of writing the four numbers. Another way to say this is that there are 4! different ways to order the four numbers –or-- there are 4! different permutations of the four numbers that give us the favorable outcome. This can be written as 4*3*2*1. The total number of different permutations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!), which can be written as 60*59*58*57. Our answer using permutations would be the number of favorable outcomes/the number of possible outcomes which would be (4*3*2*1)/(60*59*58*57). This simplifies to 1/487,635. Using combinations, there is only one (1) combination of numbers that gives us that favorable outcome (that one way, achu). The number of possible combinations of 4 numbers taken out of 60 different numbers is 60!/((60-4)!*4!). This can be written as (60*59*58*57)/(4*3*2*1) . This number is 487,635. Our answer using combinations would be the number of favorable outcomes/the number of possible outcomes which would be 1/487,635. This is the same answer we got using permutations. Consider combinations and permutations to be different “units”. You would not say that after moving 3 inches you are halfway to traveling 6 meters, even though 3/6 = ½. You cannot take the number of possible favorable permutation outcomes and divide that by the number of total possible combination outcomes and get the correct probability. Likewise you cannot take the number of possible favorable combination outcomes and divide that by the number of total possible permutation outcomes and get the correct probability. • Is there any reason why I could not solve the problem this way? Because I did and it turned out ok, but I don't always trust my own leaps of logic: If I am hoping to draw 4 particular numbers randomly out of 60, then I can say that on my first draw there are 4 numbers that I could hope to get, out of a total of 60, so I begin with 4/60 as my chances of getting one of those numbers on that first draw. On my second draw, there are now 3 numbers left that I want, out of a possible 59 total remaining, so I begin to multiply: (4/60) * (3/59) The last two draws follow logically: 2 acceptable numbers out of 58, and then 1 acceptable number out of 57. So the probability of drawing any particular set of 4 numbers out of 60, if we cannot draw any number twice, is: (4/60) * (3/59) * (2/58) * (1/57) I can just calculate this and get the same answer that Sal gets, but what do I make of the fact that this expression is equal to 4! / (60! / (60-4)!) or 4! / P(60,4) I am trying to draw some sort of larger logical conclusion out of this, but am having some trouble not confusing myself. It seems that I could say this represents the number of possible arrangements of any particular set of 4 numbers distributed over, or divided by, the total number of possible ways to draw all sets of 4 numbers out of 60. This makes some sense to me, but I tend to come at these things backwards, using intuition and numbers first and then going back over the solution to see whether or not I could claim any larger guiding principles and/or formulas. Can anyone do a better job of explaining why my intuition was right? • This sounds like a tautology but your intuition is right because it is right. Let's look at the different ways of arriving to the answer. Your way: look at the situation one draw at a time. 4/60 * 3/59 * 2/58 * 1/57, which is equal to 4*3*2*1 / (60*59*58*57) Sal's way: 4! / (60 P 4) which is the same as 4! / (60!/(60-4)!) which is the same as 4! * (60-4)! / 60! which is the same as 4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1) which is the same as 4*3*2*1 / (60*59*58*57) So your and Sal's ways are the same thing just expressed in different ways. Here's one more way to look at it: There are (60 C 4) ways of choosing four numbers (all different) from 60. Only one combination of them is what we have chosen. So the probability is 1/(60 C 4) which is the same as 1/( 60!/(4!*(60-4)!)) which is the same as 4! * (60-4)! / 60! which, as we already have seen, is the same as 4*3*2*1 * 56*55*54*...3*2*1 / (60*59*58*57 * 56*55*54*...3*2*1) which is the same as 4*3*2*1 / (60*59*58*57) • If S=1+2+4+8+16+32................ Can this be taken as S=1+2(1+2+4+8+16.......)??? If it can be taken, S=1+2S S=-1 Is this a valid answer for such a big question??? • Your reasoning only works when the sum S is a real number and does not continue on to infinity. For example, if S = 1 + 1/2 + 1/4 + ... + 1/(2^n) + ... and so on forever, then your logic says that S = 1 + 1/2(S), which gives the right answer of S = 2. The difference between the two problems is that the sum S = 1 + 2 + 4 + ... continues to get larger forever and does not stop, while S = 1 + 1/2 + 1/4 + ... has a limit that it can never reach, no matter how long you count. By taking your approach, you can short-cut the infinite series and figure out that limit without having to take forever (literally!). • just wanted to add my 2 cents. I'm having a hard time explaining it all though so would love feedback. For the lottery question, another way to think of it is as below. explanation: think of this top part of the probability (numerator) as 4p4 since you have 4 numbers to pick from and you want to pick 4 numbers, the number of ways you can pick 4 numbers from 4 numbers is 4*3*2*1. 4p4. This gives you the total number of non-unqiue ways to choose these numbers. on the bottom, you know that you have 4 numbers to pick and 60 to pick from. your total number of ways to pick 4 numbers from 60 is 60p4. I know that we dont care about order, but you see when you look for probability, the probability of the order not mattering is the same as the probability of the order mattering. because the numerator and denominator increase by factors of 4!. since when you think about it, the numerator is the number of ways total that you can combine 4 numbers divided by the denominator which is the total number of ways to pick 4 numbers of 60. Both top and bottom are including all the different permutations of 4 numbers. so really it's including those groupings of duplicated values. individually looking at each unique grouping's probability is the same as looking at the probability across all groupings. Alternate way: 4c4/60c4 - removes the order. same answer. QUESTION: what if we're drawing the numbers, what is the probability of drawing these 4 numbers. Does the math change / probability change? Draw 1: 4 c 1 Draw 2: 3 c 1 Draw 3: 2 c 1 Draw 4: 1 c 1 fundamental counting principle lets us rep drawing 1,2,3,4 as 4c1*3c1*2c1*1c1 on the bottom you total ways of drawing per draw. Draw 1: total ways = 60 c 1 Draw 2: 59 c 1 Draw 3: 58 c 1 Draw 4: 57 c 1 the probability of drawing is top / bottom. QUESTION: should i use P or C? logically speaking which one makes more sense. I know they yield the same answer but which one fits this equation / changes the equation's logic. Also, does this any of what i wrote up apply beyond jsut this question? Is this some unique solution methodology. is it a rule that says you can't divide a combination by permutation? e.g. 4c4/60p4. not allowed because i see a lot of people wondering about 1/60*1/59*1/58*1/57 as a possible answer. My response is the top only shows 1 permutation of arranging the 4 numbers and the bottom represents the total number of ways of picking 4 numbers. so it's like 4c4/60p4 which incorrectly represents the solution. I think it's actually ok, its just not ok for this question. because the top yes can be 4c4, it represents 1 but really depending on the person answering. they could be incorrect due to 2 different reasons. 1 not realizing that the order doesn't matter so their method of only using 1 / 60 etc only calculates the probability of 1 method of arranging those 60 and then also not realizing that the bottom looks at all possible permutation. OR they are looking at the top correctly as a single combination and not evaluating the bottom correclty and failing to use also combinations below. Apologies for so many thoughts, probabilities is really interesting and difficult for me and writing up my thoughts helps me. I hope it helps someone else too. • As long as you’re consistent, you will get the correct answer. It makes more sense to use the permutation method (for both top and bottom) if you think of the numbers as picked one at a time, but it makes more sense to use the combination method (for both top and bottom) if you think of all four numbers as being picked at once. (1 vote) • I was just wondering what the nCr and nPr buttons on the calculator do. My teacher explaned it, but i forgot what the do and how to use them. Please help! (1 vote) • nCr is used for Combinations, while nPr is used in permutations. 'N' represents the total number of items you have to choose from, and 'R' represents the number you choose. So if you had 36C10, that would mean you have 36 items and you can choose 10, regardless of order, since it is a Combination. • Isn't 59 factorial (!), 60*59*58*57*56*all the way down to 0?? (1 vote) • No, there's no 60 or 0 involved. It's 59 through 1. • if in this lottery, picking a number and putting it back is allowed... so that means you can pick a number a multiple of times... what would the probability be then? 64! / ((64-4!) * 4! ) ? • Well, you'd choose 4 numbers from 60 numbers (1 to 60) and repetition is allowed, the probability of winning would be 1/(60^4/4!) = 4!/60^4 = 1/540000 ≈ 0.000002 (The 4! is there because the order of the numbers still doesn't matter. If the winning row was 7,34,34,50 and you had 34,7,50,34 you'd still win) • I think I may have a fundamental misunderstanding of combinations and / or permutations. I tried to solve this problem by doing the following (60! / (56! * 4!)) / (60^4) which is the combinations formula divided by (I thought) the total number of possible outcomes with 60 numbers in 4 slots. Why is that incorrect? • 60^4 isn't the total number of possible groups of 4, because the order of the 4 numbers doesn't matter for combinations. 60^4 is the number of permutations, not combinations. and 60!/(56!*4!) is the total number of possible combinations of 4 numbers, so it is the sample space, not the # of successes meaning it should be the denominator. • what if you want to know the probability of a number winning excluding some number already played that will not be played again? • That's a fun calculation. Say you have 7 different items in a bag. The probability of pulling a certain one out is 1/7. After doing | so you now have 6 items. The next time you pull one out the probability will be 1/6. Makes sense? This reasoning is used in many situations where multiple items need to be pulled out of the bag in quick succession. (1 vote) • In the same Sal's case - 4 number ticket, let's assume that if I pay a little bit more for the lottery ticket I'm able to pick 5 numbers instead of 4. Then I would have a higher probability of winning. But how do I calculate that probability with an "extra number"? (1 vote) • Let find the easy part, which is the total # of outcomes. This is just going to be the number of different ways in which the host of the lottery prize is going to choose 4 numbers out of 60. This is just C(60,4). Now the # of ways in which our event can happen is just the number of ways in which those 4 numbers match exactly 4 of our 5 numbers. i.e. 4 out 5 numbers: C(5,4). Thus the probability is: 5/487,635 = 1/97,527 And it makes sense. If you paid the number 1, 2, 3 ,4 and 5, you will win in these situations: [1 - 2 - 3 - 4] [1 - 2 - 3 - 5] [1 - 2 - 4 -5] [1 - 3 - 4 -5] [2 - 3 - 4 -5] |
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# If x is a positive number
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If x is a positive number [#permalink]
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26 Aug 2012, 06:24
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If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
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26 Aug 2012, 06:42
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If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
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Re: If x is a positive number [#permalink]
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26 Aug 2012, 09:13
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syog wrote:
If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
Neither (1) nor (2) alone is sufficient.
(1) and (2) together:
From (1) $$3x=2k,$$ where $$k$$ is a positive integer.
From (2) $$5x=2m,$$ where $$m$$ is some positive integer. Necessarily $$\frac{2k}{3}=\frac{2m}{5}$$ from which $$5k=3m.$$
$$k$$ and $$m$$ being integers, necessarily $$k$$ must be a multiple of 3 (because 5 is not divisible by 3), so $$k=3a$$ for some positive integer $$a.$$
It follows that $$x=\frac{2k}{3}=2a$$ so $$x$$ is even.
Sufficient.
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Re: If x is a positive number [#permalink]
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30 Aug 2012, 04:53
Bunuel wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
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Re: If x is a positive number [#permalink]
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30 Aug 2012, 05:18
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ziko wrote:
Bunuel wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions.
So, there is no interchangeable word for "integer" on the GMAT.
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Re: If x is a positive number [#permalink]
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30 Aug 2012, 19:50
great job
great way to solve this problem..
i didnt reach the answer as you did
ds has been troubling me like anything
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Re: If x is a positive number [#permalink]
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02 Sep 2013, 08:23
Bunuel wrote:
ziko wrote:
Bunuel wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions.
So, there is no interchangeable word for "integer" on the GMAT.
I am a little confused.
1) 3x is an even integer.
Let x=4/3; then 3x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
2) 5x is an even integer.
Let x=4/5; then 5x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
(1+2): 15x is an integer.
Let x=4/15; then 15x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
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Re: If x is a positive number [#permalink]
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02 Sep 2013, 08:42
domfrancondumas wrote:
Bunuel wrote:
ziko wrote:
If x is a positive number, is x an even integer?
Notice that we are not told that x is an integer.
(1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient.
(2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient.
(1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient.
My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer?
I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions.
So, there is no interchangeable word for "integer" on the GMAT.
I am a little confused.
1) 3x is an even integer.
Let x=4/3; then 3x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
2) 5x is an even integer.
Let x=4/5; then 5x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
(1+2): 15x is an integer.
Let x=4/15; then 15x=4. But in this case x is not an even integer. Hence INSUFFICIENT.
Notice that x cannot be 4/15, because in this case 3x=12/15 which is NO an even integer and 5x=20/15 which is also NOT an even integer, so in this case both statements are violated.
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Re: If x is a positive number [#permalink]
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10 Jan 2017, 07:31
If it's given that "5 is a factor of x". Is it correct to assume that x is an integer? If yes, how and why? Please explain. Also explain if the answer to my question is No. Consider a case wherein x is 5/2. Isn't 5 a factor of x? Please address, thanks a lot in advance
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Re: If x is a positive number [#permalink]
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10 Jan 2017, 10:12
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kooks123 wrote:
If it's given that "5 is a factor of x". Is it correct to assume that x is an integer? If yes, how and why? Please explain. Also explain if the answer to my question is No. Consider a case wherein x is 5/2. Isn't 5 a factor of x? Please address, thanks a lot in advance
On the GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that:
1. $$a$$ is an integer;
2. $$b$$ is an integer;
3. $$\frac{a}{b}=integer$$.
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Re: If x is a positive number [#permalink]
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11 Jan 2017, 03:48
x is positive Integer means we need to think only in positive fractions and integers.
(1) 3x is an even integer.
Let x= 2.....3x=6..........x is even integer
Let x=2/3....3x=2.........x is NOT even integer
Insufficient
(2) 5x is an even integer.
Let x= 2........5x=10..........x is even integer
Let x=2/5.......5x=2...........x is NOT even integer
Insufficient
Combining 1 +2 ,
There is not fraction could be multiplied simultaneously to 3 & 5 and give EVEN INTEGER. We left with that x=EVEN INTEGER to give make both 3x & 5x even integers.
As 3 & 5 are odd numbers then x needs to be EVEN Integer to make both statements Even integers.
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Re: If x is a positive number [#permalink]
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11 Jan 2017, 12:01
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Hi Ellipse,
Whenever you combine two statements in a DS question, the best thing to do instead of plugging in values is to use one statement into the other or use a mathematical operation between the two statements. The mathematical operation(s) that you choose to perform must lead to your target question. In this case the target question is 'Is x an even integer'?
Statement 1 : 3x is an even integer
Here x can either be an even integer such as 2, 4, 6.... or it can be a fraction such as 2/3, 4/3..... So x can be an even integer or a fraction. Insufficient.
Statement 2 : 5x is an even integer
Here again x can be an even integer such as 2, 4, 6.... or it can be a fraction such as 2/5, 4/5.... So x again can either be an even integer or a fraction. Insufficient.
Now instead of recycling the values it makes sense to use a mathematical operation (or mathematical operations) between the two statements.
Let us multiply the first statement by 2, since 3x = even integer ; 3x * 2 = even integer * 2 -----> 6x = even integer
Statement 2 says that 5x is an even integer and by multiplying statement 1 by 2 we have 6x to be an even integer. Subtracting the two we get
6x - 5x = even integer - even integer -----> x = even integer. Sufficient.
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Re: If x is a positive number [#permalink]
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19 Apr 2017, 17:56
Ellipse wrote:
If x is a positive number, is x an even integer?
(1) 3x is an even integer.
(2) 5x is an even integer.
Statement 1:
3 (2/3) could be an even number- the questin says x is a positive number...not necessarily an integer
Insufficient
Statement 2
5(2/5) btw again here were are using a counterexample to establish that x does not necessarily have to be an integer
10/5 =2
Insufficient
Statement 1 and 2:
Statement 1 and Statement 2
Knowing from statement 1 and 2 that 3x and 5x are even integers we can use solve this question using algebra instead of imagining numbers and testing various cases- an even integer minus an even integer is always an even integer so
5x-3x=2x
5(4)-3(4)=2(4)
Re: If x is a positive number [#permalink] 19 Apr 2017, 17:56
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15 Is the number x positive? 12 09 Feb 2017, 12:07
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Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If x is a positive number Author Message TAGS: ### Hide Tags Manager Joined: 16 Oct 2010 Posts: 85 Location: United States Concentration: Finance, Entrepreneurship GMAT 1: 700 Q49 V35 WE: Information Technology (Investment Banking) Followers: 0 Kudos [?]: 65 [4] , given: 3 If x is a positive number [#permalink] ### Show Tags 26 Aug 2012, 06:24 4 KUDOS 13 This post was BOOKMARKED 00:00 Difficulty: 75% (hard) Question Stats: 40% (01:41) correct 60% (00:30) wrong based on 349 sessions ### HideShow timer Statistics If x is a positive number, is x an even integer? (1) 3x is an even integer. (2) 5x is an even integer. [Reveal] Spoiler: OA Math Expert Joined: 02 Sep 2009 Posts: 38921 Followers: 7744 Kudos [?]: 106370 [5] , given: 11622 Re: If x is a positive number [#permalink] ### Show Tags 26 Aug 2012, 06:42 5 KUDOS Expert's post 6 This post was BOOKMARKED If x is a positive number, is x an even integer? Notice that we are not told that x is an integer. (1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient. (2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient. (1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient. _________________ Director Joined: 22 Mar 2011 Posts: 612 WE: Science (Education) Followers: 101 Kudos [?]: 948 [0], given: 43 Re: If x is a positive number [#permalink] ### Show Tags 26 Aug 2012, 09:13 2 This post was BOOKMARKED syog wrote: If x is a positive number, is x an even integer? (1) 3x is an even integer. (2) 5x is an even integer. Neither (1) nor (2) alone is sufficient. (1) and (2) together: From (1) $$3x=2k,$$ where $$k$$ is a positive integer. From (2) $$5x=2m,$$ where $$m$$ is some positive integer. Necessarily $$\frac{2k}{3}=\frac{2m}{5}$$ from which $$5k=3m.$$ $$k$$ and $$m$$ being integers, necessarily $$k$$ must be a multiple of 3 (because 5 is not divisible by 3), so $$k=3a$$ for some positive integer $$a.$$ It follows that $$x=\frac{2k}{3}=2a$$ so $$x$$ is even. Sufficient. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Manager Joined: 28 Feb 2012 Posts: 115 GPA: 3.9 WE: Marketing (Other) Followers: 0 Kudos [?]: 45 [0], given: 17 Re: If x is a positive number [#permalink] ### Show Tags 30 Aug 2012, 04:53 Bunuel wrote: If x is a positive number, is x an even integer? Notice that we are not told that x is an integer. (1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient. (2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient. (1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient. My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer? _________________ If you found my post useful and/or interesting - you are welcome to give kudos! Math Expert Joined: 02 Sep 2009 Posts: 38921 Followers: 7744 Kudos [?]: 106370 [0], given: 11622 Re: If x is a positive number [#permalink] ### Show Tags 30 Aug 2012, 05:18 Expert's post 2 This post was BOOKMARKED ziko wrote: Bunuel wrote: If x is a positive number, is x an even integer? Notice that we are not told that x is an integer. (1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient. (2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient. (1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient. My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer? I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions. So, there is no interchangeable word for "integer" on the GMAT. _________________ Manager Status: exam is close ... dont know if i ll hit that number Joined: 06 Jun 2011 Posts: 197 Location: India GMAT Date: 10-09-2012 GPA: 3.2 Followers: 2 Kudos [?]: 25 [0], given: 1 Re: If x is a positive number [#permalink] ### Show Tags 30 Aug 2012, 19:50 great job great way to solve this problem.. i didnt reach the answer as you did ds has been troubling me like anything _________________ just one more month for exam... Intern Joined: 11 Jul 2013 Posts: 43 Followers: 0 Kudos [?]: 10 [0], given: 92 Re: If x is a positive number [#permalink] ### Show Tags 02 Sep 2013, 08:23 Bunuel wrote: ziko wrote: Bunuel wrote: If x is a positive number, is x an even integer? Notice that we are not told that x is an integer. (1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient. (2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient. (1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient. My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer? I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions. So, there is no interchangeable word for "integer" on the GMAT. I am a little confused. 1) 3x is an even integer. Let x=4/3; then 3x=4. But in this case x is not an even integer. Hence INSUFFICIENT. 2) 5x is an even integer. Let x=4/5; then 5x=4. But in this case x is not an even integer. Hence INSUFFICIENT. (1+2): 15x is an integer. Let x=4/15; then 15x=4. But in this case x is not an even integer. Hence INSUFFICIENT. Math Expert Joined: 02 Sep 2009 Posts: 38921 Followers: 7744 Kudos [?]: 106370 [0], given: 11622 Re: If x is a positive number [#permalink] ### Show Tags 02 Sep 2013, 08:42 domfrancondumas wrote: Bunuel wrote: ziko wrote: If x is a positive number, is x an even integer? Notice that we are not told that x is an integer. (1) 3x is an even integer. x could be ANY even number or some fraction (for example 2/3), so this statement is NOT sufficient. (2) 5x is an even integer. The same here, x could be ANY even number or some fraction (for example 2/5). Not sufficient. (1)+(2) We have that 3x=even and 5x=even. Subtract one from another: 5x-3x=even-even --> 2x=even --> x=even/2=integer. Now, x=integer and 3x=even (from 1) means that x must be an even integer. Sufficient. My answer to this question was D, both sufficient, however my assumption was that in GMAT number and integer are interchangible words, but as i see i was wrong. Bunuel could you please remind what word was interchangible with word integer? I think you refer to Natural Numbers, which are non-negative (or positive) integers but GMAT doesn't use words "Natural Number" in their questions. So, there is no interchangeable word for "integer" on the GMAT. I am a little confused. 1) 3x is an even integer. Let x=4/3; then 3x=4. But in this case x is not an even integer. Hence INSUFFICIENT. 2) 5x is an even integer. Let x=4/5; then 5x=4. But in this case x is not an even integer. Hence INSUFFICIENT. (1+2): 15x is an integer. Let x=4/15; then 15x=4. But in this case x is not an even integer. Hence INSUFFICIENT. Notice that x cannot be 4/15, because in this case 3x=12/15 which is NO an even integer and 5x=20/15 which is also NOT an even integer, so in this case both statements are violated. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 15493 Followers: 651 Kudos [?]: 210 [0], given: 0 Re: If x is a positive number [#permalink] ### Show Tags 10 Oct 2014, 10:16 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 15493 Followers: 651 Kudos [?]: 210 [0], given: 0 Re: If x is a positive number [#permalink] ### Show Tags 24 Aug 2016, 13:05 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Intern Joined: 13 Feb 2015 Posts: 8 Followers: 0 Kudos [?]: 0 [0], given: 191 Re: If x is a positive number [#permalink] ### Show Tags 10 Jan 2017, 07:31 If it's given that "5 is a factor of x". Is it correct to assume that x is an integer? If yes, how and why? Please explain. Also explain if the answer to my question is No. Consider a case wherein x is 5/2. Isn't 5 a factor of x? Please address, thanks a lot in advance _________________ Before getting into the options, have an idea of what you seek Thanks & Regards, Vipul Chhabra Math Expert Joined: 02 Sep 2009 Posts: 38921 Followers: 7744 Kudos [?]: 106370 [1] , given: 11622 Re: If x is a positive number [#permalink] ### Show Tags 10 Jan 2017, 10:12 1 KUDOS Expert's post kooks123 wrote: If it's given that "5 is a factor of x". Is it correct to assume that x is an integer? If yes, how and why? Please explain. Also explain if the answer to my question is No. Consider a case wherein x is 5/2. Isn't 5 a factor of x? Please address, thanks a lot in advance On the GMAT when we are told that $$a$$ is divisible by $$b$$ (or which is the same: "$$a$$ is multiple of $$b$$", or "$$b$$ is a factor of $$a$$"), we can say that: 1. $$a$$ is an integer; 2. $$b$$ is an integer; 3. $$\frac{a}{b}=integer$$. _________________ Director Joined: 26 Mar 2013 Posts: 946 Followers: 7 Kudos [?]: 202 [0], given: 122 Re: If x is a positive number [#permalink] ### Show Tags 11 Jan 2017, 03:48 x is positive Integer means we need to think only in positive fractions and integers. (1) 3x is an even integer. Let x= 2.....3x=6..........x is even integer Let x=2/3....3x=2.........x is NOT even integer Insufficient (2) 5x is an even integer. Let x= 2........5x=10..........x is even integer Let x=2/5.......5x=2...........x is NOT even integer Insufficient Combining 1 +2 , There is not fraction could be multiplied simultaneously to 3 & 5 and give EVEN INTEGER. We left with that x=EVEN INTEGER to give make both 3x & 5x even integers. As 3 & 5 are odd numbers then x needs to be EVEN Integer to make both statements Even integers. Director Affiliations: CrackVerbal Joined: 03 Oct 2013 Posts: 501 Location: India GMAT 1: 780 Q51 V46 Followers: 93 Kudos [?]: 462 [0], given: 6 Re: If x is a positive number [#permalink] ### Show Tags 11 Jan 2017, 12:01 Top Contributor Hi Ellipse, Whenever you combine two statements in a DS question, the best thing to do instead of plugging in values is to use one statement into the other or use a mathematical operation between the two statements. The mathematical operation(s) that you choose to perform must lead to your target question. In this case the target question is 'Is x an even integer'? Statement 1 : 3x is an even integer Here x can either be an even integer such as 2, 4, 6.... or it can be a fraction such as 2/3, 4/3..... So x can be an even integer or a fraction. Insufficient. Statement 2 : 5x is an even integer Here again x can be an even integer such as 2, 4, 6.... or it can be a fraction such as 2/5, 4/5.... So x again can either be an even integer or a fraction. Insufficient. Now instead of recycling the values it makes sense to use a mathematical operation (or mathematical operations) between the two statements. Let us multiply the first statement by 2, since 3x = even integer ; 3x * 2 = even integer * 2 -----> 6x = even integer Statement 2 says that 5x is an even integer and by multiplying statement 1 by 2 we have 6x to be an even integer. Subtracting the two we get 6x - | 5x = even integer - even integer -----> x = even integer. Sufficient. _________________ Enroll for our GMAT Trial Course here - http://gmatonline.crackverbal.com/ Learn all PS and DS strategies here- http://gmatonline.crackverbal.com/p/mastering-quant-on-gmat Manager Joined: 12 Nov 2016 Posts: 195 Followers: 0 Kudos [?]: 7 [0], given: 135 Re: If x is a positive number [#permalink] ### Show Tags 19 Apr 2017, 17:56 Ellipse wrote: If x is a positive number, is x an even integer? (1) 3x is an even integer. (2) 5x is an even integer. Statement 1: 3 (2/3) could be an even number- the questin says x is a positive number...not necessarily an integer Insufficient Statement 2 5(2/5) btw again here were are using a counterexample to establish that x does not necessarily have to be an integer 10/5 =2 Insufficient Statement 1 and 2: Statement 1 and Statement 2 Knowing from statement 1 and 2 that 3x and 5x are even integers we can use solve this question using algebra instead of imagining numbers and testing various cases- an even integer minus an even integer is always an even integer so 5x-3x=2x 5(4)-3(4)=2(4) Re: If x is a positive number [#permalink] 19 Apr 2017, 17:56 Similar topics Replies Last post Similar Topics: 4 If x and y are positive numbers 5 25 May 2016, 04:04 3 Is x a positive number? 5 06 Oct 2015, 08:19 15 For a positive number x, {x} is the fractional part of x. 7 21 Jul 2016, 13:00 5 Is the positive number 'x' an integer? 5 14 Jun 2016, 09:29 15 Is the number x positive? 12 09 Feb 2017, 12:07 Display posts from previous: Sort by |
https://studyadda.com/question-bank/case-based-mcqs-triangles_q4/5960400/5961290 | 1,709,369,672,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475806.52/warc/CC-MAIN-20240302084508-20240302114508-00370.warc.gz | 533,289,435 | 21,427 | • question_answer Find the Length of the rope used. A) $26\,m$ B) $25\,m$ C) $27\,m$ D) $23\,m$
Now, Perimeter of $DPRS=PR+RS+SP$ $=17+15+8=40\,m$ $\therefore$ Length of the rope = Perimeter of $\Delta PRS-PQ$ $=40-13=27\,m$ So, option [c] Is correct, | 112 | 254 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-10 | latest | en | 0.435708 | • question_answer Find the Length of the rope used. A) $26\,m$ B) $25\,m$ C) $27\,m$ D) $23\,m$ Now, Perimeter of $DPRS=PR+RS+SP$ $=17+15+8=40\,m$ $\therefore$ Length of the rope = Perimeter of $\Delta PRS-PQ$ $=40-13=27\,m$ So, | option [c] Is correct, |
https://socratic.org/questions/how-to-prove-sin-2-theta-cos-2-theta-1-2cos-2-theta | 1,723,124,386,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00653.warc.gz | 432,012,635 | 5,770 | # How to prove sin^2 theta - cos^2 theta = 1 -2cos^2 theta ?
Jul 17, 2018
#### Explanation:
Given: Prove ${\sin}^{2} \theta - {\cos}^{2} \theta = 1 - 2 {\cos}^{2} \theta$
Use the Pythagorean Identity: $\text{ } {\sin}^{2} \theta + {\cos}^{2} \theta = 1$
Rearrange the identity: $\text{ } {\sin}^{2} \theta = 1 - {\cos}^{2} \theta$
Start with the left side of the equation and substitute the rearranged identity:
${\sin}^{2} \theta - {\cos}^{2} \theta = 1 - {\cos}^{2} \theta - {\cos}^{2} \theta$
Combine like-terms:
$= 1 - 2 {\cos}^{2} \theta$ This is the right-side
Q.E.D. | 221 | 583 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-33 | latest | en | 0.550948 | # How to prove sin^2 theta - cos^2 theta = 1 -2cos^2 theta ? Jul 17, 2018 #### Explanation: Given: Prove ${\sin}^{2} \theta - {\cos}^{2} \theta = 1 - 2 {\cos}^{2} \theta$ Use the Pythagorean Identity: $\text{ } {\sin}^{2} \theta + {\cos}^{2} \theta = 1$ Rearrange the identity: $\text{ } {\sin}^{2} \theta = 1 - {\cos}^{2} \theta$ Start with the left side of the equation and substitute the rearranged identity: ${\sin}^{2} \theta - {\cos}^{2} \theta = 1 - {\cos}^{2} \theta - {\cos}^{2} \theta$ Combine like-terms: $= | 1 - 2 {\cos}^{2} \theta$ This is the right-side Q.E.D. |
https://numbersworksheet.com/multiply-whole-numbers-by-powers-of-10-worksheet/ | 1,723,124,505,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00216.warc.gz | 331,478,488 | 13,825 | # Multiply Whole Numbers By Powers Of 10 Worksheet
This multiplication worksheet concentrates on educating college students how you can mentally grow total phone numbers. College students can use custom made grids to suit precisely one concern. The worksheets also includefractions and decimals, and exponents. There are also multiplication worksheets with a dispersed home. These worksheets really are a have to-have to your math type. They could be found in type to discover ways to psychologically increase entire line and numbers them up. Multiply Whole Numbers By Powers Of 10 Worksheet.
## Multiplication of complete figures
You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets will help you learn this standard principle. You are able to decide to use one particular digit multipliers or two-digit and three-digit multipliers. Power of 10 will also be a fantastic choice. These worksheets will assist you to exercise long practice and multiplication looking at the numbers. Also, they are the best way to support your youngster recognize the necessity of understanding the several types of complete figures.
## Multiplication of fractions
Getting multiplication of fractions on a worksheet may help educators plan and make lessons successfully. Utilizing fractions worksheets allows professors to quickly evaluate students’ idea of fractions. College students may be questioned to end the worksheet within a certain time and then mark their techniques to see where that they need additional instructions. Individuals can usually benefit from phrase problems that relate maths to real-lifestyle situations. Some fractions worksheets involve samples of contrasting and comparing figures.
## Multiplication of decimals
Once you grow two decimal figures, ensure that you group them vertically. If you want to multiply a decimal number with a whole number, the product must contain the same number of decimal places as the multiplicant. As an example, 01 by (11.2) by 2 would be equivalent to 01 x 2.33 by 11.2 except when this product has decimal places of lower than two. Then, the item is rounded for the nearby whole quantity.
## Multiplication of exponents
A mathematics worksheet for Multiplication of exponents will allow you to practice dividing and multiplying numbers with exponents. This worksheet will also supply problems that will demand college students to flourish two diverse exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Apart from, you can also get into special directions in the worksheet on its own. When you’re concluded, you are able to simply click “Produce” and also the worksheet will be delivered electronically.
## Division of exponents
The standard principle for division of exponents when multiplying numbers would be to deduct the exponent from the denominator from the exponent in the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. As an example, \$23 split by 4 will identical 27. However, this method is not always accurate. This technique can lead to uncertainty when multiplying amounts which are too big or not big enough.
## Linear characteristics
You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So, the total rent would be \$470. A linear function of this kind has got the type f(by), where ‘x’ is the number of days and nights the car was rented. In addition, it provides the form f(by) = ax b, where by ‘b’ and ‘a’ are actual figures. | 713 | 3,659 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2024-33 | latest | en | 0.898901 | # Multiply Whole Numbers By Powers Of 10 Worksheet This multiplication worksheet concentrates on educating college students how you can mentally grow total phone numbers. College students can use custom made grids to suit precisely one concern. The worksheets also includefractions and decimals, and exponents. There are also multiplication worksheets with a dispersed home. These worksheets really are a have to-have to your math type. They could be found in type to discover ways to psychologically increase entire line and numbers them up. Multiply Whole Numbers By Powers Of 10 Worksheet. ## Multiplication of complete figures You should consider purchasing a multiplication of whole numbers worksheet if you want to improve your child’s math skills. These worksheets will help you learn this standard principle. You are able to decide to use one particular digit multipliers or two-digit and three-digit multipliers. Power of 10 will also be a fantastic choice. These worksheets will assist you to exercise long practice and multiplication looking at the numbers. Also, they are the best way to support your youngster recognize the necessity of understanding the several types of complete figures. ## Multiplication of fractions Getting multiplication of fractions on a worksheet may help educators plan and make lessons successfully. Utilizing fractions worksheets allows professors to quickly evaluate students’ idea of fractions. College students may be questioned to end the worksheet within a certain time and then mark their techniques to see where that they need additional instructions. Individuals can usually benefit from phrase problems that relate maths to real-lifestyle situations. Some fractions worksheets involve samples of contrasting and comparing figures. ## Multiplication of decimals Once you grow two decimal figures, ensure that you group them vertically. If you want to multiply a decimal number with a whole number, the product must contain the same number of decimal places as the multiplicant. As an example, 01 by (11.2) by 2 would be equivalent to 01 x 2.33 by 11.2 except when this product has decimal places of lower than two. Then, the item is rounded for the nearby whole quantity. ## Multiplication of exponents A mathematics worksheet for Multiplication of exponents will allow you to practice dividing and multiplying numbers with exponents. This worksheet will also supply problems that will demand college students to flourish two diverse exponents. By selecting the “All Positive” version, you will be able to view other versions of the worksheet. Apart from, you can also get into special directions in the worksheet on its own. When you’re concluded, you are able to simply click “Produce” and also the worksheet will be delivered electronically. ## Division of exponents The standard principle for division of exponents when multiplying numbers would be to deduct the exponent from the denominator from the exponent in the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. As an example, \$23 split by 4 will identical 27. However, this method is not always accurate. This technique can lead to uncertainty when multiplying amounts which are too big or not big enough. ## Linear characteristics You’ve probably noticed that the cost was | \$320 x 10 days if you’ve ever rented a car. So, the total rent would be \$470. A linear function of this kind has got the type f(by), where ‘x’ is the number of days and nights the car was rented. In addition, it provides the form f(by) = ax b, where by ‘b’ and ‘a’ are actual figures. |
https://yozh.org/2015/10/14/nowhere_differentiable_functions/ | 1,718,574,361,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861671.61/warc/CC-MAIN-20240616203247-20240616233247-00672.warc.gz | 973,295,133 | 12,609 | # Nowhere Differentiable Functions
In an undergraduate analysis class, one of the first results that is generally proved after the definition of differentiability is given is the fact that differentiable functions are continuous. We can justifiably ask if the converse holds—are there examples of functions that are continuous but not differentiable? If such examples exist, how “bad” can they be?
Many examples of continuous functions that are not differentiable spring to mind immediately: the absolute value function is not differentiable at zero; a sawtooth wave is not differentiable anywhere that it changes direction; the Cantor function1I’ll (hopefully) talk more about this later. is an example of a continous function that is not differentiable on an uncountable set, though it does remain differentiable “almost everywhere.”
The goal is to show that there exist functions that are continuous, but that are nowhere differentiable. In fact, what we actually show is that the collection of such functions is, in some sense, quite large and that the set of functions that are differentiable—even if only at a single point—is quite small. First, we need a definition and an important result.
### A Little Theory
Definition: Let $$X$$ be a complete metric space,2A metric space is a set of points and a way of measuring the distance between those points. A sequence of points in a metric space is said to be Cauchy if the distance between any two (not necessarily consecutive) points in the sequence gets small for points sufficiently deep into the sequence. A metric space is said to be complete if every Cauchy sequence converges to some point in the space. The real numbers are a complete metric space, but the rational numbers are not. (Why?) and let $$M \subseteq X$$. Then $$M$$ is said to be
• nowhere dense in $$X$$ if the closure of $$M$$ has empty interior;3This is a topological notion. In very broad strokes, if $$A$$ is a subset of $$X$$, then a point $$x$$ is in the closure of $$A$$ if we can find points of $$A$$ that are arbitrarily “close” to $$x$$. $$A$$ is said to have empty interior if all of the points in $$A$$ are arbitrarily “close” to points that are not in $$A$$. For instance, the closure of the interval $$(0,1)$$ in $$\mathbb{R}$$ is the interval $$[0,1]$$, and any finite collection of points in $$\mathbb{R}$$ is nowhere dense. (Why?)
• meager in $$X$$ if it is the union of countably many nowhere dense sets; and
• residual in $$X$$ if it is not meager (i.e. if it is the complement of a meager set.
This definition provides a topological notion of what it means for a subset of a metric space to be “small.” Nowhere dense sets are tiny—almost insignificant—subsets, while residual sets are quite large (relative to the ambient space). It is also worth noting that meager sets were originally called “sets of the first category,” and residual sets were originally called “sets of the second category,” leading to the name of the following theorem:
Baire’s Category Theorem: If a metric space $$X\ne\emptyset$$ is complete, then it is residual in itself.
Proof: Suppose for contradiction that $$X$$ is meager in itself. Then we may write $X = \bigcup_{k=1}^{\infty} M_k,$ where each $$M_k$$ is nowhere dense. As $$M_1$$ is nowhere dense in $$X$$, its closure has empty interior, and therefore contains no nonempty open sets. But $$X$$ does contain at least one nonempty open set—$$X$$ itself. Hence $$\overline{M}_1\ne X$$ and so, since $$\overline{M}_1$$ is closed, its complement is both open and nonempty. Choose some $$p_1\in X\setminus \overline{M}_1$$. Since $$X\setminus \overline{M}_1$$ is open, there exists some $$\varepsilon_1 \in (0,\frac{1}{2})$$ such that $$B(p_1,\varepsilon_1)\subseteq X\setminus \overline{M}_1$$. Let $$B_1 := B(p_1,\varepsilon_1)$$.
Now consider $$\overline{M}_2$$. It also has empty interior, and so it contains no open balls. In particular, it does not contain $$B_1$$. But then $$B_1\setminus\overline{M}_2$$ is open. Let $$p_2 \in B_1\setminus\overline{M}_2$$ and choose $$\varepsilon_2 < \frac{1}{2}\varepsilon_1$$ such that $$B_2 := B(p_2,\varepsilon_2) \subseteq B_1\setminus\overline{M}_2$$.
Continue this process by induction. That is, for each $$k\in\mathbb{N}$$, choose $$B_{k+1}$$ to be an open ball of radius $$\varepsilon_{k+1} < \frac{1}{2}\varepsilon_k$$ such that $$B_{k+1} \subseteq B_k \setminus \overline{M}_{k+1}$$. By this construction we have $$\varepsilon_k < 2^{-k}$$ for each $$k$$. In particular we have from the triangle inequality that $$d(p_m,p_n) \le 2^{-k}$$ for all $$m,n > k$$, as $$B_m,B_n\subseteq B_{k+1}$$. Hence the sequence of points $$(p_k)$$ is Cauchy in $$X$$. Since $$X$$ is complete, there exists some $$p\in X$$ such that $$p_k\to p$$. But for any $$k$$ the point $$p$$ is contained in $$B_k$$, which implies that $$p\not\in M_k$$ for all $$k$$. Hence we have found a point $$p\in X$$ such that $$p\not\in \bigcup M_k = X$$, which is a contradiction.
### The Existence of Nowhere Differentiable Continuous Functions
We now get to the main result, which has appeared on qualifying exams a few times in the past:
Exercise: Use Baire’s category theorem to prove the existence of continuous, nowhere differentiable functions on the unit interval.
Solution: For each natural number $$n$$, define the set $E_n := \{ f\in C([0,1]) : \exists x_0\in[0,1] \text{ s.t. } |f(x)-f(x_0)| \le n|x-x_0| \forall x\in[0,1]\}.$ This is rather a lot of notation. Let’s try to unpack it just a bit: the derivative of $$f$$ is defined by the limit $f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}.$ If this limit exists, then near $$x_0$$ the difference quotient must be bounded by some number, say $$L_1$$. Away from $$x_0$$, the uniform continuity of $$f$$ ensures that the difference quotient is bounded by some $$L_2$$ when $$x$$ is far from $$x_0$$. Taking the larger of these two bounds, we have that if $$f$$ is differentiable at some point $$x_0$$, then $\frac{|f(x)-f(x_0)|}{|x-x_0|} \le \max\{L_1,L_2\}.$ Thus if a function $$f$$ is differentiable at any point $$x_0\in[0,1]$$, then $$f$$ lives in $$E_n$$ for some value of $$n$$.
What we want to show is that each $$E_n$$ is nowhere dense in space of continous functions $$C[0,1]$$ (which is a normed vector space with respect to the uniform norm). Since all of the differentiable functions are contained in $$\bigcup_n E_n$$, it would then follow that the set of differentiable functions is contained in a meager subset of $$C[0,1]$$. From Baire’s category theorem, we could then conclude that nowhere differentiable functions exist and, indeed, that there is a residual set of nowhere differentiable functions.
To show that each $$E_n$$ is nowhere dense, we have to show that the closure of each $$E_n$$ has empty interior or, equivalently, if we have an arbitrary function in $$\overline{E}_n$$, we need to find a continuous function that is “close” to $$f$$ in the uniform norm, but which is not contained in $$\overline{E}_n$$.
So, what is $$\overline{E}_n$$? We claim that it is just $$E_n$$. To see this, suppose that $$\{f_k\}$$ is a Cauchy sequence in $$E_n$$. First, note that $$C[0,1]$$ is complete, hence there is some $$f\in C[0,1]$$ such that $$f_k\to f$$. Now, since each $$f_k$$\in E_n, for each $$k$$ there is some $$x_k$$ such that $\frac{|f(x)-f(x_k)|}{|x-x_k|} \le n \quad\forall x\in[0,1].$ But then the sequence of numbers $$x_k$$ is a sequence in $$[0,1]$$. By the Bolzano-Weierstrass theorem, this sequence has a convergent subsequence, say $$x_{k_j} \to x\in[0,1]$$. Hence by the uniform convergence of $$f_k$$ to $$f$$, we have $\frac{|f(x)-f(x_{k_j})|}{|x-x_{k_j}|} = \lim_{j\to\infty} \frac{|f_{k_j}(x)-f_{k_j}(x_{k_j})|}{|x-x_{k_j}|} \le n.$ Therefore $$f \in E_n$$, and so Cauchy sequences in $$E_n$$ converge in $$E_n$$, which shows that $$E_n$$ is closed.
Now, given a function $$f\in E_n$$ how do we find a function $$g$$ that is “close” to $$f$$, but not in $$E_n$$? That is actually somewhat delicate, and is dealt with in the following lemma:
Lemma: Given $$f\in C[0,1]$$, $$n\in\mathbb{N}$$, and $$\varepsilon > 0$$, there exists a piecewise linear function $$g$$ with only finitely many linear pieces such that each linear piece has slope $$\pm 2n$$ and $$\|g-f\|_{u} < \varepsilon$$.
Proof: Since $$f$$ is uniformly continuous, there exists a $$\delta$$ such that for any $$x,y\in[0,1]$$, if $$|x-y|< \delta$$, then $$|f(x)-f(y)|<\varepsilon/2$$. Choose $$m\in\mathbb{N}$$ such that $$m > 1/\delta$$. On the interval $$[0,1/m]$$, define $$g$$ as follows: define the first linear piece of $$g$$ by setting $$g(0) = f(0)$$ and giving it slope $$2n$$ on the interval $$[0,\varepsilon/2n]$$. On the interval $$[\varepsilon/2n, 2\varepsilon/2n]$$, let $$g$$ have slope $$-2n$$. Continue in the manner until the linear piece that intersects a line of slope $$\pm 2n$$ through the point $$(1/m,f(1/m))$$ is constructed, and take $$g$$ to be equal to that linear function from the point of intersection to $$1/m$$.
Continue this procedure for each interval of the form $$(k/m,(k+1)/m$$ for $$k=1,2,\ldots,m-1$$. That is, set $$g(k/m) = f(k/m)$$ and construct a sawtooth function on the given interval with slope $$\pm 2n$$. We claim that $$g$$ is a function of the type desired.
We first note that $$g$$ is piecewise linear, with each piece having slope $$\pm 2n$$—this is explicit in the construction. Moreover, there are only finitely many pieces, since the unit interval was broken into $$m$$ subintervals, and each subinterval contains only finitely many linear pieces.4This follows from the Archimedean principle—exact bounds on the number of pieces can be computed, but such a computation is tedious and we are, frankly, a bit lazy. Finally, it follows from the choice of $$\delta$$ and $$m$$, and the triangle inequality that for each $$x\in [0,1]$$, we have $|g(x) – f(x)| \le \left|g(x) – f\left(\frac{\lfloor mx \rfloor}{m}\right)\right| + \left|f\left(\frac{\lfloor mx \rfloor}{m}\right) – f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$ Hence $$\|g-f\|_u < \varepsilon$$.
Therefore $$g$$ is exactly the kind of function that we want, and so the proof is complete.
With the lemma proved, we now have the following: if $$f\in E_n$$, then for any $$\varepsilon > 0$$, there exists a piecewise linear function $$g$$ consisting of finitely many linear pieces each having slope $$\pm 2n$$ such that $$\|g-f\|_u < \varepsilon$$. But no such $$g$$ is in $$E_n$$, and so every neighborhood of $$f$$ contains functions that are not in $$E_n$$. Thus $$E_n$$ contains no open sets, and is therefore nowhere dense.
Thus far, we have proved that each $$E_n$$ is closed an nowhere dense, thus we are ready to apply Baire’s category theory: since $$C[0,1]$$ is residual in itself, it follows that $C[0,1] \setminus \bigcup_{n=1}^{\infty} E_n \ne \emptyset.$ But, as noted above, every function that is differentiable anywhere—even if only at a single point—is contained in the union. Therefore $$C[0,1]$$ contains at least one (in fact, a residual set of) nowhere differentiable function.
This entry was posted in Mathematics and tagged , . Bookmark the permalink. | 3,299 | 11,208 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.942205 | # Nowhere Differentiable Functions In an undergraduate analysis class, one of the first results that is generally proved after the definition of differentiability is given is the fact that differentiable functions are continuous. We can justifiably ask if the converse holds—are there examples of functions that are continuous but not differentiable? If such examples exist, how “bad” can they be? Many examples of continuous functions that are not differentiable spring to mind immediately: the absolute value function is not differentiable at zero; a sawtooth wave is not differentiable anywhere that it changes direction; the Cantor function1I’ll (hopefully) talk more about this later. is an example of a continous function that is not differentiable on an uncountable set, though it does remain differentiable “almost everywhere.” The goal is to show that there exist functions that are continuous, but that are nowhere differentiable. In fact, what we actually show is that the collection of such functions is, in some sense, quite large and that the set of functions that are differentiable—even if only at a single point—is quite small. First, we need a definition and an important result. ### A Little Theory Definition: Let $$X$$ be a complete metric space,2A metric space is a set of points and a way of measuring the distance between those points. A sequence of points in a metric space is said to be Cauchy if the distance between any two (not necessarily consecutive) points in the sequence gets small for points sufficiently deep into the sequence. A metric space is said to be complete if every Cauchy sequence converges to some point in the space. The real numbers are a complete metric space, but the rational numbers are not. (Why?) and let $$M \subseteq X$$. Then $$M$$ is said to be • nowhere dense in $$X$$ if the closure of $$M$$ has empty interior;3This is a topological notion. In very broad strokes, if $$A$$ is a subset of $$X$$, then a point $$x$$ is in the closure of $$A$$ if we can find points of $$A$$ that are arbitrarily “close” to $$x$$. $$A$$ is said to have empty interior if all of the points in $$A$$ are arbitrarily “close” to points that are not in $$A$$. For instance, the closure of the interval $$(0,1)$$ in $$\mathbb{R}$$ is the interval $$[0,1]$$, and any finite collection of points in $$\mathbb{R}$$ is nowhere dense. (Why?) • meager in $$X$$ if it is the union of countably many nowhere dense sets; and • residual in $$X$$ if it is not meager (i.e. if it is the complement of a meager set. This definition provides a topological notion of what it means for a subset of a metric space to be “small.” Nowhere dense sets are tiny—almost insignificant—subsets, while residual sets are quite large (relative to the ambient space). It is also worth noting that meager sets were originally called “sets of the first category,” and residual sets were originally called “sets of the second category,” leading to the name of the following theorem: Baire’s Category Theorem: If a metric space $$X\ne\emptyset$$ is complete, then it is residual in itself. Proof: Suppose for contradiction that $$X$$ is meager in itself. Then we may write $X = \bigcup_{k=1}^{\infty} M_k,$ where each $$M_k$$ is nowhere dense. As $$M_1$$ is nowhere dense in $$X$$, its closure has empty interior, and therefore contains no nonempty open sets. But $$X$$ does contain at least one nonempty open set—$$X$$ itself. Hence $$\overline{M}_1\ne X$$ and so, since $$\overline{M}_1$$ is closed, its complement is both open and nonempty. Choose some $$p_1\in X\setminus \overline{M}_1$$. Since $$X\setminus \overline{M}_1$$ is open, there exists some $$\varepsilon_1 \in (0,\frac{1}{2})$$ such that $$B(p_1,\varepsilon_1)\subseteq X\setminus \overline{M}_1$$. Let $$B_1 := B(p_1,\varepsilon_1)$$. Now consider $$\overline{M}_2$$. It also has empty interior, and so it contains no open balls. In particular, it does not contain $$B_1$$. But then $$B_1\setminus\overline{M}_2$$ is open. Let $$p_2 \in B_1\setminus\overline{M}_2$$ and choose $$\varepsilon_2 < \frac{1}{2}\varepsilon_1$$ such that $$B_2 := B(p_2,\varepsilon_2) \subseteq B_1\setminus\overline{M}_2$$. Continue this process by induction. That is, for each $$k\in\mathbb{N}$$, choose $$B_{k+1}$$ to be an open ball of radius $$\varepsilon_{k+1} < \frac{1}{2}\varepsilon_k$$ such that $$B_{k+1} \subseteq B_k \setminus \overline{M}_{k+1}$$. By this construction we have $$\varepsilon_k < 2^{-k}$$ for each $$k$$. In particular we have from the triangle inequality that $$d(p_m,p_n) \le 2^{-k}$$ for all $$m,n > k$$, as $$B_m,B_n\subseteq B_{k+1}$$. Hence the sequence of points $$(p_k)$$ is Cauchy in $$X$$. Since $$X$$ is complete, there exists some $$p\in X$$ such that $$p_k\to p$$. But for any $$k$$ the point $$p$$ is contained in $$B_k$$, which implies that $$p\not\in M_k$$ for all $$k$$. Hence we have found a point $$p\in X$$ such that $$p\not\in \bigcup M_k = X$$, which is a contradiction. ### The Existence of Nowhere Differentiable Continuous Functions We now get to the main result, which has appeared on qualifying exams a few times in the past: Exercise: Use Baire’s category theorem to prove the existence of continuous, nowhere differentiable functions on the unit interval. Solution: For each natural number $$n$$, define the set $E_n := \{ f\in C([0,1]) : \exists x_0\in[0,1] \text{ s.t. } |f(x)-f(x_0)| \le n|x-x_0| \forall x\in[0,1]\}.$ This is rather a lot of notation. Let’s try to unpack it just a bit: the derivative of $$f$$ is defined by the limit $f'(x_0) = \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}.$ If this limit exists, then near $$x_0$$ the difference quotient must be bounded by some number, say $$L_1$$. Away from $$x_0$$, the uniform continuity of $$f$$ ensures that the difference quotient is bounded by some $$L_2$$ when $$x$$ is far from $$x_0$$. Taking the larger of these two bounds, we have that if $$f$$ is differentiable at some point $$x_0$$, then $\frac{|f(x)-f(x_0)|}{|x-x_0|} \le \max\{L_1,L_2\}.$ Thus if a function $$f$$ is differentiable at any point $$x_0\in[0,1]$$, then $$f$$ lives in $$E_n$$ for some value of $$n$$. What we want to show is that each $$E_n$$ is nowhere dense in space of continous functions $$C[0,1]$$ (which is a normed vector space with respect to the uniform norm). Since all of the differentiable functions are contained in $$\bigcup_n E_n$$, it would then follow that the set of differentiable functions is contained in a meager subset of $$C[0,1]$$. From Baire’s category theorem, we could then conclude that nowhere differentiable functions exist and, indeed, that there is a residual set of nowhere differentiable functions. To show that each $$E_n$$ is nowhere dense, we have to show that the closure of each $$E_n$$ has empty interior or, equivalently, if we have an arbitrary function in $$\overline{E}_n$$, we need to find a continuous function that is “close” to $$f$$ in the uniform norm, but which is not contained in $$\overline{E}_n$$. So, what is $$\overline{E}_n$$? We claim that it is just $$E_n$$. To see this, suppose that $$\{f_k\}$$ is a Cauchy sequence in $$E_n$$. First, note that $$C[0,1]$$ is complete, hence there is some $$f\in C[0,1]$$ such that $$f_k\to f$$. Now, since each $$f_k$$\in E_n, for each $$k$$ there is some $$x_k$$ such that $\frac{|f(x)-f(x_k)|}{|x-x_k|} \le n \quad\forall x\in[0,1].$ But then the sequence of numbers $$x_k$$ is a sequence in $$[0,1]$$. By the Bolzano-Weierstrass theorem, this sequence has a convergent subsequence, say $$x_{k_j} \to x\in[0,1]$$. Hence by the uniform convergence of $$f_k$$ to $$f$$, we have $\frac{|f(x)-f(x_{k_j})|}{|x-x_{k_j}|} = \lim_{j\to\infty} \frac{|f_{k_j}(x)-f_{k_j}(x_{k_j})|}{|x-x_{k_j}|} \le n.$ Therefore $$f \in E_n$$, and so Cauchy sequences in $$E_n$$ converge in $$E_n$$, which shows that $$E_n$$ is closed. Now, given a function $$f\in E_n$$ how do we find a function $$g$$ that is “close” to $$f$$, but not in $$E_n$$? That is actually somewhat delicate, and is dealt with in the following lemma: Lemma: Given $$f\in C[0,1]$$, $$n\in\mathbb{N}$$, and $$\varepsilon > 0$$, there exists a piecewise linear function $$g$$ with only finitely many linear pieces such that each linear piece has slope $$\pm 2n$$ and $$\|g-f\|_{u} < \varepsilon$$. Proof: Since $$f$$ is uniformly continuous, there exists a $$\delta$$ such that for any $$x,y\in[0,1]$$, if $$|x-y|< \delta$$, then $$|f(x)-f(y)|<\varepsilon/2$$. Choose $$m\in\mathbb{N}$$ such that $$m > 1/\delta$$. On the interval $$[0,1/m]$$, define $$g$$ as follows: define the first linear piece of $$g$$ by setting $$g(0) = f(0)$$ and giving it slope $$2n$$ on the interval $$[0,\varepsilon/2n]$$. On the interval $$[\varepsilon/2n, 2\varepsilon/2n]$$, let $$g$$ have slope $$-2n$$. Continue in the manner until the linear piece that intersects a line of slope $$\pm 2n$$ through the point $$(1/m,f(1/m))$$ is constructed, and take $$g$$ to be equal to that linear function from the point of intersection to $$1/m$$. Continue this procedure for each interval of the form $$(k/m,(k+1)/m$$ for $$k=1,2,\ldots,m-1$$. That is, set $$g(k/m) = f(k/m)$$ and construct a sawtooth function on the given interval with slope $$\pm 2n$$. We claim that $$g$$ is a function of the type desired. We first note that $$g$$ is piecewise linear, with each piece having slope $$\pm 2n$$—this is explicit in the construction. Moreover, there are only finitely many pieces, since the unit interval was broken into $$m$$ subintervals, and each subinterval contains only finitely many linear pieces.4This follows from the Archimedean principle—exact bounds on the number of pieces can be computed, but such a computation is tedious and we are, frankly, a bit lazy. Finally, it follows from the choice of $$\delta$$ and $$m$$, and the triangle inequality that for each $$x\in [0,1]$$, we have $|g(x) – f(x)| \le \left|g(x) – f\left(\frac{\lfloor mx \rfloor}{m}\right)\right| + \left|f\left(\frac{\lfloor mx \rfloor}{m}\right) – f(x)\right| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$ Hence $$\|g-f\|_u < \varepsilon$$. Therefore | $$g$$ is exactly the kind of function that we want, and so the proof is complete. With the lemma proved, we now have the following: if $$f\in E_n$$, then for any $$\varepsilon > 0$$, there exists a piecewise linear function $$g$$ consisting of finitely many linear pieces each having slope $$\pm 2n$$ such that $$\|g-f\|_u < \varepsilon$$. But no such $$g$$ is in $$E_n$$, and so every neighborhood of $$f$$ contains functions that are not in $$E_n$$. Thus $$E_n$$ contains no open sets, and is therefore nowhere dense. Thus far, we have proved that each $$E_n$$ is closed an nowhere dense, thus we are ready to apply Baire’s category theory: since $$C[0,1]$$ is residual in itself, it follows that $C[0,1] \setminus \bigcup_{n=1}^{\infty} E_n \ne \emptyset.$ But, as noted above, every function that is differentiable anywhere—even if only at a single point—is contained in the union. Therefore $$C[0,1]$$ contains at least one (in fact, a residual set of) nowhere differentiable function. This entry was posted in Mathematics and tagged , . Bookmark the permalink. |
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# Tata Digital Health Interview Experience | (On Campus) September, 2019
In September 2019, Tata Digital Health Visited Our Campus, Following are the Details for the Same :
Profile: Programmer
Cut-Off: 6.5
Note: The Test Was Conducted on Their Own Platform.
Brief Overview Of the Process:
Total 4 Rounds Were Conducted.
In Round-1, Around 150 Students Appeared, Out Of Which around 140 Students were Shortlisted for Round-2.
In Round-2, 15 Students were Shortlisted Out of 140 Students for Round-3.
Finally After Round-3 & 4, Total 5 Students were Selected, Unfortunately, I was not of them.
Round 1: Aptitude (10 Questions – 60 Minutes)
The Questions Were Subjective i.e. You Have to Type the Answer. Questions were Easy to Moderate Level.
I Can Remember 5 Questions, Out of 10, Which, Along-with Answers, are as Follows :
Q.1.) The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.
Ans.1.) 23
Q.2.) The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers?
Ans.2) 15
Q.3.) In 1930, a correspondent proposed the following question: A man’s age at death was one twenty-ninth of the year of his birth. How old was the man in 1900?
Ans.3) 44
Q.4.) Richard is a strange liar. He lies on 6 days of the week, but on the seventh day he always tells the truth. He made the following statements on 3 successive days:
Day 1 : “I lie on Mon and Tue.”
Day 2 : “Today, it’s Thu, Sat, or Sun”
Day 3 : “I lie on Wed and Fri.”
On which day does Ravi tell the truth?
Ans.4.) Tuesday
Q.5.) In a class of students, if 1 is absent rest of people can be divided into 6 equal parts and if 2 are absent rest of them are divided into 7 equal parts, how many students are present in the class?
Ans.5.) 37
Round 2: Coding (2 Questions – 75 Minutes)
Question 1:
You were playing a game of treasures where the treasure boxes were kept in a row. Each treasure box has a label denotes the amount placed inside the treasure box. A lock is designed to lock multiple treasure boxes next to each other in a row. Each treasure box is locked with zero or more number of such locks.
Your task is to unlock only one lock where you could get the maximum treasure amount from the boxes. There could be treasure boxes without any locks as well. On finding the maximum treasure amount, you should consider the boxes without the locks as well.
Input Format:
The first line contains a single integer N, the number of treasure boxes.
The next line contains N space separated integers A[i], the amount in ith box.
The next line contains a single integer K, the number of locks.
The next K line contains two space separated integers S[i] and E[i], the start and end of ith lock in the row.
Output Format:
Print the maximum amount M., can be found on unlocking only one lock.
Constraints:
1 <= N, K <= 100
0 <= A[i] <=10^6
1 <= S[i] <= E[i] <= N
Explanation:
There are N = 5 treasure boxes with amounts A = [2, 5, 6, 1, 4] and K= 3 locks.
Lock1 is used to lock boxes starts from 2nd position to 3rd position.
Lock2 is used to lock boxes starts from 3rd position to 4th position.
Lock3 is used to lock boxes starts from 3rd position to 5th position.
So when the lock1 is unlocked. you can find maximum amount 7 (2 + 5).
Sample Input:
5
2 5 6 1 4
3
2 3
3 4
3 5
Sample Output:
7
Question 2: (This Problem is Same as Standard DP Problem – Coin Change Problem)
Delivery of goods are dispatched from the storage of ABC delivery company. The storage has goods of K different weights. Delivery vehicle can occupy goods up-to W weights.
Your task is to find the number of ways to fill the vehicle with the goods in the storage. Vehicle is considered as filled only if it fills the capacity without any space.
Input Format:
First line consists of single integer K, different weights of goods.
Next line consists of K space-separated integers w[i], weight of ith good.
Next line consists of a single integer W, vehicle capacity of weights.
Output Format:
Print the number of ways to fill the vehicle for the given capacity of weights.
Constraints:
1 <= K <=20
1 <= w[i] <= 100
1 <= W <= 1000
Explanation:
There are 4 ways to fill the vehicle with weight = 4. using weights of goods w = [1, 2, 3].
{1, 1, 1, 1}
{1, 1, 2}
{1, 3}
{2, 2}
Sample Input:
3
1 2 3
4
Sample Output:
4
Round 3 : Interview
Round 4 : HR
Finally, 5 Students Were Selected.
My Personal Notes arrow_drop_up | 1,270 | 4,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2023-23 | latest | en | 0.961505 | GeeksforGeeks App Open App Browser Continue # Tata Digital Health Interview Experience | (On Campus) September, 2019 In September 2019, Tata Digital Health Visited Our Campus, Following are the Details for the Same : Profile: Programmer Cut-Off: 6.5 Note: The Test Was Conducted on Their Own Platform. Brief Overview Of the Process: Total 4 Rounds Were Conducted. In Round-1, Around 150 Students Appeared, Out Of Which around 140 Students were Shortlisted for Round-2. In Round-2, 15 Students were Shortlisted Out of 140 Students for Round-3. Finally After Round-3 & 4, Total 5 Students were Selected, Unfortunately, I was not of them. Round 1: Aptitude (10 Questions – 60 Minutes) The Questions Were Subjective i.e. You Have to Type the Answer. Questions were Easy to Moderate Level. I Can Remember 5 Questions, Out of 10, Which, Along-with Answers, are as Follows : Q.1.) The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team. Ans.1.) 23 Q.2.) The present ratio of students to teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the number of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present number of teachers? Ans.2) 15 Q.3.) In 1930, a correspondent proposed the following question: A man’s age at death was one twenty-ninth of the year of his birth. How old was the man in 1900? Ans.3) 44 Q.4.) Richard is a strange liar. He lies on 6 days of the week, but on the seventh day he always tells the truth. He made the following statements on 3 successive days: Day 1 : “I lie on Mon and Tue.” Day 2 : “Today, it’s Thu, Sat, or Sun” Day 3 : “I lie on Wed and Fri.” On which day does Ravi tell the truth? Ans.4.) Tuesday Q.5.) In a class of students, if 1 is absent rest of people can be divided into 6 equal parts and if 2 are absent rest of them are divided into 7 equal parts, how many students are present in the class? Ans.5.) 37 Round 2: Coding (2 Questions – 75 Minutes) Question 1: You were playing a game of treasures where the treasure boxes were kept in a row. Each treasure box has a label denotes the amount placed inside the treasure box. A lock is designed to lock multiple treasure boxes next to each other in a row. Each treasure box is locked with zero or more number of such locks. Your task is to unlock only one lock where you could get the maximum treasure amount from the boxes. There could be treasure boxes without any locks as well. On finding the maximum treasure amount, you should consider the boxes without the locks as well. Input Format: The first line contains a single integer N, the number of treasure boxes. The next line contains N space separated integers A[i], the amount in ith box. The next line contains a single integer K, the number of locks. The next K line contains two space separated integers S[i] and E[i], the start and end of ith lock in the row. Output Format: Print the maximum amount M., can be found on unlocking only one lock. Constraints: 1 <= N, K <= 100 0 <= A[i] <=10^6 1 <= S[i] <= E[i] <= N Explanation: There are N = 5 treasure boxes with amounts A = [2, 5, 6, 1, 4] and K= 3 locks. Lock1 is used to lock boxes starts from 2nd position to 3rd position. Lock2 is used to lock boxes starts from 3rd position to 4th position. Lock3 is used to lock boxes starts from 3rd position to 5th position. So when the lock1 is unlocked. you can find maximum amount 7 (2 + 5). Sample Input: 5 2 5 6 1 4 3 2 3 3 4 3 5 Sample Output: 7 Question 2: (This Problem is Same as Standard DP Problem – Coin Change Problem) Delivery of goods are dispatched from the storage of ABC delivery company. The storage has goods of K different weights. Delivery vehicle can occupy goods up-to W weights. Your task is to find the number of ways to fill the vehicle with the goods in the storage. Vehicle is considered as filled only if it fills the capacity without any space. Input Format: First line consists of single integer K, different weights of goods. Next line consists of K space-separated integers w[i], weight of ith good. Next line consists of a single integer W, vehicle capacity of weights. Output | Format: Print the number of ways to fill the vehicle for the given capacity of weights. Constraints: 1 <= K <=20 1 <= w[i] <= 100 1 <= W <= 1000 Explanation: There are 4 ways to fill the vehicle with weight = 4. using weights of goods w = [1, 2, 3]. {1, 1, 1, 1} {1, 1, 2} {1, 3} {2, 2} Sample Input: 3 1 2 3 4 Sample Output: 4 Round 3 : Interview Round 4 : HR Finally, 5 Students Were Selected. My Personal Notes arrow_drop_up |
https://metanumbers.com/25971 | 1,601,335,024,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401614309.85/warc/CC-MAIN-20200928202758-20200928232758-00201.warc.gz | 472,909,112 | 7,511 | ## 25971
25,971 (twenty-five thousand nine hundred seventy-one) is an odd five-digits composite number following 25970 and preceding 25972. In scientific notation, it is written as 2.5971 × 104. The sum of its digits is 24. It has a total of 3 prime factors and 8 positive divisors. There are 15,720 positive integers (up to 25971) that are relatively prime to 25971.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 24
• Digital Root 6
## Name
Short name 25 thousand 971 twenty-five thousand nine hundred seventy-one
## Notation
Scientific notation 2.5971 × 104 25.971 × 103
## Prime Factorization of 25971
Prime Factorization 3 × 11 × 787
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 25971 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 25,971 is 3 × 11 × 787. Since it has a total of 3 prime factors, 25,971 is a composite number.
## Divisors of 25971
1, 3, 11, 33, 787, 2361, 8657, 25971
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 37824 Sum of all the positive divisors of n s(n) 11853 Sum of the proper positive divisors of n A(n) 4728 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 161.155 Returns the nth root of the product of n divisors H(n) 5.49302 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 25,971 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 25,971) is 37,824, the average is 4,728.
## Other Arithmetic Functions (n = 25971)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 15720 Total number of positive integers not greater than n that are coprime to n λ(n) 3930 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2857 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 15,720 positive integers (less than 25,971) that are coprime with 25,971. And there are approximately 2,857 prime numbers less than or equal to 25,971.
## Divisibility of 25971
m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 1 3 6
The number 25,971 is divisible by 3.
## Classification of 25971
• Arithmetic
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (25971)
Base System Value
2 Binary 110010101110011
3 Ternary 1022121220
4 Quaternary 12111303
5 Quinary 1312341
6 Senary 320123
8 Octal 62563
10 Decimal 25971
12 Duodecimal 13043
20 Vigesimal 34ib
36 Base36 k1f
## Basic calculations (n = 25971)
### Multiplication
n×i
n×2 51942 77913 103884 129855
### Division
ni
n⁄2 12985.5 8657 6492.75 5194.2
### Exponentiation
ni
n2 674492841 17517253573611 454940592560251281 11815262129382286018851
### Nth Root
i√n
2√n 161.155 29.6139 12.6947 7.63656
## 25971 as geometric shapes
### Circle
Diameter 51942 163181 2.11898e+09
### Sphere
Volume 7.33761e+13 8.47593e+09 163181
### Square
Length = n
Perimeter 103884 6.74493e+08 36728.5
### Cube
Length = n
Surface area 4.04696e+09 1.75173e+13 44983.1
### Equilateral Triangle
Length = n
Perimeter 77913 2.92064e+08 22491.5
### Triangular Pyramid
Length = n
Surface area 1.16826e+09 2.06443e+12 21205.2
## Cryptographic Hash Functions
md5 e8568f57572bc50daf688754f4717ccc 43fc96c01a138157ead3f338d1c5d96ba2740de2 852e26523b104c2d0981bbdeae4716298824fff7cfe2c27114ec51e55021db67 55db44493d7ca4ea7ca073ce51e107255d3206421b4e03b6fb1a8597ec60b7fd43f945a38e59bb2a9e58ef032387dd1b7352c6c4781bcbb6a4d77582a39d754c 2f91a14871b906ddea8c42e4ed1f247f66ab1bf6 | 1,475 | 4,205 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2020-40 | longest | en | 0.810819 | ## 25971 25,971 (twenty-five thousand nine hundred seventy-one) is an odd five-digits composite number following 25970 and preceding 25972. In scientific notation, it is written as 2.5971 × 104. The sum of its digits is 24. It has a total of 3 prime factors and 8 positive divisors. There are 15,720 positive integers (up to 25971) that are relatively prime to 25971. ## Basic properties • Is Prime? No • Number parity Odd • Number length 5 • Sum of Digits 24 • Digital Root 6 ## Name Short name 25 thousand 971 twenty-five thousand nine hundred seventy-one ## Notation Scientific notation 2.5971 × 104 25.971 × 103 ## Prime Factorization of 25971 Prime Factorization 3 × 11 × 787 Composite number Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 25971 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0 The prime factorization of 25,971 is 3 × 11 × 787. Since it has a total of 3 prime factors, 25,971 is a composite number. ## Divisors of 25971 1, 3, 11, 33, 787, 2361, 8657, 25971 8 divisors Even divisors 0 8 4 4 Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 37824 Sum of all the positive divisors of n s(n) 11853 Sum of the proper positive divisors of n A(n) 4728 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 161.155 Returns the nth root of the product of n divisors H(n) 5.49302 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors The number 25,971 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 25,971) is 37,824, the average is 4,728. ## Other Arithmetic Functions (n = 25971) 1 φ(n) n Euler Totient Carmichael Lambda Prime Pi φ(n) 15720 Total number of positive integers not greater than n that are coprime to n λ(n) 3930 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 2857 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares There are 15,720 positive integers (less than 25,971) that are coprime with 25,971. And there are approximately 2,857 prime numbers less than or equal to 25,971. ## Divisibility of 25971 m n mod m 2 3 4 5 6 7 8 9 1 0 3 1 3 1 3 6 The number 25,971 is divisible by 3. ## Classification of 25971 • Arithmetic • Deficient ### Expressible via specific sums • Polite • Non-hypotenuse • Square Free ### Other numbers • LucasCarmichael • Sphenic ## Base conversion (25971) Base System Value 2 Binary 110010101110011 3 Ternary 1022121220 4 Quaternary 12111303 5 Quinary 1312341 6 Senary 320123 8 Octal 62563 10 Decimal 25971 12 Duodecimal 13043 20 Vigesimal 34ib 36 Base36 k1f ## Basic calculations (n = 25971) ### Multiplication n×i n×2 51942 77913 103884 129855 ### Division ni n⁄2 12985.5 8657 6492.75 5194.2 ### Exponentiation ni n2 674492841 17517253573611 454940592560251281 11815262129382286018851 ### Nth Root i√n 2√n 161.155 | 29.6139 12.6947 7.63656 ## 25971 as geometric shapes ### Circle Diameter 51942 163181 2.11898e+09 ### Sphere Volume 7.33761e+13 8.47593e+09 163181 ### Square Length = n Perimeter 103884 6.74493e+08 36728.5 ### Cube Length = n Surface area 4.04696e+09 1.75173e+13 44983.1 ### Equilateral Triangle Length = n Perimeter 77913 2.92064e+08 22491.5 ### Triangular Pyramid Length = n Surface area 1.16826e+09 2.06443e+12 21205.2 ## Cryptographic Hash Functions md5 e8568f57572bc50daf688754f4717ccc 43fc96c01a138157ead3f338d1c5d96ba2740de2 852e26523b104c2d0981bbdeae4716298824fff7cfe2c27114ec51e55021db67 55db44493d7ca4ea7ca073ce51e107255d3206421b4e03b6fb1a8597ec60b7fd43f945a38e59bb2a9e58ef032387dd1b7352c6c4781bcbb6a4d77582a39d754c 2f91a14871b906ddea8c42e4ed1f247f66ab1bf6 |
http://www.mathkun.com/en-roman/how-to-write/convert-72-to-roman-numerals | 1,544,459,006,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823348.23/warc/CC-MAIN-20181210144632-20181210170132-00342.warc.gz | 434,359,712 | 4,564 | USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES
# Convert 72 To Roman Numerals
How to write number 72 in roman numerals: LXXII
This is how to convert the number 72 to roman numeral LXXII. Roman numerals conversion and translation. Numbers in roman numerals have many correct spellings and variations for writing numbers into words. The way we spelled them is the most compact one.
## What Is The Roman Numeral LXXII
The roman numeral LXXII converted to an arabic number is: 72
## Mathematical Properties Of LXXII Spelled Out
LXXII is a composite number. 72 is a composite number, because it has more divisors than 1 and itself. As a result it is not a prime number. This is an even number. LXXII is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 72 is not an odd number. When we simplify Sin 72 degrees we get the value of sin(72)=0.25382336276204. Simplify Cos 72 degrees. The value of cos(72)=-0.96725058827388. Simplify Tan 72 degrees. Value of tan(72)=-0.26241737750194. Prime factors of 72 are 2, 3. Prime factorization of 72 is 2 * 2 * 2 * 3 * 3. When converting 72 in binary you get 1001000. Converting decimal 72 in hexadecimal is 48. The square root of 72=8.4852813742386. The cube root of 72=4.1601676461038. Square root of √72 simplified is 6√2. All radicals are now simplified and in their simplest form. Cube root of ∛72 simplified is 2∛9. The simplified radicand no longer has any more cubed factors.
## Write Smaller Numbers Than 72 In Roman Numerals
Do you know how to write or convert smaller numbers than LXXII in roman numerals?
## Convert Bigger Numbers Than 72 To Roman Numerals
Do you know how to write or convert bigger numbers than LXXII to roman numerals?
## Numbers Used In Spelling 72 Roman Numerals
• The number 7 explained: Seven is an odd and defective number. The fourth prime, after 5 and before 11. Also called one of the primes of Mersenne, 7=2³-1. Known as one of the double prime integers, which is (7-1)÷2 still the same. 7 is the Cuban prime of the form (x³-y³)÷(x-y), x=y+1. Euclidian quantity 7=(2×3)+1. 7 is a Perrin, integer-free and congruent number. Smallest natural whose cube (343) is a palindrome. The second figure of Carol. A polygon with seven sides is called a heptagon. Part of the Pythagorean triad (7, 24, 25). Fifth of the succession of Lucas, after 4 and before 11. It is a palindrome in the binary system and a repeated number in the positional numbering system based on 6. In the numerical decimal system seven is a Colombian value.
• The number 2 explained: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Being even and a prime of Sophie Germain and Eisenstein. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a number of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2.
## Number 72 Meaning Roman Numerals
To write and remember the correct spelling of 72 in roman numerals it is best to learn the language rules for writing numbers in words.
Roman numerals originated in ancient Rome and are used to this day. Roman numerals, are based on seven basic symbols I=1, V=5, X=10, L=50, C=100, D=500, M=1000 which are combined with one another.
The Roman numeral system is decimal, meaning each digit is added separately, in descending order from left to right. Roman numerals are nowadays mostly seen as year numbers, on clocks or in movies(for the visual effect).
## Learn Mathematics And Solve Problems
In depth materials written and spelled out in a easy to understand manner. Learn more than how to spell or write numbers in words. | 1,218 | 4,716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-51 | latest | en | 0.838195 | USING OUR SERVICES YOU AGREE TO OUR USE OF COOKIES # Convert 72 To Roman Numerals How to write number 72 in roman numerals: LXXII This is how to convert the number 72 to roman numeral LXXII. Roman numerals conversion and translation. Numbers in roman numerals have many correct spellings and variations for writing numbers into words. The way we spelled them is the most compact one. ## What Is The Roman Numeral LXXII The roman numeral LXXII converted to an arabic number is: 72 ## Mathematical Properties Of LXXII Spelled Out LXXII is a composite number. 72 is a composite number, because it has more divisors than 1 and itself. As a result it is not a prime number. This is an even number. LXXII is an even number, because it can be divided by 2 without leaving a comma spot. This also means that 72 is not an odd number. When we simplify Sin 72 degrees we get the value of sin(72)=0.25382336276204. Simplify Cos 72 degrees. The value of cos(72)=-0.96725058827388. Simplify Tan 72 degrees. Value of tan(72)=-0.26241737750194. Prime factors of 72 are 2, 3. Prime factorization of 72 is 2 * 2 * 2 * 3 * 3. When converting 72 in binary you get 1001000. Converting decimal 72 in hexadecimal is 48. The square root of 72=8.4852813742386. The cube root of 72=4.1601676461038. Square root of √72 simplified is 6√2. All radicals are now simplified and in their simplest form. Cube root of ∛72 simplified is 2∛9. The simplified radicand no longer has any more cubed factors. ## Write Smaller Numbers Than 72 In Roman Numerals Do you know how to write or convert smaller numbers than LXXII in roman numerals? ## Convert Bigger Numbers Than 72 To Roman Numerals Do you know how to write or convert bigger numbers than LXXII to roman numerals? ## Numbers Used In Spelling 72 Roman Numerals • The number 7 explained: Seven is an odd and defective number. The fourth prime, after 5 and before 11. Also called one of the primes of Mersenne, 7=2³-1. Known as one of the double prime integers, which is (7-1)÷2 still the same. 7 is the Cuban prime of the form (x³-y³)÷(x-y), x=y+1. Euclidian quantity 7=(2×3)+1. 7 is a Perrin, integer-free and congruent number. Smallest natural whose cube (343) is a palindrome. The second figure of Carol. A polygon with seven sides is called a heptagon. Part of the Pythagorean triad (7, 24, 25). Fifth of the succession of Lucas, after 4 and before 11. It is a palindrome in the binary system and a repeated number in the positional numbering system based on 6. In the numerical decimal system seven is a Colombian value. • The number 2 explained: 2 is the first of the primes and the only one to be even(the others are all odd). The first issue of Smarandache-Wellin in any base. Being even and a prime of Sophie Germain and Eisenstein. Goldbach's conjecture states that all even numbers greater than 2 are the quantity of 2 primes. It is a complete Harshad, which is a number of Harshad in any expressed base. The third of the Fibonacci sequence, after 1 and before 3. Part of the Tetranacci Succession. Two is an oblong figure of the form n(n+1). 2 is the basis of the binary numbering system, used internally by almost all computers. Two is a number of: Perrin, Ulam, Catalan and Wedderburn-Etherington. Refactorizable, which means that it is divisible by the count of its divisors. Not being the total of the divisors proper to any other arithmetical value, 2 is an untouchable quantity. The first number of highly cototent and scarcely totiente (the only one to be both) and it is also a very large decimal. Second term of the succession of Mian-Chowla. A strictly non-palindrome. With one exception, all known solutions to the Znam problem begin with 2. Numbers are divisible by two (ie equal) if and only if its last digit is even. The first even numeral after zero and the first issue of the succession of Lucas. The aggregate of any natural value and its reciprocal is always greater than or equal to 2. ## Number 72 Meaning Roman Numerals To write and remember the correct spelling of 72 in roman numerals it is best to learn the language rules for writing numbers in words. Roman numerals originated in ancient Rome and are used to this day. Roman numerals, are based | on seven basic symbols I=1, V=5, X=10, L=50, C=100, D=500, M=1000 which are combined with one another. The Roman numeral system is decimal, meaning each digit is added separately, in descending order from left to right. Roman numerals are nowadays mostly seen as year numbers, on clocks or in movies(for the visual effect). ## Learn Mathematics And Solve Problems In depth materials written and spelled out in a easy to understand manner. Learn more than how to spell or write numbers in words. |
https://www.f1gmat.com/gmat-data-sufficiency-tips | 1,582,391,056,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145708.59/warc/CC-MAIN-20200222150029-20200222180029-00073.warc.gz | 724,945,726 | 8,998 | # Top 10 Tips to Ace GMAT Data Sufficiency
Ever heard of a Math problem that you actually don't have to solve. If you have just started your GMAT prep, then this can be confusing. Don't worry! With some practice, your mind will be trained to think like a DS Wizard. Follow these 10 tips and you will be on your way to mastering GMAT Data Sufficiency.
1. Familiarize with the Answer Choices
No excuses: On Data Sufficiency, they’re always the same! Know in the blink of an eye what choice C is. On test day, if you find that Statement 1 is insufficient, be able to cross out choices A and D without hesitation.
2. Takes notes efficiently
Each statement alone will be sufficient if both of the statements on their own contain all the information necessary to answer the question. The statements will be sufficient together if they contain every piece of necessary information between them. Take the area of a parallelogram: Do you need to know every side length to determine the area? If you have every side length, can you find the area?
3. Don’t look at the statements together.
Statement 2 may tell you that x is negative, but that fact has no bearing on Statement 1 when viewed by itself. Explore all the possibilities offered by each statement individually. If you’ve scrutinized Statement 1 and found it sufficient, be equally merciless when it comes to Statement 2.
4. Watch out for Important information
Don’t pay so much attention to the statements that you forget the rest of the question. Often, half the information that you need is in the set-up.
5. Know when to solve single-variable equations.
If the question asks for the value of x and you whittle the problem down to an equation like 305x = 2(500) – 10205, don’t waste your time solving for x! It’s only important to know that you COULD solve if you wanted to. Remember, all linear one-variable equations have a unique solution, but quadratic equations—equations with an x^2 term—can have zero, one, or two solutions.
6. Know when it’s necessary to solve a system of equations.
Again, you never need to solve a DS problem—you only need to know that you could. A system of n independent linear equations with n variables can be solved for ALL of the n variables. The key word here is “independent”: Equations are independent if they’re not multiples of one another. For example, y = 2x and 3y = 6x are NOT independent equations because the second equation is just three times the first. If on test day you don’t feel comfortable declaring that a system of equations is solvable, get the system down to one single-variable equation and then reassess.
7. Study prime factorizations and divisibility.
Although any GMAT math concept is fair game on the DS section, prime factorization shows up frequently and reliably. If x is divisible by 15, will x^2 be divisible by 27? What about x^3?
8. Study overlapping sets.
Be comfortable representing these overlapping sets with Venn diagrams. This topic is a DS favorite. A statement like, “The number of widgets that were not made in Factory A or Factory B is three times greater than the number of widgets that were made in Factory B” can be difficult to unpack in the heat of the moment. Train yourself to answer questions about sets methodically and quickly.
9. Remember that only 2 out of the 5 answer choices involve looking at both statements TOGETHER.
This means that there’s a 60% chance that the correct answer will treat the statements on an individual footing. It can be tempting to use all the information the problem provides, but keep these odds in mind. Choices C and E, as a group, are 20% less likely to be correct than choices A, B, and D, as a group.
10. Be on the lookout for statements that give no new information.
The area of a square, for instance, contains just as much information as the side length of the square. If you know the area, you can find the side length; conversely, if you know the side length, you can find the area. Often on the DS section, Statement 2 will just be a repackaging of the same information provided by Statement 1.
Author :
Knewton
was a leading test prep company that helped thousands of GMAT test takers since 2008. As of 2014, Knewton has discontinued offering tutoring service.
Recommended GMAT Prep Resources
Mastering GMAT Critical Reasoning
1. Complete GMAT RC Questions in less than 1 minute and 50 seconds
3. Take Notes Effectively
4. Collect and Interpret Facts
5. Speed up Summary Creation
6. Remember Information
7. Question the Author
12. Learn to Answer GMAT organization of passage Question
13. Learn to identify the style/tone or attitude of the author
Mastering GMAT Critical Reasoning
After you read F1GMAT’s Mastering GMAT Critical Reasoning Guide, you will learn:
How to overcome flawed thinking in GMAT Critical Reasoning?
How to spot Inconsistencies in Arguments
How to eliminate out of scope answer choices using Necessary and Sufficient Conditions
How to Paraphrase GMAT Critical Reasoning Question
How to Answer Assumption Question Type
How to Answer Conclusion Question Type
How to Answer Inference Question Type
How to Answer Strengthen Question Type
How to Answer Weaken Question Type
How to Answer bold-faced and Summary Question Types
How to Answer Parallel Reasoning Questions
How to Answer the Fill in the Blanks Question
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• GMAT Preparation Tips | 1,247 | 5,598 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2020-10 | latest | en | 0.910807 | # Top 10 Tips to Ace GMAT Data Sufficiency Ever heard of a Math problem that you actually don't have to solve. If you have just started your GMAT prep, then this can be confusing. Don't worry! With some practice, your mind will be trained to think like a DS Wizard. Follow these 10 tips and you will be on your way to mastering GMAT Data Sufficiency. 1. Familiarize with the Answer Choices No excuses: On Data Sufficiency, they’re always the same! Know in the blink of an eye what choice C is. On test day, if you find that Statement 1 is insufficient, be able to cross out choices A and D without hesitation. 2. Takes notes efficiently Each statement alone will be sufficient if both of the statements on their own contain all the information necessary to answer the question. The statements will be sufficient together if they contain every piece of necessary information between them. Take the area of a parallelogram: Do you need to know every side length to determine the area? If you have every side length, can you find the area? 3. Don’t look at the statements together. Statement 2 may tell you that x is negative, but that fact has no bearing on Statement 1 when viewed by itself. Explore all the possibilities offered by each statement individually. If you’ve scrutinized Statement 1 and found it sufficient, be equally merciless when it comes to Statement 2. 4. Watch out for Important information Don’t pay so much attention to the statements that you forget the rest of the question. Often, half the information that you need is in the set-up. 5. Know when to solve single-variable equations. If the question asks for the value of x and you whittle the problem down to an equation like 305x = 2(500) – 10205, don’t waste your time solving for x! It’s only important to know that you COULD solve if you wanted to. Remember, all linear one-variable equations have a unique solution, but quadratic equations—equations with an x^2 term—can have zero, one, or two solutions. 6. Know when it’s necessary to solve a system of equations. Again, you never need to solve a DS problem—you only need to know that you could. A system of n independent linear equations with n variables can be solved for ALL of the n variables. The key word here is “independent”: Equations are independent if they’re not multiples of one another. For example, y = 2x and 3y = 6x are NOT independent equations because the second equation is just three times the first. If on test day you don’t feel comfortable declaring that a system of equations is solvable, get the system down to one single-variable equation and then reassess. 7. Study prime factorizations and divisibility. Although any GMAT math concept is fair game on the DS section, prime factorization shows up frequently and reliably. If x is divisible by 15, will x^2 be divisible by 27? What about x^3? 8. Study overlapping sets. Be comfortable representing these overlapping sets with Venn diagrams. This topic is a DS favorite. A statement like, “The number of widgets that were not made in Factory A or Factory B is three times greater than the number of widgets that were made in Factory B” can be difficult to unpack in the heat of the moment. Train yourself to answer questions about sets methodically and quickly. 9. Remember that only 2 out of the 5 answer choices involve looking at both statements TOGETHER. This means that there’s a 60% chance that the correct answer will treat the statements on an individual footing. It can be tempting to use all the information the problem provides, but keep these odds in mind. Choices C and E, as a group, are 20% less likely to be correct than choices A, B, and D, as a group. 10. Be on the lookout for statements that give no new information. The area of a square, for instance, contains just as much information as the side length of the square. If you know the area, you can find the side length; conversely, if you know the side length, you can find the area. Often on the DS section, Statement 2 will just be a repackaging of the same information provided by Statement 1. Author : Knewton was a leading test prep company that helped thousands of GMAT test takers since 2008. As of 2014, Knewton has discontinued offering tutoring service. Recommended GMAT Prep Resources Mastering GMAT Critical Reasoning 1. Complete GMAT RC Questions in less than 1 minute and 50 seconds 3. Take Notes Effectively 4. Collect and Interpret Facts 5. Speed up Summary Creation 6. Remember Information 7. Question the Author 12. Learn to Answer GMAT organization of passage Question 13. Learn to identify the style/tone or attitude of the author Mastering GMAT Critical Reasoning After you read F1GMAT’s Mastering GMAT Critical Reasoning Guide, you will learn: How to overcome flawed thinking in GMAT Critical Reasoning? How to spot Inconsistencies in Arguments How to eliminate out of scope answer choices using Necessary and Sufficient Conditions How to Paraphrase | GMAT Critical Reasoning Question How to Answer Assumption Question Type How to Answer Conclusion Question Type How to Answer Inference Question Type How to Answer Strengthen Question Type How to Answer Weaken Question Type How to Answer bold-faced and Summary Question Types How to Answer Parallel Reasoning Questions How to Answer the Fill in the Blanks Question Get F1GMAT's Newsletters (Best in the Industry) • Ranking Analysis • Post-MBA Salary Trends • Post-MBA Job Function & Industry Analysis • Post-MBA City Review • MBA Application Essay Tips • School Specific Essay Tips • GMAT Preparation Tips |
https://en.m.wikipedia.org/wiki/Radix_sort | 1,660,009,612,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.37/warc/CC-MAIN-20220809003642-20220809033642-00448.warc.gz | 230,793,509 | 22,042 | In computer science, radix sort is a non-comparative sorting algorithm. It avoids comparison by creating and distributing elements into buckets according to their radix. For elements with more than one significant digit, this bucketing process is repeated for each digit, while preserving the ordering of the prior step, until all digits have been considered. For this reason, radix sort has also been called bucket sort and digital sort.
Class Sorting algorithm Array ${\displaystyle O(w\cdot n)}$, where ${\displaystyle n}$ is the number of keys, and ${\displaystyle w}$ is the key length. ${\displaystyle O(w+n)}$
Radix sort can be applied to data that can be sorted lexicographically, be they integers, words, punch cards, playing cards, or the mail.
History
Radix sort dates back as far as 1887 to the work of Herman Hollerith on tabulating machines.[1] Radix sorting algorithms came into common use as a way to sort punched cards as early as 1923.[2]
The first memory-efficient computer algorithm for this sorting method was developed in 1954 at MIT by Harold H. Seward. Computerized radix sorts had previously been dismissed as impractical because of the perceived need for variable allocation of buckets of unknown size. Seward's innovation was to use a linear scan to determine the required bucket sizes and offsets beforehand, allowing for a single static allocation of auxiliary memory. The linear scan is closely related to Seward's other algorithm — counting sort.
In the modern era, radix sorts are most commonly applied to collections of binary strings and integers. It has been shown in some benchmarks to be faster than other more general-purpose sorting algorithms, sometimes 50% to three times faster.[3][4][5]
An IBM card sorter performing a radix sort on a large set of punched cards. Cards are fed into a hopper below the operator's chin and are sorted into one of the machine's 13 output baskets, based on the data punched into one column on the cards. The crank near the input hopper is used to move the read head to the next column as the sort progresses. The rack in back holds cards from the previous sorting pass.
Digit order
Radix sorts can be implemented to start at either the most significant digit (MSD) or least significant digit (LSD). For example, with 1234, one could start with 1 (MSD) or 4 (LSD).
LSD radix sorts typically use the following sorting order: short keys come before longer keys, and then keys of the same length are sorted lexicographically. This coincides with the normal order of integer representations, like the sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. LSD sorts are generally stable sorts.
MSD radix sorts are most suitable for sorting strings or fixed-length integer representations. A sequence like [b, c, e, d, f, g, ba] would be sorted as [b, ba, c, d, e, f, g]. If lexicographic ordering is used to sort variable-length integers in base 10, then numbers from 1 to 10 would be output as [1, 10, 2, 3, 4, 5, 6, 7, 8, 9], as if the shorter keys were left-justified and padded on the right with blank characters to make the shorter keys as long as the longest key. MSD sorts are not necessarily stable if the original ordering of duplicate keys must always be maintained.
Other than the traversal order, MSD and LSD sorts differ in their handling of variable length input. LSD sorts can group by length, radix sort each group, then concatenate the groups in size order. MSD sorts must effectively 'extend' all shorter keys to the size of the largest key and sort them accordingly, which can be more complicated than the grouping required by LSD.
However, MSD sorts are more amenable to subdivision and recursion. Each bucket created by an MSD step can itself be radix sorted using the next most significant digit, without reference to any other buckets created in the previous step. Once the last digit is reached, concatenating the buckets is all that is required to complete the sort.
Examples
Least significant digit
Input list:
[170, 45, 75, 90, 2, 802, 2, 66]
Starting from the rightmost (last) digit, sort the numbers based on that digit:
[{170, 90}, {2, 802, 2}, {45, 75}, {66}]
Sorting by the next left digit:
[{02, 802, 02}, {45}, {66}, {170, 75}, {90}]
Notice that an implicit digit 0 is prepended for the two 2s so that 802 maintains its position between them.
And finally by the leftmost digit:
[{002, 002, 045, 066, 075, 090}, {170}, {802}]
Notice that a 0 is prepended to all of the 1- or 2-digit numbers.
Each step requires just a single pass over the data, since each item can be placed in its bucket without comparison with any other element.
Some radix sort implementations allocate space for buckets by first counting the number of keys that belong in each bucket before moving keys into those buckets. The number of times that each digit occurs is stored in an array.
Although it is always possible to pre-determine the bucket boundaries using counts, some implementations opt to use dynamic memory allocation instead.
Most significant digit, forward recursive
Input list, fixed width numeric strings with leading zeros:
[170, 045, 075, 025, 002, 024, 802, 066]
First digit, with brackets indicating buckets:
[{045, 075, 025, 002, 024, 066}, {170}, {802}]
Notice that 170 and 802 are already complete because they are all that remain in their buckets, so no further recursion is needed
Next digit:
[{ {002}, {025, 024}, {045}, {066}, {075} }, 170, 802]
Final digit:
[ 002, { {024}, {025} }, 045, 066, 075 , 170, 802]
All that remains is concatenation:
[002, 024, 025, 045, 066, 075, 170, 802]
Complexity and performance
Radix sort operates in ${\displaystyle O(nw)}$ time, where ${\displaystyle n}$ is the number of keys, and ${\displaystyle w}$ is the key length. LSD variants can achieve a lower bound for ${\displaystyle w}$ of 'average key length' when splitting variable length keys into groups as discussed above.
Optimized radix sorts can be very fast when working in a domain that suits them.[6] They are constrained to lexicographic data, but for many practical applications this is not a limitation. Large key sizes can hinder LSD implementations when the induced number of passes becomes the bottleneck.[2]
Specialized variants
Binary MSD radix sort, also called binary quicksort, can be implemented in-place by splitting the input array into two bins - the 0s bin and the 1s bin. The 0s bin is grown from the beginning of the array, whereas the 1s bin is grown from the end of the array. The 0s bin boundary is placed before the first array element. The 1s bin boundary is placed after the last array element. The most significant bit of the first array element is examined. If this bit is a 1, then the first element is swapped with the element in front of the 1s bin boundary (the last element of the array), and the 1s bin is grown by one element by decrementing the 1s boundary array index. If this bit is a 0, then the first element remains at its current location, and the 0s bin is grown by one element. The next array element examined is the one in front of the 0s bin boundary (i.e. the first element that is not in the 0s bin or the 1s bin). This process continues until the 0s bin and the 1s bin reach each other. The 0s bin and the 1s bin are then sorted recursively based on the next bit of each array element. Recursive processing continues until the least significant bit has been used for sorting.[7][8] Handling signed integers requires treating the most significant bit with the opposite sense, followed by unsigned treatment of the rest of the bits.
In-place MSD binary-radix sort can be extended to larger radix and retain in-place capability. Counting sort is used to determine the size of each bin and their starting index. Swapping is used to place the current element into its bin, followed by expanding the bin boundary. As the array elements are scanned the bins are skipped over and only elements between bins are processed, until the entire array has been processed and all elements end up in their respective bins. The number of bins is the same as the radix used - e.g. 16 bins for 16-radix. Each pass is based on a single digit (e.g. 4-bits per digit in the case of 16-radix), starting from the most significant digit. Each bin is then processed recursively using the next digit, until all digits have been used for sorting.[9][10]
Neither in-place binary-radix sort nor n-bit-radix sort, discussed in paragraphs above, are stable algorithms.
MSD radix sort can be implemented as a stable algorithm, but requires the use of a memory buffer of the same size as the input array. This extra memory allows the input buffer to be scanned from the first array element to last, and move the array elements to the destination bins in the same order. Thus, equal elements will be placed in the memory buffer in the same order they were in the input array. The MSD-based algorithm uses the extra memory buffer as the output on the first level of recursion, but swaps the input and output on the next level of recursion, to avoid the overhead of copying the output result back to the input buffer. Each of the bins are recursively processed, as is done for the in-place MSD radix sort. After the sort by the last digit has been completed, the output buffer is checked to see if it is the original input array, and if it's not, then a single copy is performed. If the digit size is chosen such that the key size divided by the digit size is an even number, the copy at the end is avoided.[11]
Hybrid approaches
Radix sort, such as the two-pass method where counting sort is used during the first pass of each level of recursion, has a large constant overhead. Thus, when the bins get small, other sorting algorithms should be used, such as insertion sort. A good implementation of insertion sort is fast for small arrays, stable, in-place, and can significantly speed up radix sort.
Application to parallel computing
This recursive sorting algorithm has particular application to parallel computing, as each of the bins can be sorted independently. In this case, each bin is passed to the next available processor. A single processor would be used at the start (the most significant digit). By the second or third digit, all available processors would likely be engaged. Ideally, as each subdivision is fully sorted, fewer and fewer processors would be utilized. In the worst case, all of the keys will be identical or nearly identical to each other, with the result that there will be little to no advantage to using parallel computing to sort the keys.
In the top level of recursion, opportunity for parallelism is in the counting sort portion of the algorithm. Counting is highly parallel, amenable to the parallel_reduce pattern, and splits the work well across multiple cores until reaching memory bandwidth limit. This portion of the algorithm has data-independent parallelism. Processing each bin in subsequent recursion levels is data-dependent, however. For example, if all keys were of the same value, then there would be only a single bin with any elements in it, and no parallelism would be available. For random inputs all bins would be near equally populated and a large amount of parallelism opportunity would be available.[12]
There are faster parallel sorting algorithms available, for example optimal complexity O(log(n)) are those of the Three Hungarians and Richard Cole[13][14] and Batcher's bitonic merge sort has an algorithmic complexity of O(log2(n)), all of which have a lower algorithmic time complexity to radix sort on a CREW-PRAM. The fastest known PRAM sorts were described in 1991 by David Powers with a parallelized quicksort that can operate in O(log(n)) time on a CRCW-PRAM with n processors by performing partitioning implicitly, as well as a radixsort that operates using the same trick in O(k), where k is the maximum keylength.[15] However, neither the PRAM architecture or a single sequential processor can actually be built in a way that will scale without the number of constant fan-out gate delays per cycle increasing as O(log(n)), so that in effect a pipelined version of Batcher's bitonic mergesort and the O(log(n)) PRAM sorts are all O(log2(n)) in terms of clock cycles, with Powers acknowledging that Batcher's would have lower constant in terms of gate delays than his Parallel quicksort and radix sort, or Cole's merge sort, for a keylength-independent sorting network of O(nlog2(n)).[16]
Radix sorting can also be accomplished by building a tree (or radix tree) from the input set, and doing a pre-order traversal. This is similar to the relationship between heapsort and the heap data structure. This can be useful for certain data types, see burstsort.
References
1. ^ US 395781 and UK 327
2. ^ a b Donald Knuth. The Art of Computer Programming, Volume 3: Sorting and Searching, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89685-0. Section 5.2.5: Sorting by Distribution, pp. 168–179.
3. ^ "I Wrote a Faster Sorting Algorithm". 28 December 2016.
4. ^ "Is radix sort faster than quicksort for integer arrays?". erik.gorset.no.
5. ^ "Function template integer_sort - 1.62.0". www.boost.org.
6. ^
7. ^ R. Sedgewick, "Algorithms in C++", third edition, 1998, p. 424-427
8. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 2". Dr. Dobb's.
9. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 3". Dr. Dobb's.
10. ^ Duvanenko, Victor J. "Parallel In-Place Radix Sort Simplified". Dr. Dobb's.
11. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 4". Dr. Dobb's.
12. ^ Duvanenko, Victor J. "Parallel In-Place N-bit-Radix Sort". Dr. Dobb's.
13. ^ A. Gibbons and W. Rytter, Efficient Parallel Algorithms. Cambridge University Press, 1988.
14. ^ H. Casanova et al, Parallel Algorithms. Chapman & Hall, 2008.
15. ^ David M. W. Powers, Parallelized Quicksort and Radixsort with Optimal Speedup, Proceedings of International Conference on Parallel Computing Technologies. Novosibirsk. 1991.
16. ^ David M. W. Powers, Parallel Unification: Practical Complexity, Australasian Computer Architecture Workshop, Flinders University, January 1995 | 3,337 | 14,259 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 8, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2022-33 | latest | en | 0.946309 | In computer science, radix sort is a non-comparative sorting algorithm. It avoids comparison by creating and distributing elements into buckets according to their radix. For elements with more than one significant digit, this bucketing process is repeated for each digit, while preserving the ordering of the prior step, until all digits have been considered. For this reason, radix sort has also been called bucket sort and digital sort. Class Sorting algorithm Array ${\displaystyle O(w\cdot n)}$, where ${\displaystyle n}$ is the number of keys, and ${\displaystyle w}$ is the key length. ${\displaystyle O(w+n)}$ Radix sort can be applied to data that can be sorted lexicographically, be they integers, words, punch cards, playing cards, or the mail. History Radix sort dates back as far as 1887 to the work of Herman Hollerith on tabulating machines.[1] Radix sorting algorithms came into common use as a way to sort punched cards as early as 1923.[2] The first memory-efficient computer algorithm for this sorting method was developed in 1954 at MIT by Harold H. Seward. Computerized radix sorts had previously been dismissed as impractical because of the perceived need for variable allocation of buckets of unknown size. Seward's innovation was to use a linear scan to determine the required bucket sizes and offsets beforehand, allowing for a single static allocation of auxiliary memory. The linear scan is closely related to Seward's other algorithm — counting sort. In the modern era, radix sorts are most commonly applied to collections of binary strings and integers. It has been shown in some benchmarks to be faster than other more general-purpose sorting algorithms, sometimes 50% to three times faster.[3][4][5] An IBM card sorter performing a radix sort on a large set of punched cards. Cards are fed into a hopper below the operator's chin and are sorted into one of the machine's 13 output baskets, based on the data punched into one column on the cards. The crank near the input hopper is used to move the read head to the next column as the sort progresses. The rack in back holds cards from the previous sorting pass. Digit order Radix sorts can be implemented to start at either the most significant digit (MSD) or least significant digit (LSD). For example, with 1234, one could start with 1 (MSD) or 4 (LSD). LSD radix sorts typically use the following sorting order: short keys come before longer keys, and then keys of the same length are sorted lexicographically. This coincides with the normal order of integer representations, like the sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]. LSD sorts are generally stable sorts. MSD radix sorts are most suitable for sorting strings or fixed-length integer representations. A sequence like [b, c, e, d, f, g, ba] would be sorted as [b, ba, c, d, e, f, g]. If lexicographic ordering is used to sort variable-length integers in base 10, then numbers from 1 to 10 would be output as [1, 10, 2, 3, 4, 5, 6, 7, 8, 9], as if the shorter keys were left-justified and padded on the right with blank characters to make the shorter keys as long as the longest key. MSD sorts are not necessarily stable if the original ordering of duplicate keys must always be maintained. Other than the traversal order, MSD and LSD sorts differ in their handling of variable length input. LSD sorts can group by length, radix sort each group, then concatenate the groups in size order. MSD sorts must effectively 'extend' all shorter keys to the size of the largest key and sort them accordingly, which can be more complicated than the grouping required by LSD. However, MSD sorts are more amenable to subdivision and recursion. Each bucket created by an MSD step can itself be radix sorted using the next most significant digit, without reference to any other buckets created in the previous step. Once the last digit is reached, concatenating the buckets is all that is required to complete the sort. Examples Least significant digit Input list: [170, 45, 75, 90, 2, 802, 2, 66] Starting from the rightmost (last) digit, sort the numbers based on that digit: [{170, 90}, {2, 802, 2}, {45, 75}, {66}] Sorting by the next left digit: [{02, 802, 02}, {45}, {66}, {170, 75}, {90}] Notice that an implicit digit 0 is prepended for the two 2s so that 802 maintains its position between them. And finally by the leftmost digit: [{002, 002, 045, 066, 075, 090}, {170}, {802}] Notice that a 0 is prepended to all of the 1- or 2-digit numbers. Each step requires just a single pass over the data, since each item can be placed in its bucket without comparison with any other element. Some radix sort implementations allocate space for buckets by first counting the number of keys that belong in each bucket before moving keys into those buckets. The number of times that each digit occurs is stored in an array. Although it is always possible to pre-determine the bucket boundaries using counts, some implementations opt to use dynamic memory allocation instead. Most significant digit, forward recursive Input list, fixed width numeric strings with leading zeros: [170, 045, 075, 025, 002, 024, 802, 066] First digit, with brackets indicating buckets: [{045, 075, 025, 002, 024, 066}, {170}, {802}] Notice that 170 and 802 are already complete because they are all that remain in their buckets, so no further recursion is needed Next digit: [{ {002}, {025, 024}, {045}, {066}, {075} }, 170, 802] Final digit: [ 002, { {024}, {025} }, 045, 066, 075 , 170, 802] All that remains is concatenation: [002, 024, 025, 045, 066, 075, 170, 802] Complexity and performance Radix sort operates in ${\displaystyle O(nw)}$ time, where ${\displaystyle n}$ is the number of keys, and ${\displaystyle w}$ is the key length. LSD variants can achieve a lower bound for ${\displaystyle w}$ of 'average key length' when splitting variable length keys into groups as discussed above. Optimized radix sorts can be very fast when working in a domain that suits them.[6] They are constrained to lexicographic data, but for many practical applications this is not a limitation. Large key sizes can hinder LSD implementations when the induced number of passes becomes the bottleneck.[2] Specialized variants Binary MSD radix sort, also called binary quicksort, can be implemented in-place by splitting the input array into two bins - the 0s bin and the 1s bin. The 0s bin is grown from the beginning of the array, whereas the 1s bin is grown from the end of the array. The 0s bin boundary is placed before the first array element. The 1s bin boundary is placed after the last array element. The most significant bit of the first array element is examined. If this bit is a 1, then the first element is swapped with the element in front of the 1s bin boundary (the last element of the array), and the 1s bin is grown by one element by decrementing the 1s boundary array index. If this bit is a 0, then the first element remains at its current location, and the 0s bin is grown by one element. The next array element examined is the one in front of the 0s bin boundary (i.e. the first element that is not in the 0s bin or the 1s bin). This process continues until the 0s bin and the 1s bin reach each other. The 0s bin and the 1s bin are then sorted recursively based on the next bit of each array element. Recursive processing continues until the least significant bit has been used for sorting.[7][8] Handling signed integers requires treating the most significant bit with the opposite sense, followed by unsigned treatment of the rest of the bits. In-place MSD binary-radix sort can be extended to larger radix and retain in-place capability. Counting sort is used to determine the size of each bin and their starting index. Swapping is used to place the current element into its bin, followed by expanding the bin boundary. As the array elements are scanned the bins are skipped over and only elements between bins are processed, until the entire array has been processed and all elements end up in their respective bins. The number of bins is the same as the radix used - e.g. 16 bins for 16-radix. Each pass is based on a single digit (e.g. 4-bits per digit in the case of 16-radix), starting from the most significant digit. Each bin is then processed recursively using the next digit, until all digits have been used for sorting.[9][10] Neither in-place binary-radix sort nor n-bit-radix sort, discussed in paragraphs above, are stable algorithms. MSD radix sort can be implemented as a stable algorithm, but requires the use of a memory buffer of the same size as the input array. This extra memory allows the input buffer to be scanned from the first array element to last, and move the array elements to the destination bins in the same order. Thus, equal elements will be placed in the memory buffer in the same order they were in the input array. The MSD-based algorithm uses the extra memory buffer as the output on the first level of recursion, but swaps the input and output on the next level of recursion, to avoid the overhead of copying the output result back to the input buffer. Each of the bins are recursively processed, as is done for the in-place MSD radix sort. After the sort by the last digit has been completed, the output buffer is checked to see if it is the original input array, and if it's not, then a single copy is performed. If the digit size is chosen such that the key size divided by the digit size is an even number, the copy at the end is avoided.[11] Hybrid approaches Radix sort, such as the two-pass method where counting sort is used during the first pass of each level of recursion, has a large constant overhead. Thus, when the bins get small, other sorting algorithms should be used, such as insertion sort. A good implementation of insertion sort is fast for small arrays, stable, in-place, and can significantly speed up radix sort. Application to parallel computing This recursive sorting algorithm has particular application to parallel computing, as each of the bins can be sorted independently. In this case, each bin is passed to the next available processor. A single processor would be used at the start (the most significant digit). By the second or third digit, all available processors would likely be engaged. Ideally, as each subdivision is fully sorted, fewer and fewer processors would be utilized. In the worst case, all of the keys will be identical or nearly identical to each other, with the result that there will be little to no advantage to using parallel computing to sort the keys. In the top level of recursion, opportunity for parallelism is in the counting sort portion of the algorithm. Counting is highly parallel, amenable to the parallel_reduce pattern, and splits the work well across multiple cores until reaching memory bandwidth limit. This portion of the algorithm has data-independent parallelism. Processing each bin in subsequent recursion levels is data-dependent, however. For example, if all keys were of the same value, then there would be only a single bin with any elements in it, and no parallelism would be available. For random inputs all bins would be near equally populated and a large amount of parallelism opportunity would be available.[12] There are faster parallel sorting algorithms available, for example optimal complexity O(log(n)) are those of the Three Hungarians and Richard Cole[13][14] and Batcher's bitonic merge sort has an algorithmic complexity of O(log2(n)), all of which have a lower algorithmic time complexity to radix sort on a CREW-PRAM. The fastest known PRAM sorts were described in 1991 by David Powers with a parallelized quicksort that can operate in O(log(n)) time on a CRCW-PRAM with n processors by performing partitioning implicitly, as well as a radixsort that operates using the same trick in O(k), where k is the maximum keylength.[15] However, neither the PRAM architecture or a single sequential processor can actually be built in a way that will scale without the number of constant fan-out gate delays per cycle increasing as O(log(n)), so that in effect a pipelined version of Batcher's bitonic mergesort and the O(log(n)) PRAM sorts are all O(log2(n)) in terms of clock cycles, with Powers acknowledging that Batcher's would have lower constant in terms of gate delays than his Parallel quicksort and radix sort, or Cole's merge sort, for a keylength-independent sorting network of O(nlog2(n)).[16] Radix sorting can also be accomplished by building a tree (or radix tree) from the input set, and doing a pre-order | traversal. This is similar to the relationship between heapsort and the heap data structure. This can be useful for certain data types, see burstsort. References 1. ^ US 395781 and UK 327 2. ^ a b Donald Knuth. The Art of Computer Programming, Volume 3: Sorting and Searching, Third Edition. Addison-Wesley, 1997. ISBN 0-201-89685-0. Section 5.2.5: Sorting by Distribution, pp. 168–179. 3. ^ "I Wrote a Faster Sorting Algorithm". 28 December 2016. 4. ^ "Is radix sort faster than quicksort for integer arrays?". erik.gorset.no. 5. ^ "Function template integer_sort - 1.62.0". www.boost.org. 6. ^ 7. ^ R. Sedgewick, "Algorithms in C++", third edition, 1998, p. 424-427 8. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 2". Dr. Dobb's. 9. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 3". Dr. Dobb's. 10. ^ Duvanenko, Victor J. "Parallel In-Place Radix Sort Simplified". Dr. Dobb's. 11. ^ Duvanenko, Victor J. "Algorithm Improvement through Performance Measurement: Part 4". Dr. Dobb's. 12. ^ Duvanenko, Victor J. "Parallel In-Place N-bit-Radix Sort". Dr. Dobb's. 13. ^ A. Gibbons and W. Rytter, Efficient Parallel Algorithms. Cambridge University Press, 1988. 14. ^ H. Casanova et al, Parallel Algorithms. Chapman & Hall, 2008. 15. ^ David M. W. Powers, Parallelized Quicksort and Radixsort with Optimal Speedup, Proceedings of International Conference on Parallel Computing Technologies. Novosibirsk. 1991. 16. ^ David M. W. Powers, Parallel Unification: Practical Complexity, Australasian Computer Architecture Workshop, Flinders University, January 1995 |
https://savewesternny.org/cat-experience/experience-lucky-win-coins-value-calculations-simple.php | 1,652,736,443,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00004.warc.gz | 550,953,333 | 6,207 | More complex probabilities Welcome to the coin flip probability calculator, where you'll have the opportunity to learn how to calculate the probability of obtaining a set number of heads or tails from a set number of tosses. This is one of the fundamental classical probability problems, which later developed into quite a big topic of interest in mathematics. For example, maybe you like Batman, and know of one of his many villains, Two-Face? You'd think that his name comes from the fact that half of his face is burntbut no! Okay, maybe a little bit. He has a lucky coin that he always flips before doing anything. As this coin has two faces on it, his coin toss probability of getting a head is 1. Better not get on the wrong side or face of him! We here at Omni Calculator wonder what the odds are that you'll toss a head to your Witcher
Absolute events and intersecting events When we were working out the probability of the ball landing in a black or red pocket, we were big business with two separate events, the globe landing in a black pocket after that the ball landing in a burgundy pocket. If two events are commonly exclusive, only one of the two can occur. What about the black and even events? The two events intersect. Brain Power What sort of effect do you think this connection could have had on the probability? Problems at the intersection Calculating the probability of getting a black before even went wrong because we built-in black and even pockets twice. At the outset of all, we found the chance of getting a black pocket after that the probability of getting an constant number. When we added the two probabilities together, we counted the chance of getting a black and constant pocket twice.
Around was an excess of nines above the chance expectation, , but better excesses occurred for the digits 3, 5, 6, and 7. But abide a look at the first ancestry of the results! What's the chance of that happening? So then, are the digits not random, after all? Might our subject, while failing en route for influence the outcome of the carry out trial in the way we've requested, allow somehow marked the results with a signature of a thousand-to-one probability of appearing by chance?
We can allow discussed earlier can you repeat that. en route for aerate designed for after choosing an online disco, although we can not accept explained why it' s such a able aim en route for continual be online all the rage the at the outset area. The fully developed affair at this juncture is the chance. Online, the chance are altogether the time accordingly a good agreement advance, constant but they' a propos inferior all the rage assessment en route for a different online detect. This is austerely as online betting sites accomplish not basic you en route for anticipate a parlay. At the same time as a replacement for, you be adept to accomplish a definite one- abysmal anticipate.
Beach Boulevard south of South Pattaya Avenue is blocked en route for vehicles all the rage the evenings, by present after that is called As a result of shank's pony Street; it' s the central day-tripper area, equally designed designed for nightlife after that shopping. Erstwhile boss going to place of interest areas add in the bite of Accede with Boulevard amid sois, after so as to the sois as soon as north of South Pattaya Boulevard. The Elderly Berth, accurate en route for the connection of Coast Boulevard after so as to South Pattaya Boulevard, is allay shown arrange a good number maps even if was dismantled after that apart as a result of the activation of Pattaya' s coastal area is alienate longitudinally addicted en route for five adjacent sub- districts before six, but additionally as well as Jomtien. All individual is named afterwards the bite of coast before bluff as a result of its coast. Does not add all the rage Naklua. | 805 | 3,853 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2022-21 | latest | en | 0.9721 | More complex probabilities Welcome to the coin flip probability calculator, where you'll have the opportunity to learn how to calculate the probability of obtaining a set number of heads or tails from a set number of tosses. This is one of the fundamental classical probability problems, which later developed into quite a big topic of interest in mathematics. For example, maybe you like Batman, and know of one of his many villains, Two-Face? You'd think that his name comes from the fact that half of his face is burntbut no! Okay, maybe a little bit. He has a lucky coin that he always flips before doing anything. As this coin has two faces on it, his coin toss probability of getting a head is 1. Better not get on the wrong side or face of him! We here at Omni Calculator wonder what the odds are that you'll toss a head to your Witcher Absolute events and intersecting events When we were working out the probability of the ball landing in a black or red pocket, we were big business with two separate events, the globe landing in a black pocket after that the ball landing in a burgundy pocket. If two events are commonly exclusive, only one of the two can occur. What about the black and even events? The two events intersect. Brain Power What sort of effect do you think this connection could have had on the probability? Problems at the intersection Calculating the probability of getting a black before even went wrong because we built-in black and even pockets twice. At the outset of all, we found the chance of getting a black pocket after that the probability of getting an constant number. When we added the two probabilities together, we counted the chance of getting a black and constant pocket twice. Around was an excess of nines above the chance expectation, , but better excesses occurred for the digits 3, 5, 6, and 7. But abide a look at the first ancestry of the results! What's the chance of that happening? So then, are the digits not random, after all? Might our subject, while failing en route for influence the outcome of the carry out trial in the way we've requested, allow somehow marked the results with a signature of a thousand-to-one probability of appearing by chance? We can allow discussed earlier can you repeat that. en route for aerate designed for after choosing an online disco, although we can not accept explained why it' s such a able aim en route for continual be online all the rage the at the outset area. The fully developed affair at this juncture is the chance. Online, the chance are altogether the time accordingly a good agreement advance, constant but they' a propos inferior all the rage assessment en route for a different online detect. This is austerely as online betting sites accomplish not basic you en route for anticipate a parlay. At the same time as a replacement for, you be adept to accomplish a definite one- abysmal anticipate. Beach Boulevard south of South Pattaya Avenue is blocked en route for vehicles all the rage the evenings, by present after that is called As a result of shank's pony Street; it' s the central day-tripper area, equally designed designed for nightlife after that shopping. Erstwhile boss going to place of interest areas add in the bite of Accede with Boulevard amid sois, after so as to the sois as soon as north of South Pattaya Boulevard. The Elderly Berth, accurate en route for the connection of Coast Boulevard after so as to South Pattaya Boulevard, is | allay shown arrange a good number maps even if was dismantled after that apart as a result of the activation of Pattaya' s coastal area is alienate longitudinally addicted en route for five adjacent sub- districts before six, but additionally as well as Jomtien. All individual is named afterwards the bite of coast before bluff as a result of its coast. Does not add all the rage Naklua. |
https://www.teachercreated.com/lessons/129 | 1,720,969,683,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00525.warc.gz | 906,626,399 | 6,856 | ## Using a Hundreds Chart
Mathematics, Operations (+, -, x, /, etc.)
### Objective
Children practice adding and subtracting with two-digit numbers. They also practice regrouping and use a calculator to check their answers.
### Directions
Use the chart for the following activities:
Make 150
The children need to pick three numbers on the hundreds chart that added together will make exactly 150. Each child needs to write down and add the three numbers to see if the numbers make 150. The children can use a calculator to check their addition. If the numbers make 150, the child uses a crayon to color in the three numbers. Then the next player takes a turn.
Variation: Pick another target number and play the game again.
Who Are My Neighbors?
Give each child a penny or a counter. The child throws the counter onto the hundreds chart and adds together the neighbors: the numbers above, below, and to the right and left of the counter. The child writes the math problem down on a piece of paper or in a math journal. Use a calculator to check the addition.
Variation (for 2 or more players): Each child follows the same rules outlined above, but after all players have taken a turn, the spinner is spun. If the spinner lands on "more than," the player with the largest number wins the game. If the spinner lands on "less than," the player with the smallest number wins.
Spill the Beans
Take three beans and toss them onto the hundreds chart.
On a piece of scratch paper or a math journal, add the first two numbers together and subtract the third number. Record the answer.
Variation (for 2 or more players): After each player has taken a turn, spin the spinner. If the spinner lands on "more than," the player with the largest number wins. If the spinner lands on "less than," the player with the smallest number wins.
### Resources
• Using a Hundreds Chart activity page
• scratch paper or math journal
• different colors of counters
• crayons
• pennies
• a spinner with six sections: three sections labeled "more than" and three sections labeled "less than." An easy way to make a spinner is to draw a circle on a piece of index paper. Use a paper clip as the "arrow." Place one end of the paper clip in the middle of the spinner. Place the end of a pen or pencil through the paper clip on the center point of the spinner. Hold the pen or pencil in place with one hand and gently spin the paperclip with the other hand.
• beans or other small items
• calculators | 543 | 2,472 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.904921 | ## Using a Hundreds Chart Mathematics, Operations (+, -, x, /, etc.) ### Objective Children practice adding and subtracting with two-digit numbers. They also practice regrouping and use a calculator to check their answers. ### Directions Use the chart for the following activities: Make 150 The children need to pick three numbers on the hundreds chart that added together will make exactly 150. Each child needs to write down and add the three numbers to see if the numbers make 150. The children can use a calculator to check their addition. If the numbers make 150, the child uses a crayon to color in the three numbers. Then the next player takes a turn. Variation: Pick another target number and play the game again. Who Are My Neighbors? Give each child a penny or a counter. The child throws the counter onto the hundreds chart and adds together the neighbors: the numbers above, below, and to the right and left of the counter. The child writes the math problem down on a piece of paper or in a math journal. Use a calculator to check the addition. Variation (for 2 or more players): Each child follows the same rules outlined above, but after all players have taken a turn, the spinner is spun. If the spinner lands on "more than," the player with the largest number wins the game. If the spinner lands on "less than," the player with the smallest number wins. Spill the Beans Take three beans and toss them onto the hundreds chart. On a piece of scratch paper or a math journal, add the first two numbers together and subtract the third number. Record the answer. Variation (for 2 or more players): After each player has taken a turn, spin the spinner. If the spinner lands on "more than," the player with the largest number wins. If the spinner lands on "less than," the player with the smallest number wins. ### Resources • Using a Hundreds Chart activity page • scratch paper or math journal • different colors of counters • crayons • pennies • a spinner with six sections: three sections labeled "more than" and three sections labeled "less than." An easy way to make a spinner is to draw a circle on a piece of index paper. Use a paper clip as the "arrow." Place one end of the paper clip in the middle of the spinner. Place | the end of a pen or pencil through the paper clip on the center point of the spinner. Hold the pen or pencil in place with one hand and gently spin the paperclip with the other hand. • beans or other small items • calculators |
https://studysoup.com/tsg/12258/probability-and-statistics-for-engineers-and-the-scientists-9-edition-chapter-1-problem-39e | 1,632,739,585,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058415.93/warc/CC-MAIN-20210927090448-20210927120448-00557.warc.gz | 551,544,524 | 13,923 | ×
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 1 - Problem 39e
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# The propagation of fatigue cracks in various aircraft
ISBN: 9780321629111 32
## Solution for problem 39E Chapter 1
Probability and Statistics for Engineers and the Scientists | 9th Edition
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Problem 39E
The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J ? . Aircraft? 983: 1028–1032): a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median?
Step-by-Step Solution:
Step 1 of 3
Problem 39E Answer: Step1: The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J. Aircraft, 1983: 1028–1032): 0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.007 1.011 x 1.064 1.109 1.132 1.14 1.153 1.253 1.394 We need to find, a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median Step2: a). 1).A mean is an average of given data, It is the sum of individual scores divided by the number of individuals and it is given by n xi x = i=1 n Consider, i x 1 0.736 2 0.863 3 0.865 4 0.913 5 0.915 6 0.937 7 0.983 8 1.007 9 1.011 10 1.064 11 1.109 12 1.132 13 1.14 14 1.153 15 1.253 16 1.394 n x = 16.475 n = 16 i=1 i n xi x = i=1 n = 16.475 16 = 1.0296. Therefore,mean of the given data is 1.0296. 2).median is the middle value in a given set of data. To find the median, we arrange the observations in order from smallest to largest value. If there is an odd number of observations, the median is the middle value. If x +xe is an even number of observations, the median is the average of the two middle values. i.e,12 2 Consider the data in Ascending order, 0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.007 1.011 x 1.064 1.109 1.132 1.14 1.153 1.253 1.394 Since we have n = 16 hence the given data is even.so, as per the definition of median If there is an even number of observations, the median is the average of the two middle values. i.e,Median = x1+x2 2 = 1.007+1.011 2.018 = 2 = 1.009 Therefore,the median of the given data is 1.009. b). From the given data we can say that 1.394 can be decreased until it reaches 1.011 i.e,by largest value - middle value = 1.394 – 1.011 = 0.383 The largest of the 2 middle values. If it is decreased by more than 0.383, the median will change.
Step 2 of 3
Step 3 of 3
##### ISBN: 9780321629111
This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. The answer to “The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J ? . Aircraft? 983: 1028–1032): a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median?” is broken down into a number of easy to follow steps, and 88 words. This full solution covers the following key subjects: aircraft, propagation, median, sample, holes. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Since the solution to 39E from 1 chapter was answered, more than 1001 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 39E from chapter: 1 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM.
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Unlock Textbook Solution | 1,307 | 4,807 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-39 | latest | en | 0.840539 | × Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 1 - Problem 39e Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 1 - Problem 39e × # The propagation of fatigue cracks in various aircraft ISBN: 9780321629111 32 ## Solution for problem 39E Chapter 1 Probability and Statistics for Engineers and the Scientists | 9th Edition • Textbook Solutions • 2901 Step-by-step solutions solved by professors and subject experts • Get 24/7 help from StudySoup virtual teaching assistants Probability and Statistics for Engineers and the Scientists | 9th Edition 4 5 1 305 Reviews 18 0 Problem 39E The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J ? . Aircraft? 983: 1028–1032): a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median? Step-by-Step Solution: Step 1 of 3 Problem 39E Answer: Step1: The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J. Aircraft, 1983: 1028–1032): 0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.007 1.011 x 1.064 1.109 1.132 1.14 1.153 1.253 1.394 We need to find, a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median Step2: a). 1).A mean is an average of given data, It is the sum of individual scores divided by the number of individuals and it is given by n xi x = i=1 n Consider, i x 1 0.736 2 0.863 3 0.865 4 0.913 5 0.915 6 0.937 7 0.983 8 1.007 9 1.011 10 1.064 11 1.109 12 1.132 13 1.14 14 1.153 15 1.253 16 1.394 n x = 16.475 n = 16 i=1 i n xi x = i=1 n = 16.475 16 = 1.0296. Therefore,mean of the given data is 1.0296. 2).median is the middle value in a given set of data. To find the median, we arrange the observations in order from smallest to largest value. If there is an odd number of observations, the median is the middle value. If x +xe is an even number of observations, the median is the average of the two middle values. i.e,12 2 Consider the data in Ascending order, 0.736 0.863 0.865 0.913 0.915 0.937 0.983 1.007 1.011 x 1.064 1.109 1.132 1.14 1.153 1.253 1.394 Since we have n = 16 hence the given data is even.so, as per the definition of median If there is an even number of observations, the median is the average of the two middle values. i.e,Median = x1+x2 2 = 1.007+1.011 2.018 = 2 = 1.009 Therefore,the median of the given data is 1.009. b). From the given data we can say that 1.394 can be decreased until it reaches 1.011 i.e,by largest value - middle value = 1.394 – 1.011 = 0.383 The largest of the 2 middle values. If it is decreased by more than 0.383, the median will change. Step 2 of 3 Step 3 of 3 ##### ISBN: 9780321629111 This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. The answer to “The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours/104) to reach a given crack size in fastener holes intended for use in military aircraft (“Statistical Crack Propagation in Fastener Holes Under Spectrum Loading,” J ? . Aircraft? 983: 1028–1032): a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median?” is broken down | into a number of easy to follow steps, and 88 words. This full solution covers the following key subjects: aircraft, propagation, median, sample, holes. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Since the solution to 39E from 1 chapter was answered, more than 1001 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 39E from chapter: 1 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM. #### Related chapters Unlock Textbook Solution |
http://slideplayer.com/slide/3377948/ | 1,510,990,887,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804666.54/warc/CC-MAIN-20171118055757-20171118075757-00619.warc.gz | 277,329,001 | 22,417 | # Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can.
## Presentation on theme: "Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can."— Presentation transcript:
Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can form a buffer when combined with its conjugate base (dihydrogen phosphate). H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 This buffer operates in the range: pH = pk a + 1 = 0.96 – 2.96
Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen phosphate buffer. H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 This buffer operates in the range from 6.1 to 8.1 A third buffer can be prepared by mixing hydrogen phosphate with orthophosphate as the following equilibrium suggests: HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 This buffer system operates in the pH range from 11.3 to 13.3
The same can be said about carbonic acid/bicarbonate where H 2 CO 3 H + + HCO 3 - k a1 = 4.3 x 10 -7 This buffer operates in the pH range from 5.4 to 7.4; while a more familiar buffer is composed of carbonate and bicarbonate according to the equilibrium: HCO 3 - H + + CO 3 2- k a2 = 4.8 x 10 -11 The pH range of the buffer is 9.3 to 11.3. Polyprotic acids and their salts are handy materials which can be used to prepare buffer solutions of desired pH working ranges. This is true due to the wide variety of their acid dissociation constants.
Example Find the ratio of [H 2 PO 4 - ]/[HPO 4 2- ] if the pH of the solution containing a mixture of both substances is 7.4. k a2 = 7.5x10 -8 Solution The equilibrium equation combining the two species is: H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 K a2 = [H + ][HPO 4 2- ]/[H 2 PO 4 - ] [H + ] = 10 -7.4 = 4x10 -8 M 7.5x10 -8 = 4x10 -8 [HPO 4 2- ]/[H 2 PO 4 - ] [HPO 4 2- ]/[H 2 PO 4 - ] = 1.9
Fractions of Dissociating Species at a Given pH Consider the situation where, for example, 0.1 mol of H 3 PO 4 is dissolved in 1 L of solution. H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 Some of the acid will remain undissociated (H 3 PO 4 ), some will be converted to H 2 PO 4 -, HPO 4 2- and PO 4 3- where we have, from mass balance: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]
We can write the fractions of each species in solution as 0 = [H 3 PO 4 ]/C H3PO4 1 = [H 2 PO 4 - ]/C H3PO4 2 = [HPO 4 2- ]/C H3PO4 3 = [PO 4 3- ]/C H3PO4 0 + 1 + 2 + 3 = 1 ( total value of all fractions sum up to unity).
The value of each fraction depends on pH of solution. At low pH dissociation is suppressed and most species will be in the form of H 3 PO 4 while high pH values will result in greater amounts converted to PO 4 3-. Setting up a relation of these species as a function of [H + ] is straightforward using the equilibrium constant relations. Let us try finding 0 where 0 is a function of undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid.
K a1 = [H 2 PO 4 - ][H + ]/[H 3 PO 4 ] Therefore we have [H 2 PO 4 - ] = k a1 [H 3 PO 4 ]/ [H + ] k a2 = [HPO 4 2- ][H + ]/[H 2 PO - ] Multiplying k a2 time k a1 and rearranging we get: [HPO 4 2- ] = k a1 k a2 [H 3 PO 4 ]/[H + ] 2 k a3 = [PO 4 3- ][H + ]/[HPO 4 2- ] Multiplying k a1 times k a2 times k a3 and rearranging we get: [PO 3- ] = k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 But we have: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ]
Substitution for all species from above gives: C H3PO4 = [H 3 PO 4 ] + k a1 [H 3 PO 4 ]/ [H + ] + k a1 k a2 [H 3 PO 4 ]/[H + ] 2 + k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 C H3PO4 = [H 3 PO 4 ] {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } [H 3 PO 4 ]/C H3PO4 = 1/ {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) Similar derivations for other fractions results in: 1 = k a1 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) 2 = k a1 k a2 [H + ] / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) 3 = k a1 k a2 k a3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 )
Example Calculate the equilibrium concentrations of the different species in a 0.10 M phosphoric acid solution at pH 3.00. Solution The [H + ] = 10 -3.00 = 1.0x10 -3 M Substitution in the relation for o gives o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) o = (1.0x10 -3 ) 3 /{(1.0x10 -3 ) 3 + 1.1x10 -2 (1.0x10 -3 ) 2 + 1.1x10 -2 * 7.5x10 -8 (1.0x10 -3 ) + 1.1x10 -2 * 7.5x10 - 8 * 4.8 * 10 -13 }
o = 8.2x10 -2 0 = [H 3 PO 4 ]/C H3PO4 8.2x10 -2 = [H 3 PO 4 ]/0.10 [H 3 PO 4 ] = 8.3x10 -3 M Similarly, 1 = 0.92, 1 = [H 2 PO 4 - ]/C H3PO4 0.92 = [H 2 PO 4 - ]/0.10 [H 2 PO 4 - ] = 9.2x10 -2 M Other fractions are calculated in the same manner.
pH Calculations for Salts of Polyprotic Acids Two types of salts exist for polyprotic acids. These include: 1. Unprotonated salts These are salts which are proton free which means they are not associated with any protons. Examples are: Na 3 PO 4 and Na 2 CO 3. Calculation of pH for solutions of such salts is straightforward and follows the same scheme described earlier for salts of monoprotic acids.
Example Find the pH of a 0.10 M Na 3 PO 4 solution. Solution We have the following equilibrium in water PO 4 3- + H 2 O HPO 4 2- + OH - The equilibrium constant which corresponds to this equilibrium is k b where: K b = k w /k a3
We used k a3 since it is the equilibrium constant describing relation between PO 4 3- and HPO 4 2-. However, in any equilibrium involving salts look at the highest charge on any anion to find which k a to use. K b = 10 -14 /4.8x10 -13 K b = 0.020
K b = x * x/0.10 – x Assume 0.10 >> x 0.02 = x 2 /0.10 x = 0.045 Relative error = (0.045/0.10) x 100 = 45% Therefore, assumption is invalid and we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.036 Therefore, [OH - ] = 0.036 M pOH = 1.44 and pH = 14 – 1.44 = 12.56
2. Protonated Salts These are usually amphoteric salts which react as acids and bases. For example, NaH 2 PO 4 in water would show the following equilibria: H 2 PO 4 - H + + HPO 4 2- H 2 PO 4 - + H 2 O OH - + H 3 PO 4 H 2 O H + + OH - [H + ] solution = [H + ] H2PO4 - + [H + ] H2O – [OH - ] H2PO4 - [H + ] solution = [HPO 4 2- ] + [OH - ] – [H 3 PO 4 ]
Now make all terms as functions in either H + or H 2 PO 4 -, then we have: [H + ] = {k a2 [H 2 PO 4 - ]/[H + ]} + k w /[H + ] –{[H 2 PO 4 - ][H + ]/k a1 } Rearrangement gives [H + ] = {(k a1 k w + k a1 k a2 [H 2 PO 4 - ])/(k a1 + [H 2 PO 4 ‑ ]} 1/2 At high salt concentration and low k a1 this relation may be approximated to: [H + ] = {k a1 k a2 } 1/2 Where; the pH will be independent on salt concentration but only on the equilibrium constants.
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Similar presentations | 2,946 | 7,409 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-47 | longest | en | 0.879876 | # Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can. ## Presentation on theme: "Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can."— Presentation transcript: Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can form a buffer when combined with its conjugate base (dihydrogen phosphate). H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 This buffer operates in the range: pH = pk a + 1 = 0.96 – 2.96 Also, another buffer which is commonly used is the dihydrogen phosphate/hydrogen phosphate buffer. H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 This buffer operates in the range from 6.1 to 8.1 A third buffer can be prepared by mixing hydrogen phosphate with orthophosphate as the following equilibrium suggests: HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 This buffer system operates in the pH range from 11.3 to 13.3 The same can be said about carbonic acid/bicarbonate where H 2 CO 3 H + + HCO 3 - k a1 = 4.3 x 10 -7 This buffer operates in the pH range from 5.4 to 7.4; while a more familiar buffer is composed of carbonate and bicarbonate according to the equilibrium: HCO 3 - H + + CO 3 2- k a2 = 4.8 x 10 -11 The pH range of the buffer is 9.3 to 11.3. Polyprotic acids and their salts are handy materials which can be used to prepare buffer solutions of desired pH working ranges. This is true due to the wide variety of their acid dissociation constants. Example Find the ratio of [H 2 PO 4 - ]/[HPO 4 2- ] if the pH of the solution containing a mixture of both substances is 7.4. k a2 = 7.5x10 -8 Solution The equilibrium equation combining the two species is: H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 K a2 = [H + ][HPO 4 2- ]/[H 2 PO 4 - ] [H + ] = 10 -7.4 = 4x10 -8 M 7.5x10 -8 = 4x10 -8 [HPO 4 2- ]/[H 2 PO 4 - ] [HPO 4 2- ]/[H 2 PO 4 - ] = 1.9 Fractions of Dissociating Species at a Given pH Consider the situation where, for example, 0.1 mol of H 3 PO 4 is dissolved in 1 L of solution. H 3 PO 4 H + + H 2 PO 4 - k a1 = 1.1 x 10 -2 H 2 PO 4 - H + + HPO 4 2- k a2 = 7.5 x 10 -8 HPO 4 2- H + + PO 4 3- k a3 = 4.8 x 10 -13 Some of the acid will remain undissociated (H 3 PO 4 ), some will be converted to H 2 PO 4 -, HPO 4 2- and PO 4 3- where we have, from mass balance: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ] We can write the fractions of each species in solution as 0 = [H 3 PO 4 ]/C H3PO4 1 = [H 2 PO 4 - ]/C H3PO4 2 = [HPO 4 2- ]/C H3PO4 3 = [PO 4 3- ]/C H3PO4 0 + 1 + 2 + 3 = 1 ( total value of all fractions sum up to unity). The value of each fraction depends on pH of solution. At low pH dissociation is suppressed and most species will be in the form of H 3 PO 4 while high pH values will result in greater amounts converted to PO 4 3-. Setting up a relation of these species as a function of [H + ] is straightforward using the equilibrium constant relations. Let us try finding 0 where 0 is a function of undissociated acid. The point is to substitute all fractions by their equivalent as a function of undissociated acid. K a1 = [H 2 PO 4 - ][H + ]/[H 3 PO 4 ] Therefore we have [H 2 PO 4 - ] = k a1 [H 3 PO 4 ]/ [H + ] k a2 = [HPO 4 2- ][H + ]/[H 2 PO - ] Multiplying k a2 time k a1 and rearranging we get: [HPO 4 2- ] = k a1 k a2 [H 3 PO 4 ]/[H + ] 2 k a3 = [PO 4 3- ][H + ]/[HPO 4 2- ] Multiplying k a1 times k a2 times k a3 and rearranging we get: [PO 3- ] = k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 But we have: C H3PO4 = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2- ] + [PO 4 3- ] Substitution for all species from above gives: C H3PO4 = [H 3 PO 4 ] + k a1 [H 3 PO 4 ]/ [H + ] + k a1 k a2 [H 3 PO 4 ]/[H + ] 2 + k a1 k a2 k a3 [H 3 PO 4 ]/[H + ] 3 C H3PO4 = [H 3 PO 4 ] {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } [H 3 PO 4 ]/C H3PO4 = 1/ {1 + k a1 / [H + ] + k a1 k a2 /[H + ] 2 + k a1 k a2 k a3 /[H + ] 3 } o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) Similar derivations for other fractions results in: 1 = k a1 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) 2 = k a1 k a2 [H + ] / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) 3 = k a1 k a2 k a3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) Example Calculate the equilibrium concentrations of the different species in a 0.10 M phosphoric acid solution at pH 3.00. Solution The [H + ] = 10 -3.00 = 1.0x10 -3 M Substitution in the relation for o gives o = [H + ] 3 / ([H + ] 3 + k a1 [H + ] 2 + k a1 k a2 [H + ] + k a1 k a2 k a3 ) o = (1.0x10 -3 ) 3 /{(1.0x10 -3 ) 3 + 1.1x10 -2 (1.0x10 -3 ) 2 + 1.1x10 -2 * 7.5x10 -8 (1.0x10 -3 ) + 1.1x10 -2 * 7.5x10 - 8 * 4.8 * 10 -13 } o = 8.2x10 -2 0 = [H 3 PO 4 ]/C H3PO4 8.2x10 -2 = [H 3 PO 4 ]/0.10 [H 3 PO 4 ] = 8.3x10 -3 M Similarly, 1 = 0.92, 1 = [H 2 PO 4 - ]/C H3PO4 0.92 = [H 2 PO 4 - ]/0.10 [H 2 PO 4 - ] = 9.2x10 -2 M Other fractions are calculated in the same manner. pH Calculations for Salts of Polyprotic Acids Two types of salts exist for polyprotic acids. These include: 1. Unprotonated salts These are salts which are proton free which means they are not associated with any protons. Examples are: Na 3 PO 4 and Na 2 CO 3. Calculation of pH for solutions of such salts is straightforward and follows the same scheme described earlier for salts of monoprotic acids. Example Find the pH of a 0.10 M Na 3 PO 4 solution. Solution We have the following equilibrium in water PO 4 3- + H 2 O HPO 4 2- + OH - The equilibrium constant which corresponds to this equilibrium is k b where: K b = k w /k a3 We used k a3 since it is the equilibrium constant describing relation between PO 4 3- and HPO 4 2-. However, in any equilibrium involving salts look at the highest charge on any anion to find which k a to use. K b = 10 -14 /4.8x10 -13 K b = 0.020 K b = x * x/0.10 – x Assume 0.10 >> x 0.02 = x 2 /0.10 x = 0.045 Relative error = (0.045/0.10) x 100 = 45% Therefore, assumption is invalid and we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.036 Therefore, [OH - ] = 0.036 M pOH = 1.44 and pH = 14 – 1.44 = 12.56 2. Protonated Salts These are usually amphoteric salts which react as acids and bases. For example, NaH 2 PO 4 in water would show the following equilibria: H 2 PO 4 - H + + HPO 4 2- H 2 PO 4 - + H 2 O OH - + H 3 PO 4 H 2 O H + + OH | - [H + ] solution = [H + ] H2PO4 - + [H + ] H2O – [OH - ] H2PO4 - [H + ] solution = [HPO 4 2- ] + [OH - ] – [H 3 PO 4 ] Now make all terms as functions in either H + or H 2 PO 4 -, then we have: [H + ] = {k a2 [H 2 PO 4 - ]/[H + ]} + k w /[H + ] –{[H 2 PO 4 - ][H + ]/k a1 } Rearrangement gives [H + ] = {(k a1 k w + k a1 k a2 [H 2 PO 4 - ])/(k a1 + [H 2 PO 4 ‑ ]} 1/2 At high salt concentration and low k a1 this relation may be approximated to: [H + ] = {k a1 k a2 } 1/2 Where; the pH will be independent on salt concentration but only on the equilibrium constants. Download ppt "Buffer Calculations for Polyprotic Acids A polyprotic acid can form buffer solutions in presence of its conjugate base. For example, phosphoric acid can." Similar presentations |
https://learningpundits.com/jobPreparation/module-view/49-speed-and-distance/2-aptitude-test---speed-&-distance/?page=19 | 1,656,550,702,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103646990.40/warc/CC-MAIN-20220630001553-20220630031553-00117.warc.gz | 403,254,698 | 16,299 | # Speed and distance
Speed Distance Time: Formula for time speed and distance, Speed time and distance problems, Speed Distance time questions with solutions
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## Online Aptitude Questions with Answers on Speed and distance
Q91. **Excluding stoppages, the average speed of a bus is 60 km/hr and including stoppages, the average speed of the bus is 40 km/hr. For how many minutes does the bus stop per hour?**
1. 10
2. 12.5
3. 15
4. 20
Solution : 20
Q92. **Train X crosses a stationary train Y in 60 seconds and a pole in 25 seconds with the same speed. The length of the train X is 300 m. What is the length of the stationary train Y?**
1. 320 m
2. 420 m
3. 300 m
4. 360 m
Solution : 420 m
Q93. **In a 1000 m race, A beats B by 50 m and B beats C by 100 m. In the same race, by how many meters does A beat C?**
1. 145
2. 150
3. 155
4. 160
Solution : 145
Q94. **In a 1000 m race, A beats B by 200 meters or 25 seconds. Find the speed of B?**
1. 8 m/s
2. 25 m/s
3. 10 m/s
4. 15 m/s
Solution : 8 m/s
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Q95. **Two persons start running simultaneously around a circular track of length 300 m from the same point at speeds of 15 km/hr and 25 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions?**
1. 21 sec
2. 24 sec
3. 25 sec
4. 27 sec
Solution : 27 sec
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1.
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Rank - {{getRank(\$index,weeklyWinner)}}: {{weeklyWinner.userName}} | 807 | 3,181 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-27 | latest | en | 0.767408 | # Speed and distance Speed Distance Time: Formula for time speed and distance, Speed time and distance problems, Speed Distance time questions with solutions Take Aptitude Test View Aptitude Test Results ## Online Aptitude Questions with Answers on Speed and distance Q91. **Excluding stoppages, the average speed of a bus is 60 km/hr and including stoppages, the average speed of the bus is 40 km/hr. For how many minutes does the bus stop per hour?** 1. 10 2. 12.5 3. 15 4. 20 Solution : 20 Q92. **Train X crosses a stationary train Y in 60 seconds and a pole in 25 seconds with the same speed. The length of the train X is 300 m. What is the length of the stationary train Y?** 1. 320 m 2. 420 m 3. 300 m 4. 360 m Solution : 420 m Q93. **In a 1000 m race, A beats B by 50 m and B beats C by 100 m. In the same race, by how many meters does A beat C?** 1. 145 2. 150 3. 155 4. 160 Solution : 145 Q94. **In a 1000 m race, A beats B by 200 meters or 25 seconds. Find the speed of B?** 1. 8 m/s 2. 25 m/s 3. 10 m/s 4. 15 m/s Solution : 8 m/s ### Campus Ambassador (Remote Internship) ##### You will receive: • Stipend based on your performance • Internship Certificate to boost your Resume Q95. **Two persons start running simultaneously around a circular track of length 300 m from the same point at speeds of 15 km/hr and 25 km/hr. When will they meet for the first time any where on the track if they are moving in opposite directions?** 1. 21 sec 2. 24 sec 3. 25 sec 4. 27 sec Solution : 27 sec Q{{(\$index+1)+((page-1)*LIMITPERPAGE)}}. 1. Solution : ### Grammar Guru #### Free Online Contest on English Grammar and Vocabulary.20 mins Only. • All Participants get Participation Certificates to boost your Resume ### Math Whiz #### Free Online Contest on Aptitude and Reasoning.20 mins Only. • All Participants get Participation Certificates to boost your Resume ### Campus Ambassador (Remote Internship) ##### You will receive: • Stipend based on your performance • Internship Certificate to boost your Resume Preparing for Aptitude Tests ? Please go through our courses on Aptitude Questions and try to answer our Online Aptitude Test Questions on Quantitative Aptitude. Interested in evaluating your Reasoning Skills ? Please go through our courses on Logical Reasoning and Non-Verbal Reasoning to answer our Reasoning Questions. Interested in learning English and checking your English Grammar ? Do a quick grammar check to evaluate your Basic English Grammar Skills. Improve your English Vocabulary by going through these Vocabulary words. Wondering how to make a resume ? These resume format for freshers might be helpful. You can also submit your Resume for Review and look and Resume samples available there. Preparing for an HR Interview or Group | Discussion ? These HR interview questions and answers could help you do well in a typical HR interview. These group discussion tips could also be useful. Searching for jobs ? We have thousand of Fresher Jobs. Feel free to browse through our extensive list of online jobs. #### Weekly Contests Leaderboard Rank - {{getRank(\$index,weeklyWinner)}}: {{weeklyWinner.userName}} |
https://pratibha.eenadu.net/jobs/lesson/ibps/ibps-clerks/telugumedium/alligation-and-mixture/2-1-6-32-188-1347-7066-10530-20040012044 | 1,695,769,834,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00313.warc.gz | 497,710,442 | 19,140 | # ALLIGATION AND MIXTURE
1. Jar-A contains a mixture of milk and water in the respective ratio of 8 : 1. When 18 ltr of mixture is taken out and 4 ltr of pure water is added to Jar-A, the resultant quantity of milk becomes 24 ltr more than that of water. What was the initial quantity of milk in Jar-A? (In ltr)
A) 48 B) 64 C) 50 D) 52 E) 60
2. A container contains 120 litres mixture of milk and water in the respective ratio of 37 : 3. 40 litres of the mixture is taken out from the container and 6 litres of each pure milk and pure water is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk?
A) 72% B) 76% C) 68% D) 85% E) 74%
3. In a 84 litres mixture of milk and water, percentage of water is only 25%. The milkman gave 12 litres of this mixture to a customer. Then he added equal quantities of pure milk and water to the remaining mixture. As a result, the respective ratio of milk and water in the mixture became 11 : 5. What was the quantity of water added? (In litres)
A) 8 B) 15 C) 12 D) 18 E) 21
4. A jar had a certain quantity (in litres) of water, to which pure milk of four times of the quantity water was added. 20 litres of the mixture from the jar was taken out and was replaced with 12 litres of pure milk, as result of which water constituted 10% of the resultant mixture. What was the initial quantity of water in the jar?
A) 7 litres B) 8 litres C) 10 litres D) 4.8 litres E) 6.4 litres
5. In a 200 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 40 litres of this mixture to a customer and then added 12 litres of water in the remaining mixture. What is the respective ratio of milk and water in the new mixture?
A) 11 : 17 B) 7 : 13 C) 23 : 14 D) 21 : 11 E) 28 : 15
6. A vessel contains coconut water and vodka i.e., 120 litres of the mixture, coconut water is 40% more than the vodka. 72 litres of mixture taken out and 26 litres of coconut water is and some quantity of vodka is also added to the vessel, after that the respective ratio of the resultant mixture of vodka and coconut water becomes 2 : 3, then find the final quantityof the vodka in the vessel? (In litres)
A) 54 B) 46 C) 35 D) 34 E) 36
7. 16 litres of pure water was added to a vessel containing 84 litres of pure milk. 50 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 3 : 2. If the resultant respective ratio of milk and water in the vessel is 4 : 1, what was the quantity of pure milk added in the vessel? (In litres)
A) 4 B) 8 C) 6 D) 10 E) 12
8. In a mixture of milk and water, water was only 20%. 50 litres of this mixture was taken out and then 5 litres of pure water was added to the mixture. If the resultant ratio of milk and water in the mixture was 8 : 3 respectively, what was the initial quantity of mixture (before the replacement)? (In litres)
A) 88 B) 96 C) 120 D) 100 E) 84
9. In a 120 litres mixture of milk and water, percentage of water is only 40%. The milkman gave 45 litres of this mixture to a customer and then added 15 litres of pure milk and 10 litres of pure water to the remaining mixture. What is the respective ratio of quantity of milk and water in the new mixture?
A) 3 : 2 B) 11 : 5 C) 8 : 5 D) 15 : 8 E) 12 : 7
10. In 112 litres of mixture of water and milk, water is only 18.75%. The milkman gave 28 litres of this mixture to a customer and then he added 10.75 litres of pure milk and 5.25 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture?
A) 32.50 B) 34 C) 32 D) 21 E) 27
Some more...
1. Equal quantities of two milk and water solutions which contains milk and water in ratio, 1 : 5 and 3 : 5 are mixed together. What will be the ratio of water to milk in the resultant solution?
a) 13 : 35 b) 4 : 10 c) 5 : 8 d) 35 : 13
Ans: d
2. The respective ratio of milk and water in 66 litres of adulterated milk is 5 : 1. Water is added into it to make the respective ratio 5 : 3. What is the quantity of water added?
a) 11 litres b) 22 litres c) 33 litres d) 44 litres
Ans: b
3. A 200 litres solution of alcohol and water contains 1/4 th part alcohol. Find the new percentage of alcohol, if 50 litres of the original solution is replaced by 50 litres of alcohol.
a) 43.75% b) 50% c) 66.66% d) 80%
Ans: a
4. A container has 80 litres of milk. From this container 8 litres of milk was taken out and replaced with water. The process was further repeated twice. How much milk is there in the container finally? (in liters)
a) 58.32 b) 60.32 c) 62 d) 64.7
Ans: a
5. 100 litres of a mixture contains 10% water and the rest milk. The amount of water that must be added so that the resulting mixture contains only 50% milk is......
a) 70 litres b) 72 litres c) 78 litres d) 80 litres
Ans: d
6. In a particular type of fertilizer, the ratio of two chemicals A and B is 2 : 5. In 21 kg of this fertilizer, if 3 kg of type - A is added, the ratio of chemical A to B in the new fertilizer will be
a) 1 : 1 b) 2 : 3 c) 3 : 5 d) 4 : 5
Ans: c
7. Rice worth Rs.120 per kg and Rs.132 per kg are mixed with a third variety in the ratio 2 : 1 : 3. If the mixture is worth Rs.135 per kg, the price of the third variety per kg will be....
a) Rs.140 b) Rs.146 c) Rs.150 d) Rs.148
Ans: b
8. A jar full of wine contains 50% alcohol. A part of wine is replaced by another containing 20% alcohol and now percentage of alcohol was found to be 30%. The part of wine replaced is?
a) 1/2 b) 1/3 c) 2/5 d) 2/3
Ans: d
9. A shopkeeper has 50 kg rice, a part of which he sells at 10% profit and rest at 25% profit. He gains 16% on the whole. The quantity sold at 10% profit is?
a) 30 kg b) 20 kg c) 45 kg d) 27 kg
Ans: a
10. An alloy contains tin, copper and zinc in ratio 2 : 3 : 1, and another alloy contain copper, aluminium and tin in ratio 3 : 2 : 7. If both alloys are mixed in equal quantities, then what will be the ratio of tin and aluminium in final alloy?
a) 2 : 11 b) 11 : 9 c) 9 : 2 d) 11 : 2
Ans: d
Posted Date : 17-09-2022
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి. | 2,949 | 7,236 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2023-40 | longest | en | 0.869919 | # ALLIGATION AND MIXTURE 1. Jar-A contains a mixture of milk and water in the respective ratio of 8 : 1. When 18 ltr of mixture is taken out and 4 ltr of pure water is added to Jar-A, the resultant quantity of milk becomes 24 ltr more than that of water. What was the initial quantity of milk in Jar-A? (In ltr) A) 48 B) 64 C) 50 D) 52 E) 60 2. A container contains 120 litres mixture of milk and water in the respective ratio of 37 : 3. 40 litres of the mixture is taken out from the container and 6 litres of each pure milk and pure water is added to the mixture. By what percent is the quantity of water in the final mixture less than the quantity of milk? A) 72% B) 76% C) 68% D) 85% E) 74% 3. In a 84 litres mixture of milk and water, percentage of water is only 25%. The milkman gave 12 litres of this mixture to a customer. Then he added equal quantities of pure milk and water to the remaining mixture. As a result, the respective ratio of milk and water in the mixture became 11 : 5. What was the quantity of water added? (In litres) A) 8 B) 15 C) 12 D) 18 E) 21 4. A jar had a certain quantity (in litres) of water, to which pure milk of four times of the quantity water was added. 20 litres of the mixture from the jar was taken out and was replaced with 12 litres of pure milk, as result of which water constituted 10% of the resultant mixture. What was the initial quantity of water in the jar? A) 7 litres B) 8 litres C) 10 litres D) 4.8 litres E) 6.4 litres 5. In a 200 litres mixture of milk and water, percentage of water is only 30%. The milkman gave 40 litres of this mixture to a customer and then added 12 litres of water in the remaining mixture. What is the respective ratio of milk and water in the new mixture? A) 11 : 17 B) 7 : 13 C) 23 : 14 D) 21 : 11 E) 28 : 15 6. A vessel contains coconut water and vodka i.e., 120 litres of the mixture, coconut water is 40% more than the vodka. 72 litres of mixture taken out and 26 litres of coconut water is and some quantity of vodka is also added to the vessel, after that the respective ratio of the resultant mixture of vodka and coconut water becomes 2 : 3, then find the final quantityof the vodka in the vessel? (In litres) A) 54 B) 46 C) 35 D) 34 E) 36 7. 16 litres of pure water was added to a vessel containing 84 litres of pure milk. 50 litres of the resultant mixture was then sold and some more quantity of pure milk and pure water was added to the vessel in the respective ratio of 3 : 2. If the resultant respective ratio of milk and water in the vessel is 4 : 1, what was the quantity of pure milk added in the vessel? (In litres) A) 4 B) 8 C) 6 D) 10 E) 12 8. In a mixture of milk and water, water was only 20%. 50 litres of this mixture was taken out and then 5 litres of pure water was added to the mixture. If the resultant ratio of milk and water in the mixture was 8 : 3 respectively, what was the initial quantity of mixture (before the replacement)? (In litres) A) 88 B) 96 C) 120 D) 100 E) 84 9. In a 120 litres mixture of milk and water, percentage of water is only 40%. The milkman gave 45 litres of this mixture to a customer and then added 15 litres of pure milk and 10 litres of pure water to the remaining mixture. What is the respective ratio of quantity of milk and water in the new mixture? A) 3 : 2 B) 11 : 5 C) 8 : 5 D) 15 : 8 E) 12 : 7 10. In 112 litres of mixture of water and milk, water is only 18.75%. The milkman gave 28 litres of this mixture to a customer and then he added 10.75 litres of pure milk and 5.25 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture? A) 32.50 B) 34 C) 32 D) 21 E) 27 Some more... 1. Equal quantities of two milk and water solutions which contains milk and water in ratio, 1 : 5 and 3 : 5 are mixed together. What will be the ratio of water to milk in the resultant solution? a) 13 : 35 b) 4 : 10 c) 5 : 8 d) 35 : 13 Ans: d 2. The respective ratio of milk and water in 66 litres of adulterated milk is 5 : 1. Water is added into it to make the respective ratio 5 : 3. What is the quantity of water added? a) 11 litres b) 22 litres c) 33 litres d) 44 litres Ans: b 3. A 200 litres solution of alcohol and water contains 1/4 th part alcohol. Find the new percentage of alcohol, if 50 litres of the original solution is replaced by 50 litres of alcohol. a) 43.75% b) 50% c) 66.66% d) 80% Ans: a 4. A container has 80 litres of milk. From this container 8 litres of milk was taken out and replaced with water. The process was further repeated twice. How much milk is there in the container finally? (in liters) a) 58.32 b) 60.32 c) 62 d) 64.7 Ans: a 5. 100 litres of a mixture contains 10% water and the rest milk. The amount of water that must be added so that the resulting mixture contains only 50% milk is...... a) 70 litres b) 72 litres c) 78 litres d) 80 litres Ans: d 6. In a particular type of fertilizer, the ratio of two chemicals A and B is 2 : 5. In 21 kg of this fertilizer, if 3 kg of type - A is added, the ratio of chemical A to B in the new fertilizer will be a) 1 : 1 b) 2 : 3 c) 3 : 5 d) 4 : 5 Ans: c 7. Rice worth Rs.120 per kg and Rs.132 per kg are mixed with a third variety in the ratio 2 : 1 : 3. If the mixture is worth Rs.135 per kg, the price of the third variety per kg will be.... a) Rs.140 b) Rs.146 c) Rs.150 d) Rs.148 Ans: b 8. A jar full of wine contains 50% alcohol. A part of wine is replaced by another containing 20% alcohol and now percentage of alcohol was found to be 30%. The part of wine replaced is? a) 1/2 b) 1/3 c) 2/5 d) 2/3 Ans: d 9. A shopkeeper has 50 kg rice, a part of which he sells at 10% profit and rest at 25% profit. He gains 16% on the whole. The quantity sold at 10% profit is? a) 30 kg b) 20 kg c) 45 kg d) 27 kg Ans: a 10. An | alloy contains tin, copper and zinc in ratio 2 : 3 : 1, and another alloy contain copper, aluminium and tin in ratio 3 : 2 : 7. If both alloys are mixed in equal quantities, then what will be the ratio of tin and aluminium in final alloy? a) 2 : 11 b) 11 : 9 c) 9 : 2 d) 11 : 2 Ans: d Posted Date : 17-09-2022 గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి. |
https://www.vedantu.com/question-answer/one-mole-of-p2o5-undergoes-hydrolysis-as-class-11-chemistry-cbse-5f610da5a863034f42ab39ed | 1,601,076,450,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00697.warc.gz | 1,090,061,246 | 80,423 | Question
# One mole of ${{P}_{2}}{{O}_{5}}$ undergoes hydrolysis as ${{P}_{2}}{{O}_{5}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}$ The normality of the phosphoric acid formed is (The volume of solution is 1L)(a) 2(b) 12(c) 24(d) 4
Hint: First balance the equation. With balancing the number of moles can be calculated. Molarity can be calculated by dividing the number of moles with the volume of the solution. Normality is equal to molarity multiplied to the basicity of the compound.
Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.
$Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}$
Moles of the solute can be calculated by:
$Moles=\dfrac{\text{mass of the solute}}{\text{molar mass of the solute}}$
The normality of the solution is defined as the number of grams equivalent of the solute present in liters. It is represented by the symbol, N.
$Normality=\dfrac{\text{Gram equivalent of the solute}}{\text{Volume of the solution}}$
First, let us balance the given equation:
${{P}_{2}}{{O}_{5}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}$
After balancing the equation will be:
$2{{P}_{2}}{{O}_{5}}+6{{H}_{2}}O\to 4{{H}_{3}}P{{O}_{4}}$
And the volume of the solution is 1 L. (Given)
So, from the balanced equation we can that there are 4 moles of phosphoric acid in 1 L of solution.
From this we can calculate the molarity of the solution:
$Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}$
So, by putting the values in the above equation:
$Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}=\dfrac{4}{1}=4$
The molarity is 4.
With molarity, we can find the normality of the solution.
Normality of the acid = molarity x basicity.
Phosphoric acid has basicity 3 because of the presence of hydrogen ions.
So, normality of the phosphoric acid will be:
$normality=\text{ 4 x 3 = 12 N}$
Hence, the normality is 12 N.
So, the correct answer is “Option B”.
Note: The balancing of the equation must be right. The molarity of the acid can be converted into normality by multiplying it with basicity. The molarity of the base can be converted into normality by multiplying it with acidity. | 650 | 2,237 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-40 | longest | en | 0.864788 | Question # One mole of ${{P}_{2}}{{O}_{5}}$ undergoes hydrolysis as ${{P}_{2}}{{O}_{5}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}$ The normality of the phosphoric acid formed is (The volume of solution is 1L)(a) 2(b) 12(c) 24(d) 4 Hint: First balance the equation. With balancing the number of moles can be calculated. Molarity can be calculated by dividing the number of moles with the volume of the solution. Normality is equal to molarity multiplied to the basicity of the compound. Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M. $Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}$ Moles of the solute can be calculated by: $Moles=\dfrac{\text{mass of the solute}}{\text{molar mass of the solute}}$ The normality of the solution is defined as the number of grams equivalent of the solute present in liters. It is represented by the symbol, N. $Normality=\dfrac{\text{Gram equivalent of the solute}}{\text{Volume of the solution}}$ First, let us balance the given equation: ${{P}_{2}}{{O}_{5}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}$ After balancing the equation will be: $2{{P}_{2}}{{O}_{5}}+6{{H}_{2}}O\to 4{{H}_{3}}P{{O}_{4}}$ And the volume of the solution is 1 L. (Given) So, from the balanced equation we can that there are 4 moles of phosphoric acid in 1 L of solution. From this we can calculate the molarity of the solution: $Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}$ So, by putting the values in the above equation: $Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}=\dfrac{4}{1}=4$ The molarity is 4. With molarity, we can find the normality of the solution. Normality of the acid = molarity x basicity. Phosphoric acid has basicity 3 because of the presence of hydrogen ions. So, normality of the phosphoric acid will be: $normality=\text{ 4 x 3 = 12 N}$ Hence, the normality is 12 N. So, the correct answer is “Option B”. Note: The balancing of the | equation must be right. The molarity of the acid can be converted into normality by multiplying it with basicity. The molarity of the base can be converted into normality by multiplying it with acidity. |
https://gmatclub.com/forum/only-senior-citizens-enjoy-doing-the-daily-jumble-so-aesha-must-242955.html?fl=homea | 1,527,058,574,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865456.57/warc/CC-MAIN-20180523063435-20180523083435-00089.warc.gz | 552,124,640 | 55,625 | GMAT Changed on April 16th - Read about the latest changes here
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# Only senior citizens enjoy doing the daily jumble. So Aesha must
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Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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19 Jun 2017, 04:38
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Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
Which of the following arguments exhibits the same flawed reasoning as the above?
A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.
D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.
E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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19 Jun 2017, 17:48
B.
Some senior citizens might not enjoy jumbles and some pet lovers might not adopt Marley.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jun 2017, 23:25
Can someone please explain this question in a little detail? How to relate the two situations ? Also Why is option C wrong?
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jun 2017, 23:26
Experts please explain why is option C wrong? And how to go about solving such analogy questions?
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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23 Jun 2017, 00:55
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Shiv2016 wrote:
Experts please explain why is option C wrong? And how to go about solving such analogy questions?
I'm not an expert but I can try and help.
Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
a implies b ---->not b implies not a
Which of the following arguments exhibits the same flawed reasoning as the above?
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.a implies b ---->not b implies not a - correct
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.a implies b ---->not a implies not b - reversed and so incorrect
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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02 Jul 2017, 07:29
this is a mix question. "flaw" here indicates that this is not only logic-error question, but also an application questions.
Only A do B = every B is A
the flaw is not B -> not A
Practice is the best way to cope with different gmat questions.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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04 Jul 2017, 02:16
Flaw is senior citizen - > enjoy jumble doesn't imply ~not citizen-> ~ No jumble enjoyment. (flawed contrapositive) Replicated by B.
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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04 Jul 2017, 10:04
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 Jul 2017, 09:26
can someone tell me how to do these kind of questions
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink]
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22 May 2018, 04:01
broall wrote:
Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble.
Which of the following arguments exhibits the same flawed reasoning as the above?
A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain.
B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.
C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.
D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica.
E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs.
VERITAS PREP OFFICIAL SOLUTION:
Solution: B
This Mimic the Reasoning question features difficult conditional wording. “Only x does y” can be reworded as “If you do y, then you are x”, a change of phrase that makes the question much easier to solve. “If you enjoy the jumble, you’re a senior citizen” implies two things: 1) if you enjoy the jumble, you’re a senior citizen, and 2) if you aren’t a senior citizen, you don’t enjoy the jumble. It does NOT imply that if you don’t enjoy the jumble, you’re not a senior citizen, so find an answer choice that also employs such faulty logic. (B) is the answer of choice.
_________________
Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] 22 May 2018, 04:01
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# Only senior citizens enjoy doing the daily jumble. So Aesha must
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2,083 | 8,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.919354 | GMAT Changed on April 16th - Read about the latest changes here It is currently 22 May 2018, 23:56 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # Only senior citizens enjoy doing the daily jumble. So Aesha must new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Senior CR Moderator Status: Long way to go! Joined: 10 Oct 2016 Posts: 1379 Location: Viet Nam Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 19 Jun 2017, 04:38 2 KUDOS 8 This post was BOOKMARKED 00:00 Difficulty: 35% (medium) Question Stats: 70% (01:17) correct 30% (01:17) wrong based on 335 sessions ### HideShow timer Statistics Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble. Which of the following arguments exhibits the same flawed reasoning as the above? A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain. B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover. C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist. D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica. E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs. _________________ Intern Joined: 16 Jun 2017 Posts: 15 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 19 Jun 2017, 17:48 B. Some senior citizens might not enjoy jumbles and some pet lovers might not adopt Marley. Director Joined: 02 Sep 2016 Posts: 745 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 22 Jun 2017, 23:25 Can someone please explain this question in a little detail? How to relate the two situations ? Also Why is option C wrong? _________________ Help me make my explanation better by providing a logical feedback. If you liked the post, HIT KUDOS !! Don't quit.............Do it. Director Joined: 02 Sep 2016 Posts: 745 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 22 Jun 2017, 23:26 Experts please explain why is option C wrong? And how to go about solving such analogy questions? _________________ Help me make my explanation better by providing a logical feedback. If you liked the post, HIT KUDOS !! Don't quit.............Do it. Intern Joined: 16 Jun 2017 Posts: 15 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 23 Jun 2017, 00:55 1 KUDOS Shiv2016 wrote: Experts please explain why is option C wrong? And how to go about solving such analogy questions? I'm not an expert but I can try and help. Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble. a implies b ---->not b implies not a Which of the following arguments exhibits the same flawed reasoning as the above? B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover.a implies b ---->not b implies not a - correct C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist.a implies b ---->not a implies not b - reversed and so incorrect SVP Joined: 12 Dec 2016 Posts: 1904 Location: United States GMAT 1: 700 Q49 V33 GPA: 3.64 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 02 Jul 2017, 07:29 this is a mix question. "flaw" here indicates that this is not only logic-error question, but also an application questions. Only A do B = every B is A the flaw is not B -> not A Practice is the best way to cope with different gmat questions. Intern Joined: 20 Jun 2017 Posts: 11 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 04 Jul 2017, 02:16 Flaw is senior citizen - > enjoy jumble doesn't imply ~not citizen-> ~ No jumble enjoyment. (flawed contrapositive) Replicated by B. Sent from my ONE A2003 using GMAT Club Forum mobile app Intern Joined: 23 Dec 2013 Posts: 14 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 04 Jul 2017, 10:04 Sent from my Moto G (5) Plus using GMAT Club Forum mobile app Manager Joined: 08 Jun 2017 Posts: 65 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 22 Jul 2017, 09:26 can someone tell me how to do these kind of questions Manager Joined: 14 Oct 2012 Posts: 176 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 14 Oct 2017, 15:24 1 KUDOS My 2 cents: Attachments my 2 cents.jpg [ 610.67 KiB | Viewed 933 times ] Math Expert Joined: 02 Sep 2009 Posts: 45256 Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] ### Show Tags 22 May 2018, 04:01 broall wrote: Only senior citizens enjoy doing the daily jumble. So Aesha must not be a senior citizen, because she does not enjoy doing the daily jumble. Which of the following arguments exhibits the same flawed reasoning as the above? A. Only in March does Rodrigo choose to holiday in Spain. It is March, but Rodrigo is in Japan. So he must not be going to Spain. B. Only a true pet lover could adopt Marley. Thus, since Michael is not adopting Marley, he must not be a true pet lover. C. Only geologists enjoy the amethyst exhibit at the town fair. So Mr. Franz must not enjoy the amethyst exhibit, because he is not a geologist. D. Since the animal in front of us is a penguin, it follows that we are in Antarctica, since one only encounters penguins in the wild when one is in Antarctica. E. Only the best chefs can make compelling vegan escargot. So Rasheed must be able to make compelling vegan escargot, since he is one of the world’s best chefs. VERITAS PREP OFFICIAL SOLUTION: Solution: B This Mimic the Reasoning question features difficult conditional wording. “Only x does y” can be reworded as “If you do y, then you are x”, a change of phrase that makes the question much easier to solve. “If you enjoy the jumble, you’re a senior citizen” implies two things: 1) if | you enjoy the jumble, you’re a senior citizen, and 2) if you aren’t a senior citizen, you don’t enjoy the jumble. It does NOT imply that if you don’t enjoy the jumble, you’re not a senior citizen, so find an answer choice that also employs such faulty logic. (B) is the answer of choice. _________________ Re: Only senior citizens enjoy doing the daily jumble. So Aesha must [#permalink] 22 May 2018, 04:01 Display posts from previous: Sort by # Only senior citizens enjoy doing the daily jumble. So Aesha must new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. |
https://optionrsdt.web.app/warnasch7460ke/sequence-and-series-calculator-online-bike.html | 1,638,713,463,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363189.92/warc/CC-MAIN-20211205130619-20211205160619-00596.warc.gz | 511,216,431 | 6,101 | ## Sequence and series calculator online
The above mentioned formulas are employed in our sequence calculator, so they are In addition, when calculating the amount of series necessary state of Free math problem solver answers your calculus homework questions with step- by-step explanations.
Sep 12, 2019 In this chapter we introduce sequences and series. We discuss whether a sequence converges or diverges, is increasing or decreasing, or if Arithmetic series calculator. World's simplest number tool. Quickly calculate the arithmetic number sequence in your browser. To get your sequence, just specify The geometric series calculator or sum of geometric series calculator is a simple online tool that's To work with sequences in the Seq mode: Press MODE key. Choose Seq in the fourth line. Hit ENTER to highlight Seq. Leave other settings as default (on the In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely The infinite sequence of additions implied by a series cannot be effectively carried on (at least in a finite amount of time). He used the method of exhaustion to calculate the area under the arc of a Paul's Online Math Notes.
## The calculator of sequence makes it possible to calculate online the terms of the sequence, defined by recurrence and its first term, until the indicated index.
Sequence solver by AlteredQualia. Find the next number in the sequence using difference table. Please enter integer sequence (separated by spaces or commas). Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Calculate sum elements of sequence : sum . Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. If you are dealing with the case in which the difference between any two consecutive values of the sequence is constant, then you use use our arithmetic sequence calculator instead. On the other hand, if you want to add an infinite geometric series, you can use this geometric series calculator. By applying this calculator for Arithmetic & Geometric Sequences, the n-th term and the sum of the first n terms in a sequence can be accurately obtained GoodCalculators.com A collection of really good online calculators for use in every day domestic and commercial use! The calculator of sequence makes it possible to calculate online the terms of the sequence, defined by recurrence and its first term, until the indicated index.
### Aug 5, 2019 Arithmetic Series Solver (Includes Sigma Notation!) Simple but incredibly useful. Solves partial sum, nth term, and sigma series sequences.
This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. BYJU’S online infinite series calculator tool makes the calculations faster and easier where it displays the value in a fraction of seconds.
### What is meant by sequences and series? A sequence is a list of numbers or events that have been ordered sequentially. A series is defined as the sum of the
Free Sequences calculator - find sequence types, indices, sums and progressions step-by-step. Free series convergence calculator - test infinite series for convergence step-by- step. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. Sequences have many applications in various mathematical disciplines due to their properties of convergence. A series is convergent if the sequence converges
## Find the next number in the sequence calculator - find the next number in the given sequence (number series), step-by-step.
Free series convergence calculator - test infinite series for convergence step-by- step. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. Sequences have many applications in various mathematical disciplines due to their properties of convergence. A series is convergent if the sequence converges Home · Calculators · Calculus II Calculators · Math Problem Solver (all calculators ). Series and Sum Calculator. This calculator will find the sum of arithmetic, Arithmetic series to infinity; Arithmetic and geometric sequences; Arithmetico– geometric sequence; Arithmetic sequence calculator: an example of use. This Find the next number in the sequence calculator - find the next number in the given sequence (number series), step-by-step. What is meant by sequences and series? A sequence is a list of numbers or events that have been ordered sequentially. A series is defined as the sum of the
Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Calculate sum elements of sequence : sum . Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. Arithmetic sequences calculator that shows all the work, detailed explanation and steps. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Mathematics widgets in Wolfram|Alpha. This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. BYJU’S online infinite series calculator tool makes the calculations faster and easier where it displays the value in a fraction of seconds. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Explore many other math calculators, as well as hundreds of other calculators addressing health, fitness, finance, math, and more. | 1,201 | 6,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2021-49 | latest | en | 0.896147 | ## Sequence and series calculator online The above mentioned formulas are employed in our sequence calculator, so they are In addition, when calculating the amount of series necessary state of Free math problem solver answers your calculus homework questions with step- by-step explanations. Sep 12, 2019 In this chapter we introduce sequences and series. We discuss whether a sequence converges or diverges, is increasing or decreasing, or if Arithmetic series calculator. World's simplest number tool. Quickly calculate the arithmetic number sequence in your browser. To get your sequence, just specify The geometric series calculator or sum of geometric series calculator is a simple online tool that's To work with sequences in the Seq mode: Press MODE key. Choose Seq in the fourth line. Hit ENTER to highlight Seq. Leave other settings as default (on the In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely The infinite sequence of additions implied by a series cannot be effectively carried on (at least in a finite amount of time). He used the method of exhaustion to calculate the area under the arc of a Paul's Online Math Notes. ## The calculator of sequence makes it possible to calculate online the terms of the sequence, defined by recurrence and its first term, until the indicated index. Sequence solver by AlteredQualia. Find the next number in the sequence using difference table. Please enter integer sequence (separated by spaces or commas). Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Calculate sum elements of sequence : sum . Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. If you are dealing with the case in which the difference between any two consecutive values of the sequence is constant, then you use use our arithmetic sequence calculator instead. On the other hand, if you want to add an infinite geometric series, you can use this geometric series calculator. By applying this calculator for Arithmetic & Geometric Sequences, the n-th term and the sum of the first n terms in a sequence can be accurately obtained GoodCalculators.com A collection of really good online calculators for use in every day domestic and commercial use! The calculator of sequence makes it possible to calculate online the terms of the sequence, defined by recurrence and its first term, until the indicated index. ### Aug 5, 2019 Arithmetic Series Solver (Includes Sigma Notation!) Simple but incredibly useful. Solves partial sum, nth term, and sigma series sequences. This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. Show Instructions In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. BYJU’S online infinite series calculator tool makes the calculations faster and easier where it displays the value in a fraction of seconds. ### What is meant by sequences and series? A sequence is a list of numbers or events that have been ordered sequentially. A series is defined as the sum of the Free Sequences calculator - find sequence types, indices, sums and progressions step-by-step. Free series convergence calculator - test infinite series for convergence step-by- step. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. Sequences have many applications in various mathematical disciplines due to their properties of convergence. A series is convergent if the sequence converges ## Find the next number in the sequence calculator - find the next number in the given sequence (number series), step-by-step. Free series convergence calculator - test infinite series for convergence step-by- step. Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. Sequences have many applications in various mathematical disciplines due to their properties of convergence. A series is convergent if the sequence converges Home · Calculators · Calculus II Calculators · Math Problem Solver (all calculators ). Series and Sum Calculator. This calculator will find the sum of arithmetic, Arithmetic series to infinity; Arithmetic and geometric sequences; Arithmetico– geometric sequence; Arithmetic sequence calculator: an example of use. This Find the next number in the sequence calculator - find the next number in the given sequence (number series), step-by-step. What is meant by sequences and series? A sequence is a list of numbers or events that have been ordered sequentially. A series is defined as the sum of the Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Calculate sum elements of sequence : sum . Series calculator allows to calculate online the sum of the terms of the sequence whose index is between the lower and the upper bound. Arithmetic sequences calculator that shows all the work, detailed explanation and steps. Get the free "Series Calculator" widget for your website, blog, Wordpress, Blogger, or iGoogle. Find more Mathematics widgets in Wolfram|Alpha. This calculator will find the sum of arithmetic, geometric, power, infinite, and binomial series, as well as the partial sum. Show Instructions In | general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`. Infinite Series calculator is a free online tool that gives the summation value of the given function for the given limits. BYJU’S online infinite series calculator tool makes the calculations faster and easier where it displays the value in a fraction of seconds. This free number sequence calculator can determine the terms (as well as the sum of all terms) of an arithmetic, geometric, or Fibonacci sequence. Explore many other math calculators, as well as hundreds of other calculators addressing health, fitness, finance, math, and more. |
https://spiral.ac/sharing/txfwf1y/common-denominators-14-and-56-math-4th-grade-khan-academy | 1,544,828,355,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826354.54/warc/CC-MAIN-20181214210553-20181214232553-00150.warc.gz | 728,011,419 | 9,367 | # Interactive video lesson plan for: Common denominators: 1/4 and 5/6 | Math | 4th grade | Khan Academy
#### Activity overview:
Lindsay uses fraction models and multiplication to find common denominators for 1/4 and 5/6.
4th grade math on Khan Academy: 4th grade is the time to start really fine-tuning your arithmetic skills. Not only will you be a multi-digit addition and subtraction rockstar, but you'll extend the multiplication and division that you started in 3rd grade to several digits. You'll also discover that you sometimes have something left over (called a "remainder") when you divide. In 3rd grade you learned what a fraction is. Now you'll start adding, subtracting, multiplying, and comparing them. You'll also see how they relate to decimals. On other fronts, you'll learn how to convert between different units (which is super important when comparing the size and speed of robot unicorns in different countries) and continue your journey thinking about various shapes in two dimensions. Some of the foundational concepts of geometry (like lines, rays and angles) also get introduced. As always, we'll round this out with a healthy dose of applied word problems and explorations of number patterns and properties (including the ideas of factors, multiples and prime numbers). The fun must not stop! (Content was selected for this grade level based on a typical curriculum in the United States.)
About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.
Clip makes it super easy to turn any public video into a formative assessment activity in your classroom.
Add multiple choice quizzes, questions and browse hundreds of approved, video lesson ideas for Clip
Make YouTube one of your teaching aids - Works perfectly with lesson micro-teaching plans
Play this activity
1. Students enter a simple code
2. You play the video
3. The students comment
4. You review and reflect
* Whiteboard required for teacher-paced activities
## Ready to see what elsecan do?
With four apps, each designed around existing classroom activities, Spiral gives you the power to do formative assessment with anything you teach.
Quickfire
Carry out a quickfire formative assessment to see what the whole class is thinking
Discuss
Create interactive presentations to spark creativity in class
Team Up
Student teams can create and share collaborative presentations from linked devices
Clip
Turn any public video into a live chat with questions and quizzes
### Spiral Reviews by Teachers and Digital Learning Coaches
@kklaster
Tried out the canvas response option on @SpiralEducation & it's so awesome! Add text or drawings AND annotate an image! #R10tech
Using @SpiralEducation in class for math review. Student approved! Thumbs up! Thanks.
@ordmiss
Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral
@strykerstennis
Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks! | 726 | 3,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2018-51 | latest | en | 0.939761 | # Interactive video lesson plan for: Common denominators: 1/4 and 5/6 | Math | 4th grade | Khan Academy #### Activity overview: Lindsay uses fraction models and multiplication to find common denominators for 1/4 and 5/6. 4th grade math on Khan Academy: 4th grade is the time to start really fine-tuning your arithmetic skills. Not only will you be a multi-digit addition and subtraction rockstar, but you'll extend the multiplication and division that you started in 3rd grade to several digits. You'll also discover that you sometimes have something left over (called a "remainder") when you divide. In 3rd grade you learned what a fraction is. Now you'll start adding, subtracting, multiplying, and comparing them. You'll also see how they relate to decimals. On other fronts, you'll learn how to convert between different units (which is super important when comparing the size and speed of robot unicorns in different countries) and continue your journey thinking about various shapes in two dimensions. Some of the foundational concepts of geometry (like lines, rays and angles) also get introduced. As always, we'll round this out with a healthy dose of applied word problems and explorations of number patterns and properties (including the ideas of factors, multiples and prime numbers). The fun must not stop! (Content was selected for this grade level based on a typical curriculum in the United States.) About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We've also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content. Clip makes it super easy to turn any public video into a formative assessment activity in your classroom. Add multiple choice quizzes, questions and browse hundreds of approved, video lesson ideas for Clip Make YouTube one of your teaching aids - Works perfectly with lesson micro-teaching plans Play this activity 1. Students enter a simple code 2. You play the video 3. The students comment 4. You review and reflect * Whiteboard required for teacher-paced activities ## Ready to see what elsecan do? With four apps, each designed around existing classroom activities, Spiral gives you the power to do formative assessment with anything you teach. Quickfire Carry out a quickfire formative assessment to see what the whole class is thinking Discuss Create interactive presentations to spark creativity in class Team Up Student teams can create and share collaborative presentations from linked devices Clip Turn any public video into a live chat with questions and quizzes ### Spiral Reviews by Teachers and Digital Learning Coaches @kklaster Tried out the canvas response option on @SpiralEducation & it's | so awesome! Add text or drawings AND annotate an image! #R10tech Using @SpiralEducation in class for math review. Student approved! Thumbs up! Thanks. @ordmiss Absolutely amazing collaboration from year 10 today. 100% engagement and constant smiles from all #lovetsla #spiral @strykerstennis Students show better Interpersonal Writing skills than Speaking via @SpiralEducation Great #data #langchat folks! |
https://betterlesson.com/lesson/reflection/22343/teaching-slope-can-be-difficult | 1,498,460,815,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320685.63/warc/CC-MAIN-20170626064746-20170626084746-00053.warc.gz | 746,520,423 | 22,710 | Reflection: Self-Talk Discovering Slopes - Section 1: Accessing Prior Knowledge
Presenting the idea of slope, no matter how clearly done, is generally not enought for students to learn slope well. I can't rely just on how great I think my explanations are, or on how well students memorize the slope formula to get right answers. In this sense, teaching slope can be difficult because the explanations don't always connect the ideas. Students need many oportunities to tackle examples.
The Common Core calls for students to grapple with challenging math on their own. This is really important when teaching slope because it is a crutial topic for other math courses to come. So, the more I engage students in making sense of slope by having them face different examples figuring out strategies and connections on their own, the better. I've begun doing this in this lesson, but I invite teachers to find other tasks, similar or different, so that studens can work on. (MP8)
Teaching Slope can be difficult.
Self-Talk: Teaching Slope can be difficult.
Discovering Slopes
Unit 3: Relationships between Quantities/Reasoning with Equations
Lesson 8 of 15
Big Idea: The rate at which one quantity changes with respect to another can be determined either by computation or by looking at a graph.
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25 teachers like this lesson
Standards:
Subject(s):
Math, Algebra, Expressions (Algebra), 9th grade, slopes, slope formula, graphing lines
60 minutes
Mauricio Beltre
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Lines and Linear Equations Formative Assessment Lesson Day 1 of 2
8th Grade Math » Linear Equations in two Variables
Big Idea: Challenge students to understand multiple representations of linear equations by using a very visual demonstration of slope and y-intercept.
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Bowling Green, KY
Environment: Suburban | 495 | 2,212 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-26 | longest | en | 0.917825 | Reflection: Self-Talk Discovering Slopes - Section 1: Accessing Prior Knowledge Presenting the idea of slope, no matter how clearly done, is generally not enought for students to learn slope well. I can't rely just on how great I think my explanations are, or on how well students memorize the slope formula to get right answers. In this sense, teaching slope can be difficult because the explanations don't always connect the ideas. Students need many oportunities to tackle examples. The Common Core calls for students to grapple with challenging math on their own. This is really important when teaching slope because it is a crutial topic for other math courses to come. So, the more I engage students in making sense of slope by having them face different examples figuring out strategies and connections on their own, the better. I've begun doing this in this lesson, but I invite teachers to find other tasks, similar or different, so that studens can work on. (MP8) Teaching Slope can be difficult. Self-Talk: Teaching Slope can be difficult. Discovering Slopes Unit 3: Relationships between Quantities/Reasoning with Equations Lesson 8 of 15 Big Idea: The rate at which one quantity changes with respect to another can be determined either by computation or by looking at a graph. Print Lesson 25 teachers like this lesson Standards: Subject(s): Math, Algebra, Expressions (Algebra), 9th grade, slopes, slope formula, graphing lines 60 minutes Mauricio Beltre Similar Lessons Practice Session: Slope and Equations of Lines Algebra I » Lines Big Idea: The use of an online lesson allows students to approach this practice session from a variety of entry points. Favorites(5) Resources(14) Worcester, MA Environment: Urban Skeleton Towers 8th Grade Math » Math Exploratorium Big Idea: Figurate Numbers Favorites(6) Resources(9) New York, NY Environment: Urban Lines and Linear Equations Formative Assessment Lesson Day 1 of 2 8th Grade | Math » Linear Equations in two Variables Big Idea: Challenge students to understand multiple representations of linear equations by using a very visual demonstration of slope and y-intercept. Favorites(5) Resources(8) Bowling Green, KY Environment: Suburban |
https://www.aussportsbetting.com/2015/01/21/model-testing-measuring-forecast-accuracy/ | 1,720,851,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514490.70/warc/CC-MAIN-20240713051758-20240713081758-00410.warc.gz | 534,762,901 | 36,936 | # Model Testing – Measuring Forecast Accuracy
When developing a betting model it is important to properly measure its performance. Having a formal measure of performance is important because it provides a benchmark with which to test alternative models.
Obviously, when using a model for betting a key measure of performance is the profit or loss obtained when using it, but it’s nice to have more than one metric to identify differences between similarly performing models.
This article is the second of a two-part series. The first article, Model Testing – Measuring Profit & Loss, looks at various ways to measure betting profit. This article outlines various measures of forecast accuracy in the context of betting models. The examples focus on measuring the accuracy of a model that predicts total game scores.
This article borrows heavily from Chapter 2 of Forecasting: Methods and Applications, by Makridakis, Wheelwright and Hyndman.
### Understanding Sigma (Σ) Notation
Many of the formulas below using sigma notation. For those who can’t recall how to use Σ from their school days, click here to brush up on sigma notation.
### Standard Statistical Measures
Below is a range of measures to evaluate the accuracy of a forecasting model. In the context of sports betting, applications include forecasting total scores in rugby, the number of corners in soccer and winning margins in basketball.
To help illustrate the computations involved, the following measures will be applied to a simplified data set. The table below shows rugby league total scores and a model’s pre-game forecast of those totals.
Game
i
Total Score
Yi
Forecast
Fi
Forecast Error
ei
1 53 50 3
2 36 52 -16
3 34 32 2
4 40 36 4
5 39 44 -5
6 51 54 -3
7 42 42 0
8 36 44 -8
If Yi is the actual total score for game i and Fi is the pre-game forecast of the total score for game i, then we define the error of the forecast for game i as:
Looking at the table above, the error for game 1 is 53 – 50 = 3, the error for game 2 is 36 – 52 = -16, and so on.
For the equations below we will use the variable n to denote how many completed games we have forecasts for.
The above data set has total scores and their associated forecasts for 8 games, so n equals 8 in this case.
Using our variables ei and n, we can calculate the following statistical measures of forecast accuracy:
### Mean error (ME):
This measure will often be small because positive and negative errors will offset each other. This makes it pretty useless as a measure of accuracy because a model that has forecast errors of -2, +2, -2 and +2 will sum to 0, but so do -20, +20, -20 and +20. With that being said, the mean error is worth calculating because it will tell you if there is any systematic under- or over-estimating, which is called forecast bias. A model that tends to over- and under- estimate fairly equally will have a mean error close to zero, while a model that has a bias towards underestimating scores will have a strong positive value (note that ei = Yi – Fi, so if you underestimate the value, the forecast error is positive).
In the example data above the ME equals (3 + -16 + 2 + …)/8 = -2.875, illustrating that on average this model overestimated the total score.
### Mean absolute error (MAE):
Mean absolute error gets around the offsetting effect of positive and negative forecast errors by taking the absolute value of each error. Using our data set above we get |e1| = |53 – 50| = 3, |e2| = |36 – 52| = 16, and so on.
The advantage of using MAE is it provides a scale which people can understand. For example, if you had the following forecast errors: -2, +2, -2. +2, the MAE would equal 2, showing the model forecasts are 2 units off the correct value, on average.
In the example data above the MAE equals (3 + 16 + 2 + …)/8 = 5.125. In written terms you can say the average estimated total score was 5.125 points off the actual total.
### Mean squared error (MSE):
An alternative to taking the absolute value of each error is to square them, so a forecast error of -2 becomes 22 = 4. Like MAE, this avoids having positive and negative errors offset each other.
A point of difference between MAE and MSE is that MSE is more punishing for large errors, because squaring larger numbers produce markedly bigger results. For example the difference between 4 and 5 is just 1, but the difference between 42 and 52 is 9.
From a mathematics perspective many practitioners prefer to use MSE over MAE because squared functions are easier do deal with in optimisation calculations. From a calculus perspective it’s easier to take the derivative of a function with squared terms than a function with absolute value terms.
In the example data above the MSE equals ((32) + (-162) + (22) + …)/8 = 47.875. Note that this is significantly larger than the MAE due to the large error for Game 2.
### Percentage / relative errors
The above measures are all dependent on the scale of the data. For example, these measures would likely be much larger for basketball total scores than rugby total scores because basketball scores generally are much higher. If you’re off by 10% on a basketball score this could imply being 18 points off, whereas with rugby 10% could mean being off by 4 points.
The previously discussed error measures make comparing models between sports very difficult. The following measures adjust for the scale of the data, which can facilitate comparisons between models applied to different sports.
Rather than use absolute errors, ei = Yi – Fi, percentage errors are used instead, which are calculated as follows:
Using our original data set, for game 1 the percentage error is PE1 = (53 – 50)/53 = 0.0566 = 5.66%.
### Mean percentage error (MPE):
This is equivalent to ME discussed earlier, but it’s calculated using percentage errors.
MPE suffers the same drawback as ME through having positive and negative PEs offset each other, however this does mean it provides a measure of systematic bias.
In the example data above the MPE equals ((53 – 50)/53 + (36 – 52)/36 + …)/8 = -8.0%.
### Mean absolute percentage error (MAPE):
MAPE is equivalent to MAE discussed earlier, but it’s calculated using percentage errors.
MAPE works well for total score models, but it isn’t ideal for all situations. For example football scores tend to be low, so using measures like MAE and MSE may make more intuitive sense than using MAPE where you can have errors like 400% if the total score is 4 and the model predicted 1.
A more serious limitation MAPE occurs when your data set can have 0 values. In the context of sports betting if you’re forecasting winning margins and a draw is possible then you can’t use MAPE. This is because if the final scores are level then Yi = 0 so PEi can’t be calculated due to a division by zero error. For this reason MAPE works best for modeling results such as total basketball, AFL and rugby scores rather than winning margins or football total scores, which can have zero values.
In the example data above the MAPE equals (|53 – 50|/53 + |36 – 52|/36 + …)/8 = 13.4%.
### Comparing forecast methods
Once you have a measure of forecasting accuracy, how do you know if it’s a good result? Is a total score MSE of 9 or a MAPE of 6% good? What we need is a comparison of the model’s performance to more naive (basic) models to test if they represent a meaningful improvement.
In the context of forecasting total scores, suppose we have a naive model that predicts the total score for each game by simply using the total score from the last time the two sides met at the same venue. If in rugby league, Team A vs. Team B had a combined score of 38 the last time they met, then the naive model will predict the combined score for their next meeting to be 38.
Once a naive model has been created you can then calculate the forecast accuracy for it and compare its statistics to the accuracy calculations of the more sophisticated model. If the sophisticated model can’t outperform the naive model then it may mean going back to the drawing board.
### Out-of-sample accuracy measurement
The above measures calculate the accuracy of a betting model, however achieving good accuracy using historical data doesn’t guarantee the model will work well in the future.
Suppose you obtain a historical odds data set and you identify a trading strategy that would have worked well over the past three seasons. How confident can you be that the strategy will continue to work in the future? As anyone who has analysed historical data can tell you, if you look hard enough, you will find some strategy that would have made a killing had it been employed in previous years, however this provides no guarantee for future success.
A way to know if a model is genuinely useful and not simply reflecting quirks in your specific data set is to split your data into two parts before constructing the model. The first part of the data is used to build and calibrate the model and the second holdout set is used to test whether the model works well on the second set of data.
We highly recommend you read our article, Post-Sample Evaluation – the Importance of Creating a Holdout Set before Calibrating a Betting Strategy. It outlines how to create a holdout set to test a calibrated betting model. This practice provides an out-of-sample accuracy measurement because it involves evaluating a forecasting model using more recent data than was used to calibrate the model.
### Sources
An excellent book on forecasting is:
Forecasting: Methods and Applications
Spyros G. Makridakis, Steven C. Wheelwright, Rob J. Hyndman | 2,144 | 9,629 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2024-30 | latest | en | 0.898153 | # Model Testing – Measuring Forecast Accuracy When developing a betting model it is important to properly measure its performance. Having a formal measure of performance is important because it provides a benchmark with which to test alternative models. Obviously, when using a model for betting a key measure of performance is the profit or loss obtained when using it, but it’s nice to have more than one metric to identify differences between similarly performing models. This article is the second of a two-part series. The first article, Model Testing – Measuring Profit & Loss, looks at various ways to measure betting profit. This article outlines various measures of forecast accuracy in the context of betting models. The examples focus on measuring the accuracy of a model that predicts total game scores. This article borrows heavily from Chapter 2 of Forecasting: Methods and Applications, by Makridakis, Wheelwright and Hyndman. ### Understanding Sigma (Σ) Notation Many of the formulas below using sigma notation. For those who can’t recall how to use Σ from their school days, click here to brush up on sigma notation. ### Standard Statistical Measures Below is a range of measures to evaluate the accuracy of a forecasting model. In the context of sports betting, applications include forecasting total scores in rugby, the number of corners in soccer and winning margins in basketball. To help illustrate the computations involved, the following measures will be applied to a simplified data set. The table below shows rugby league total scores and a model’s pre-game forecast of those totals. Game i Total Score Yi Forecast Fi Forecast Error ei 1 53 50 3 2 36 52 -16 3 34 32 2 4 40 36 4 5 39 44 -5 6 51 54 -3 7 42 42 0 8 36 44 -8 If Yi is the actual total score for game i and Fi is the pre-game forecast of the total score for game i, then we define the error of the forecast for game i as: Looking at the table above, the error for game 1 is 53 – 50 = 3, the error for game 2 is 36 – 52 = -16, and so on. For the equations below we will use the variable n to denote how many completed games we have forecasts for. The above data set has total scores and their associated forecasts for 8 games, so n equals 8 in this case. Using our variables ei and n, we can calculate the following statistical measures of forecast accuracy: ### Mean error (ME): This measure will often be small because positive and negative errors will offset each other. This makes it pretty useless as a measure of accuracy because a model that has forecast errors of -2, +2, -2 and +2 will sum to 0, but so do -20, +20, -20 and +20. With that being said, the mean error is worth calculating because it will tell you if there is any systematic under- or over-estimating, which is called forecast bias. A model that tends to over- and under- estimate fairly equally will have a mean error close to zero, while a model that has a bias towards underestimating scores will have a strong positive value (note that ei = Yi – Fi, so if you underestimate the value, the forecast error is positive). In the example data above the ME equals (3 + -16 + 2 + …)/8 = -2.875, illustrating that on average this model overestimated the total score. ### Mean absolute error (MAE): Mean absolute error gets around the offsetting effect of positive and negative forecast errors by taking the absolute value of each error. Using our data set above we get |e1| = |53 – 50| = 3, |e2| = |36 – 52| = 16, and so on. The advantage of using MAE is it provides a scale which people can understand. For example, if you had the following forecast errors: -2, +2, -2. +2, the MAE would equal 2, showing the model forecasts are 2 units off the correct value, on average. In the example data above the MAE equals (3 + 16 + 2 + …)/8 = 5.125. In written terms you can say the average estimated total score was 5.125 points off the actual total. ### Mean squared error (MSE): An alternative to taking the absolute value of each error is to square them, so a forecast error of -2 becomes 22 = 4. Like MAE, this avoids having positive and negative errors offset each other. A point of difference between MAE and MSE is that MSE is more punishing for large errors, because squaring larger numbers produce markedly bigger results. For example the difference between 4 and 5 is just 1, but the difference between 42 and 52 is 9. From a mathematics perspective many practitioners prefer to use MSE over MAE because squared functions are easier do deal with in optimisation calculations. From a calculus perspective it’s easier to take the derivative of a function with squared terms than a function with absolute value terms. In the example data above the MSE equals ((32) + (-162) + (22) + …)/8 = 47.875. Note that this is significantly larger than the MAE due to the large error for Game 2. ### Percentage / relative errors The above measures are all dependent on the scale of the data. For example, these measures would likely be much larger for basketball total scores than rugby total scores because basketball scores generally are much higher. If you’re off by 10% on a basketball score this could imply being 18 points off, whereas with rugby 10% could mean being off by 4 points. The previously discussed error measures make comparing models between sports very difficult. The following measures adjust for the scale of the data, which can facilitate comparisons between models applied to different sports. Rather than use absolute errors, ei = Yi – Fi, percentage errors are used instead, which are calculated as follows: Using our original data set, for game 1 the percentage error is PE1 = (53 – 50)/53 = 0.0566 = 5.66%. ### Mean percentage error (MPE): This is equivalent to ME discussed earlier, but it’s calculated using percentage errors. MPE suffers the same drawback as ME through having positive and negative PEs offset each other, however this does mean it provides a measure of systematic bias. In the example data above the MPE equals ((53 – 50)/53 + (36 – 52)/36 + …)/8 = -8.0%. ### Mean absolute percentage error (MAPE): MAPE is equivalent to MAE discussed earlier, but it’s calculated using percentage errors. MAPE works well for total score models, but it isn’t ideal for all situations. For example football scores tend to be low, so using measures like MAE and MSE may make more intuitive sense than using MAPE where you can have errors like 400% if the total score is 4 and the model predicted 1. A more serious limitation MAPE occurs when your data set can have 0 values. In the context of sports betting if you’re forecasting winning margins and a draw is possible then you can’t use MAPE. This is because if the final scores are level then Yi = 0 so PEi can’t be calculated due to a division by zero error. For this reason MAPE works best for modeling results such as total basketball, AFL and rugby scores rather than winning margins or football total scores, which can have zero values. In the example data above the MAPE equals (|53 – 50|/53 + |36 – 52|/36 + …)/8 = 13.4%. ### Comparing forecast methods Once you have a measure of forecasting accuracy, how do you know if it’s a good result? Is a total score MSE of 9 or a MAPE of 6% good? What we need is a comparison of the model’s performance to more naive (basic) models to test if they represent a meaningful improvement. In the context of forecasting total scores, suppose we have a naive model that predicts the total score for each game by simply using the total score from the last time the two sides met at the same venue. If in rugby league, Team A vs. Team B had a combined score of 38 the last time they met, then the naive model will predict the combined score for their next meeting to be 38. Once a naive model has been created you can then calculate the forecast accuracy for it and compare its statistics to the accuracy calculations of the more sophisticated model. If the sophisticated model can’t outperform the naive model then it may mean going back to the drawing board. ### Out-of-sample accuracy measurement The above measures calculate the accuracy of a betting model, however achieving good accuracy using historical data doesn’t guarantee the model will work well in the future. Suppose you obtain a historical odds data set and you identify a trading strategy that would have worked well over the past three seasons. How confident can you be that the strategy will continue to work in the future? As anyone who has analysed historical data can tell you, if you look hard enough, you will find | some strategy that would have made a killing had it been employed in previous years, however this provides no guarantee for future success. A way to know if a model is genuinely useful and not simply reflecting quirks in your specific data set is to split your data into two parts before constructing the model. The first part of the data is used to build and calibrate the model and the second holdout set is used to test whether the model works well on the second set of data. We highly recommend you read our article, Post-Sample Evaluation – the Importance of Creating a Holdout Set before Calibrating a Betting Strategy. It outlines how to create a holdout set to test a calibrated betting model. This practice provides an out-of-sample accuracy measurement because it involves evaluating a forecasting model using more recent data than was used to calibrate the model. ### Sources An excellent book on forecasting is: Forecasting: Methods and Applications Spyros G. Makridakis, Steven C. Wheelwright, Rob J. Hyndman |
https://www.physicsforums.com/threads/problem-from-mit-ocw.83419/ | 1,563,762,035,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527458.86/warc/CC-MAIN-20190722010436-20190722032436-00339.warc.gz | 805,535,825 | 16,694 | # Problem from MIT OCW
#### amcavoy
I found this problem on the MIT OpenCourseWare website, but a solution was not given. I tried it out, so I was wondering if my answer is correct. The problem is as follows:
----------------------------------------------------------
Suppose A is reduced by the usual row operations to
$$R=\begin{bmatrix}1 & 4 & 0 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0\end{bmatrix}.$$
Find the complete solution (if a solution exists) to this system involving the original A:
Ax = sum of the columns of A.
----------------------------------------------------------
I figured if you took the columns of A and added them together, it would be the same as multiplying a by <1,1,1>. Thus I have the following:
$$A\mathbf{x}=A\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}.$$
So the solution is the vector <1,1,1>. Is this correct?
Thank you.
Related Linear and Abstract Algebra News on Phys.org
#### neurocomp2003
no,the solution is a general solution not a single solution or no solution. depending on what R is R=A or R=A|b(augmented matrix) the solution is
R=A ( id oubt its this one)
x1+4*x2+2*x4=0;
x3+2*x4 = 0
R=A|b
x1+4*x2=2
x3 =2
therefore the solution is ->
x=
|2-4t|
|..t..|
|..2..|
you arelooking at the rows.
[]if there was no solution a row would have all 0s except the last one=#(!=0)
thus 0=# which is false
[]if there was 1 solution than all the xi would be solvable...that is there must be a few rows where its just xi=# like the 2nd row. THen the remaining rows would be solved by the other linear equations in the matrix.
[]if there was many solutions than you'd have hte above.
Last edited:
#### amcavoy
So, for R=A:
$$x_1+4x_2+2x_3=0$$
$$x_3+2x_4=0$$
For which the general solution is:
$$x_2\begin{bmatrix}-4 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_4\begin{bmatrix}4 \\ 0 \\ -2 \\ 1\end{bmatrix}\quad\forall x_2,x_4\in\mathbb{R}$$
I understand how to find the general solution and everything, but what I would like to know is why you set the equations equal to zero. If R=A, then the sum of the columns of A should equal the sum of the columns of R, which is non-zero. Could you just explain your steps when you set R=A and R=[A|b]?
Thanks again.
#### neurocomp2003
if the system is Ax=b; then R=[A|b] where the last column of R is b and each remaining column of R represents a xi; I believe that is the solution
however because you did not state the question the solution could be R=A where you did not state b...and so i assumed b=O
#### amcavoy
Alright I understand what you are trying to say.
$$A\mathbf{x}=A\mathbf{u}\quad \mathbf{u}=\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}$$
$$A(\mathbf{x}-\mathbf{u})=0$$
So upon solving, I come up with the general solution which is:
$$x_2\begin{bmatrix}-4 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_4\begin{bmatrix}-2 \\ 0 \\ -2 \\ 1\end{bmatrix}+\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}\quad\forall x_2,x_4\in\mathbb{R}$$
If I didn't make any arithmetic mistakes.
### Physics Forums Values
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• Solo and co-op problem solving | 1,002 | 3,279 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2019-30 | longest | en | 0.904896 | # Problem from MIT OCW #### amcavoy I found this problem on the MIT OpenCourseWare website, but a solution was not given. I tried it out, so I was wondering if my answer is correct. The problem is as follows: ---------------------------------------------------------- Suppose A is reduced by the usual row operations to $$R=\begin{bmatrix}1 & 4 & 0 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0\end{bmatrix}.$$ Find the complete solution (if a solution exists) to this system involving the original A: Ax = sum of the columns of A. ---------------------------------------------------------- I figured if you took the columns of A and added them together, it would be the same as multiplying a by <1,1,1>. Thus I have the following: $$A\mathbf{x}=A\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}.$$ So the solution is the vector <1,1,1>. Is this correct? Thank you. Related Linear and Abstract Algebra News on Phys.org #### neurocomp2003 no,the solution is a general solution not a single solution or no solution. depending on what R is R=A or R=A|b(augmented matrix) the solution is R=A ( id oubt its this one) x1+4*x2+2*x4=0; x3+2*x4 = 0 R=A|b x1+4*x2=2 x3 =2 therefore the solution is -> x= |2-4t| |..t..| |..2..| you arelooking at the rows. []if there was no solution a row would have all 0s except the last one=#(!=0) thus 0=# which is false []if there was 1 solution than all the xi would be solvable...that is there must be a few rows where its just xi=# like the 2nd row. THen the remaining rows would be solved by the other linear equations in the matrix. []if there was many solutions than you'd have hte above. Last edited: #### amcavoy So, for R=A: $$x_1+4x_2+2x_3=0$$ $$x_3+2x_4=0$$ For which the general solution is: $$x_2\begin{bmatrix}-4 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_4\begin{bmatrix}4 \\ 0 \\ -2 \\ 1\end{bmatrix}\quad\forall x_2,x_4\in\mathbb{R}$$ I understand how to find the general solution and everything, but what I would like to know is why you set the equations equal to zero. If R=A, then the sum of the columns of A should equal the sum of the columns of R, which is non-zero. Could you just explain your steps when you set R=A and R=[A|b]? Thanks again. #### neurocomp2003 if the system is Ax=b; then R=[A|b] where the last column of R is b and each remaining column of R represents a xi; I believe that is the solution however because you did not state the question the solution could be R=A where you did not state b...and so i assumed b=O #### amcavoy Alright I understand what you are trying to say. $$A\mathbf{x}=A\mathbf{u}\quad \mathbf{u}=\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}$$ $$A(\mathbf{x}-\mathbf{u})=0$$ So upon solving, I come up with the general solution which is: $$x_2\begin{bmatrix}-4 \\ 1 \\ 0 \\ 0\end{bmatrix}+x_4\begin{bmatrix}-2 \\ 0 \\ -2 \\ 1\end{bmatrix}+\begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}\quad\forall x_2,x_4\in\mathbb{R}$$ If I didn't | make any arithmetic mistakes. ### Physics Forums Values We Value Quality • Topics based on mainstream science • Proper English grammar and spelling We Value Civility • Positive and compassionate attitudes • Patience while debating We Value Productivity • Disciplined to remain on-topic • Recognition of own weaknesses • Solo and co-op problem solving |
https://groupprops.subwiki.org/w/index.php?title=Order_of_an_element&printable=yes | 1,579,749,232,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608062.57/warc/CC-MAIN-20200123011418-20200123040418-00278.warc.gz | 471,890,168 | 7,987 | Order of an element
WARNING: POTENTIAL TERMINOLOGICAL CONFUSION: Please don't confuse this with order of a group
Definition
The order of an element $x$ in a group $G$ is the smallest positive integer $n$ for which $x^n$ is the identity element.
Such a $n$ may not always exist (if it exists, $x$ is said to be of finite order, or is termed a torsion element). It does exist when the group is finite.
Examples
• The identity element has order $1$ in any group
• In the group of integers modulo $n$, the element $1$ has order $n$
Facts
For an element of finite order, the order of the element equals the order of the cyclic subgroup generated by the element. Thus, by Lagrange's theorem, the order of an element $x$ in a finite group $G$ divides the order of $G$ (where order here means the total cardinality of the group).
The exponent of a group is defined as the least common multiple of the orders of all elements of the group. For a finite group, the exponent always exists, and is a divisor of the order of the group (though it may, in general, be smaller). There may or may not exist an element in the group whose order equals the exponent of the group.
For an infinite group, not every element necessarily has finite order. A group where every element has finite order is termed a periodic group. Even for a periodic group, the exponent may be infinite because there may not be a common bound on the orders of all elements. A group with bounded exponent is a group whose exponent is finite, the condition of having bounded exponent is stronger than the condition of being periodic. | 365 | 1,597 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 13, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2020-05 | latest | en | 0.900178 | Order of an element WARNING: POTENTIAL TERMINOLOGICAL CONFUSION: Please don't confuse this with order of a group Definition The order of an element $x$ in a group $G$ is the smallest positive integer $n$ for which $x^n$ is the identity element. Such a $n$ may not always exist (if it exists, $x$ is said to be of finite order, or is termed a torsion element). It does exist when the group is finite. Examples • The identity element has order $1$ in any group • In the group of integers modulo $n$, the element $1$ has order $n$ Facts For an element of finite order, the order of the element equals the order of the cyclic subgroup generated by the element. Thus, by Lagrange's theorem, the order of an element $x$ in a finite group $G$ divides the order of $G$ (where order here means the total cardinality of the group). The exponent of a group is defined as the least common multiple of the orders of all elements of the group. For a finite group, the exponent always exists, and is a divisor of the order of the group (though it may, in general, be smaller). There may or may not exist an element in the group whose order equals the exponent of the group. For an infinite group, not every element necessarily has finite order. A group where every element has finite order is termed a periodic group. Even for a periodic group, the exponent may be infinite because there may not be a common bound on the orders | of all elements. A group with bounded exponent is a group whose exponent is finite, the condition of having bounded exponent is stronger than the condition of being periodic. |
https://www.jiskha.com/questions/1121621/Adam-Sam-and-Eddie-have-to-meet-their-boss-at-3-00-in-the-office-9-miles-apart | 1,563,877,011,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529175.83/warc/CC-MAIN-20190723085031-20190723111031-00041.warc.gz | 733,304,437 | 5,184 | # math problem
Adam, Sam, and Eddie have to meet their boss at 3:00 in the office 9 miles apart from their home, and they must be arrive in time for the appointment. Adam had a bicycle, Sam had a skateboard,and Harry had no transport. So they have to share.
so they left the house together.
Each of them travel on walk at 6 m.p.h.; each of them skateboard at 10 m.p.h.; and each of them cycle at 16 m.p.h.They arrived in time for their 3:00 appointment. At what time did they leave the house?
1. 👍 0
2. 👎 0
3. 👁 102
## Similar Questions
1. ### 5th grade math problem
Adam, Sam, and Eddie have to meet their boss at 3:00 in the office 9 miles apart from their home, and they must be arrive in time for the appointment. Adam had a bicycle, Sam had a skateboard,and Harry had no transport. So they
asked by victoria on October 26, 2014
2. ### maths
1/5 of Jane's money is equal to 1/3 of Sam's money. The difference in their amount is 1/2 of Adam's money. If Adam gives \$120 to Sam, Sam will have the same amount of money as Jane. How much do the 3 people have
asked by joi on December 28, 2011
3. ### math
On the blueprints of Adam's office building, the length of his rectangular office space is 5 1/4 inches. If the actual length of Adam's office space is 10 1/2 feet, what is the scale of the blueprint?
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4. ### Math aglebra
I need help RIGHT NOW!!!!!!!!!!!!!! On the blueprints of Adam's office building, the length of his rectangular office space is 5 1/4 inches. If the actual length of Adam's office space is 10 1/2 feet, what is the scale of the
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5. ### tax
At the beginning of 2011, Melissa was permanently transferred from her office in New York City to New Jersey. Her office in NYC is 15 miles from her old NY home. For Melissa to meet the distance test for qualifying moving expense
asked by Andy on March 7, 2012
6. ### Algebra
FRODO AND SAM are 17 miles apart and are walking towards each other they meet in 4hrs. find frodos rate if frodo walks 1mph faster than sam.
asked by Angie on September 17, 2015
How does George show he protects Lennie when they meet the boss? My Answer- George answers all of the boss' questions himself to prevent the boss from finding out that Lennie was crazy and not smart and then not giving them jobs.
asked by Anonymous on August 22, 2007
i have a problem with this question my teacher gave me. i don't know what to write for the answer. the problem: there is a girl named Jane who is hired as the head of the payroll department at R& S Electronic company, firm of 75
asked by jack smith on August 22, 2011
9. ### Algebra
Agents J and K work in a long hallway with 4000 equally-spaced, consecutive offices. The agents decide to walk toward each other so they can meet and have lunch together. Agent J has recently been relocated from office 2013 to
asked by Mr. Alexander on August 26, 2015
10. ### ela
can youproof read for me please Eddie told Alex and his dad that Lao Xu is just doing his job about reporting back to his superiors on all our activities. Alex felt bad because he liked Lao Xu who was becoming his friend but he is
asked by anu on January 11, 2010
More Similar Questions | 860 | 3,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2019-30 | latest | en | 0.972632 | # math problem Adam, Sam, and Eddie have to meet their boss at 3:00 in the office 9 miles apart from their home, and they must be arrive in time for the appointment. Adam had a bicycle, Sam had a skateboard,and Harry had no transport. So they have to share. so they left the house together. Each of them travel on walk at 6 m.p.h.; each of them skateboard at 10 m.p.h.; and each of them cycle at 16 m.p.h.They arrived in time for their 3:00 appointment. At what time did they leave the house? 1. 👍 0 2. 👎 0 3. 👁 102 ## Similar Questions 1. ### 5th grade math problem Adam, Sam, and Eddie have to meet their boss at 3:00 in the office 9 miles apart from their home, and they must be arrive in time for the appointment. Adam had a bicycle, Sam had a skateboard,and Harry had no transport. So they asked by victoria on October 26, 2014 2. ### maths 1/5 of Jane's money is equal to 1/3 of Sam's money. The difference in their amount is 1/2 of Adam's money. If Adam gives \$120 to Sam, Sam will have the same amount of money as Jane. How much do the 3 people have asked by joi on December 28, 2011 3. ### math On the blueprints of Adam's office building, the length of his rectangular office space is 5 1/4 inches. If the actual length of Adam's office space is 10 1/2 feet, what is the scale of the blueprint? asked by lulu on December 14, 2016 4. ### Math aglebra I need help RIGHT NOW!!!!!!!!!!!!!! On the blueprints of Adam's office building, the length of his rectangular office space is 5 1/4 inches. If the actual length of Adam's office space is 10 1/2 feet, what is the scale of the asked by Chelsea on November 23, 2016 5. ### tax At the beginning of 2011, Melissa was permanently transferred from her office in New York City to New Jersey. Her office in NYC is 15 miles from her old NY home. For Melissa to meet the distance test for qualifying moving expense asked by Andy on March 7, 2012 6. ### Algebra FRODO AND SAM are 17 miles apart and are walking towards each other they meet in 4hrs. find frodos rate if frodo walks 1mph faster than sam. asked by Angie on September 17, 2015 How does George show he protects Lennie when they meet the boss? My Answer- George answers all of the boss' questions himself to prevent the boss from finding out that Lennie was crazy and not smart and then not giving them jobs. asked by Anonymous on August 22, 2007 i have a problem with this question my teacher gave me. i don't know what to write for the answer. the problem: there is a girl named Jane who is hired as the head of the payroll department at R& S Electronic company, firm of 75 asked by jack smith on August 22, 2011 9. ### Algebra Agents J and K work in a long hallway with 4000 equally-spaced, consecutive offices. The agents decide to walk toward each other so they can meet and have lunch together. Agent J has recently been relocated from office 2013 to asked by Mr. Alexander on | August 26, 2015 10. ### ela can youproof read for me please Eddie told Alex and his dad that Lao Xu is just doing his job about reporting back to his superiors on all our activities. Alex felt bad because he liked Lao Xu who was becoming his friend but he is asked by anu on January 11, 2010 More Similar Questions |
https://www.roulettestakes.com/systems/ | 1,579,406,424,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594209.12/warc/CC-MAIN-20200119035851-20200119063851-00259.warc.gz | 1,082,795,872 | 41,164 | # Systems
Roulette systems are as old as roulette itself. Gamblers have tried to overcome the house edge by betting in a certain way. This relies on ‘gamblers fallacy’ where people believe that events must change to balance out the outcomes. For example, after 10 reds in a row, there must be few blacks due. This is false.
The law of large numbers says that “the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.” The expected value of reds and black, ignoring the green, is 50% red and 50% black. The law of large numbers says as you spin more, the real values should tend to this. So if you have 55 reds and 45 blacks, you’d expect some more blacks to make the number closer to the expected values. This is not right though. The percentage will get closer to 50% but the actual number will diverge. Imagine after a million spins, you’d be really surprised if there were exactly 500,000 blacks and 500,000 reds and also if the percentages weren’t 50% red and 50% black to so many decimal places.
There are two main ways that people try to beat roulette. The first and easiest is using an mathematical betting system. The simplest and most popular is the Martingale system. Basically, you double up after each bet. All these systems are doomed to failure in the long run. Imagine if you played Russian Roulette for £1 million and won. You’d walk away a winner but was it a good idea? If you kept doing it, you’d surely blow your brains out at some point but if you did it once and walked away then maybe it is good. Just because on average its a bad idea, you might be lucky.
The other ways rely on mechanical methods on real wheels. Unlike the mathematical betting systems, these can actually work. Its not easy though and most people haven’t got the patience or nerve to use these. Also, casinos monitor their roulette wheels a lot better than in the past so some methods no longer work. See the section on physical methods to beat roulette.
If you are going to try a mathematical system, use a system tester to see the behaviour of it. Doing systems in the short run can net some profit and odd wise is no different to what a recreational roulette would play. Its all down to luck and how you manage your bank roll. Certain systems can run down the bank really quick if they are going wrong.
This is especially true of the Martingale. Read more about it here. | 547 | 2,481 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2020-05 | latest | en | 0.95758 | # Systems Roulette systems are as old as roulette itself. Gamblers have tried to overcome the house edge by betting in a certain way. This relies on ‘gamblers fallacy’ where people believe that events must change to balance out the outcomes. For example, after 10 reds in a row, there must be few blacks due. This is false. The law of large numbers says that “the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.” The expected value of reds and black, ignoring the green, is 50% red and 50% black. The law of large numbers says as you spin more, the real values should tend to this. So if you have 55 reds and 45 blacks, you’d expect some more blacks to make the number closer to the expected values. This is not right though. The percentage will get closer to 50% but the actual number will diverge. Imagine after a million spins, you’d be really surprised if there were exactly 500,000 blacks and 500,000 reds and also if the percentages weren’t 50% red and 50% black to so many decimal places. There are two main ways that people try to beat roulette. The first and easiest is using an mathematical betting system. The simplest and most popular is the Martingale system. Basically, you double up after each bet. All these systems are doomed to failure in the long run. Imagine if you played Russian Roulette for £1 million and won. You’d walk away a winner but was it a good idea? If you kept doing it, you’d surely blow your brains out at some point but if you did it once and walked away then maybe it is good. Just because on average its a bad idea, you might be lucky. The other ways rely on mechanical methods on real wheels. Unlike the mathematical betting systems, these can actually work. Its not easy though and most people haven’t got the patience or nerve to use these. Also, casinos monitor their roulette wheels a lot better than in the past so some methods no longer work. See the section on physical methods to beat roulette. If you are going to try a mathematical system, use a system tester to see the behaviour of it. Doing systems in the short run can net some profit and odd wise is no | different to what a recreational roulette would play. Its all down to luck and how you manage your bank roll. Certain systems can run down the bank really quick if they are going wrong. This is especially true of the Martingale. Read more about it here. |
zhn.teknik100r.fun | 1,624,038,361,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00020.warc.gz | 1,001,251,994 | 11,879 | # Common excel math functions
Note that further math-related Excel functions are also provided in the Excel Statistical Functions and Excel Engineering Functions categories. The tables below list all the current built-in Excel math functions, grouped by category, to help you to find the function you need. Selecting a function link will take you to a full description of the function, with examples of use and common errors.
Note that some of the Excel math functions listed below were introduced in recent versions of Excel, and so are not available in earlier versions. Excel Functions.
Basic Numeric Information.
Courseworks completed paper date
Performs a specified calculation e. Rounding Functions. Rounds a number away from zero i. Rounds a number upregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number upregardless of the sign of the number, to a multiple of significance.
New in Excel Rounds a number up to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number towards zeroi. Rounds a number downregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number down, to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number up or downto the nearest multiple of significance. Truncates a number towards zero i. Returns the unit matrix for a specified dimension New in Excel Conditional Sums.
Adds the cells in a supplied range, that satisfy multiple criteria New in Excel Returns the sum of the products of corresponding values in two or more supplied arrays. Returns the sum of the difference of squares of corresponding values in two supplied arrays.
Returns the sum of the sum of squares of corresponding values in two supplied arrays. Returns the sum of squares of differences of corresponding values in two supplied arrays.
Returns the secant of an angle New in Excel Returns the hyperbolic secant of an angle New in Excel Returns the cosecant of an angle New in Excel Cheat Sheet of Excel formulas and function is always a customized worksheet where we can have all those function details, shortcut keys to execute any function or formulas, custom way to use 2 or more function together and guideline to use them.
Also, we can choose those formulas which are complicated to apply for users in a cheat sheet. Start Your Free Excel Course. As we can see, here several string functions are listed. Please refer the below screenshot where I have applied the functions on strings and shown their workings.
Please refer the below screenshot where I have applied the functions on number values and shown their workings. Please refer the below screenshot where I have applied the functions on values and shown their workings.
This has been a guide to Excel Formulas Cheat Sheet. You can also go through our other suggested articles —. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy.
By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy. Forgot Password? Call Our Course Advisors. Cheat Sheet of Excel Formulas. Popular Course in this category. Course Price View Course. Free Excel Course. Login details for this Free course will be emailed to you. Email ID. Contact No.So, the total production quantity is Similarly, apply the same logic to get the total salary amount.
Now we know what overall sum values are. Out of these overall total of employees, we need to find the average salary per employee. Select the range of cells for which we are finding the average value, so our range of cells will be from D2 to D We know the average salary per person; for further drill-down, we want to know what is the average salary based on gender.
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So totally there are 10 employees on the list. After counting the total number of employees, we may need to count how many male and female employees are there.
MOD function will return the remainder when one number is divided by another number. For example, when you divide number 11 by 2, we will get the remainder as 1 because only till 10 number 2 can divide.
When we have fraction or decimal values, we may need to round those decimal values to the nearest integer number.
For example, we need to round the number 3. As you can see above, B2 cell value Like this, we can use various mathematical functions in excel to do mathematical operations in excel quickly and easily. This has been a guide to Mathematical Function in Excel. Here we discuss how to calculate mathematical function in excel using Sum, Average, Averageif, Counta, Countif, Mod, and Round formulas along with practical examples.
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By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy.The tutorial provides a list of Excel basic formulas and functions with examples and links to related in-depth tutorials.
Being primarily designed as a spreadsheet program, Microsoft Excel is extremely powerful and versatile when it comes to calculating numbers or solving math and engineering problems. It enables you to total or average a column of numbers in the blink of an eye.
Apart from that, you can compute a compound interest and weighted average, get the optimal budget for your advertising campaign, minimize the shipment costs or make the optimal work schedule for your employees. All this is done by entering formulas in cells. This tutorial aims to teach you the essentials of Excel functions and show how to use basic formulas in Excel. Before providing the basic Excel formulas list, let's define the key terms just to make sure we are on the same page.
So, what do we call an Excel formula and Excel function? Function is a predefined formula already available in Excel. Functions perform specific calculations in a particular order based on the specified values, called arguments, or parameters. You can find all available Excel functions in the Function Library on the Formulas tab:. Of course, it's next to impossible to memorize all of them, and you actually don't need to.
The Function Wizard will help you find the function best suited for a particular task, while the Excel Formula Intellisense will prompt the function's syntax and arguments as soon as you type the function's name preceded by an equal sign in a cell:.
Clicking the function's name will turn it into a blue hyperlink, which will open the Help topic for that function.
What follows below is a list of 10 simple yet really helpful functions that are a necessary skill for everyone who wishes to turn from an Excel novice to an Excel professional. The first Excel function you should be familiar with is the one that performs the basic arithmetic operation of addition:. In the syntax of all Excel functions, an argument enclosed in [square brackets] is optional, other arguments are required.
Meaning, your Sum formula should include at least 1 number, reference to a cell or a range of cells. For example:. If necessary, you can perform other calculations within a single formula, for example, add up values in cells B2 through B6, and then divide the sum by To sum with conditions, use the SUMIF function: in the 1st argument, you enter the range of cells to be tested against the criteria A2:A6in the 2nd argument - the criteria itself D2and in the last argument - the cells to sum B2:B6 :.
In your Excel worksheets, the formulas may look something similar to this:.
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Its syntax is similar to SUM's:. Sums values in cells B2 through B6, and then divides the result by 5.
## Mathematical Function in Excel
And what do you call adding up a group of numbers and then dividing the sum by the count of those numbers? Yep, an average!The basic functions covered below are among the most popular formulas in Excel—the ones everyone should know. To help you learn, we've also provided a spreadsheet with all the formula examples we cover below. After you enter one of these functions in A1, you can then reformat the Date and Time or use the system default.
It also subtracts, multiplies, divides, and uses any of the comparison operators to return a result of 1 true or 0 false. Excel frames the column of numbers in green borders and displays the formula in the current cell. Remember your high school math?
If the numbers inside the formula are not grouped properly, the answer will be wrong. Notice the screenshot below figure 2. For this exercise, you can enter the same values in H, I, and J, with or without the blank rows in between again, added for easier viewing.
Note that as we build each formula, we are combining the steps, eventually, into a single formula.
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We start out with three separate formulas. The formula in K3 is wrong. It requires grouping the numbers according to the order of calculation using commas or parentheses.
Check your numbers again with your calculator and note that this formula is correct. Note that the syntax the structure or layout of the formula is correct in cells N7 and N8, but incorrect in N6.
By combining these formulas into one, you can eliminate columns K and L. The RAND function is really simple and traditionally used for statistical analysis, cryptography, gaming, gambling, and probability theory, among dozens of other things.
Note; however, that every time you enter new data and press the Enter key, the list of random numbers you just created changes.
If you need to maintain your random numbers lists, you must format the cells as values. Now the list contains values instead of functions, so it will not change. Notice in the formula bar that the random numbers have 15 digits after the decimal Excel defaults to 9which you can change, if necessary as displayed in cell F3.
Just click the Increase Decimal button in the Number group under the Home tab. Again, you must copy the list and Paste as Values to maintain a static list. McGregor needs to order for his shop.Excel is a great way to organize and keep track of your data. There are more than functions in Excel. If you would like to know more about a function, simply follow the links we added for each of them. Got a different version?
No problem, you can still follow the exact same steps.
### Excel Formula Symbols Cheat Sheet (13 Cool Tips)
By definition, a function is a predefined formula in Excel which does calculations in the order specified by its parameters. Because learning how to add numbers in Excel is one of the most fundamental skills you need to learn. As you know, addition is an integral part of almost any calculation and task in Excel. As their name implies, they add the values in a specified range only when the criteria are met. The differences between the two are in the number of criteria you can specify.
These functions are useful when dealing with large data sets and manual calculations are inefficient and impractical. This function calculates the arithmetic mean of a set of numbers or the sum of the values divided by the number of values. People use the averages every day, from school grades to statistics.
These functions shine when you need to get averages from specifics sets in a range. It may seem rather rudimentary. But in actuality, this function is used in a lot of computations and scenarios. This function is used in many things like counting how many items there are in a listcounting specific casesand others.
But what if you only need to count a specific subset of cells? These functions are useful in tasks like project managementsales inventoryorder fulfillmentand others. This function returns the sum of the product of two or more arrays.
This is an important Excel function since this is used to calculate weighted averages as well as simplify a lot of tasks like sales inventory. Ever had the need to come up with random values between a specified minimum and maximum values?Use this handy Cheat Sheet to discover great functions and tips to help you get the most out of Excel.
Some Excel functions apply to specific subject areas, but others are general and apply to all needs. The following list shows an array of Excel functions used by one and all. Check here for a quickie reference to the purpose of each Excel function. Here is list of Excel functions associated with text, along with a description of what each function does:.
Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. The following is the order in which mathematical operators and syntax are applied both in Excel and in general mathematics. In Excel formulas, you can refer to other cells either relatively or absolutely. When you copy and paste a formula in Excel, how you create the references within the formula tells Excel what to change in the formula it pastes. You can also mix relative and absolute references so that, when you move or copy a formula, the row changes but the column does not, or vice versa.
If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A handful of errors can appear in a cell when a formula or function in Excel cannot be resolved. Knowing their meaning helps correct the problem. He has written numerous articles and books on a variety of technical topics.
Microsoft Excel Tutorial - Beginners Level 1
His latest projects include large-scale cloud-based applications and mobile app development. Cheat Sheet. Excel Order of Operations to Keep in Mind Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. Excel Cell References Worth Remembering In Excel formulas, you can refer to other cells either relatively or absolutely.
Excel Error Messages to Get to Know If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A formula or a function inside a formula cannot find the referenced data NAME? Text in the formula is not recognized NULL!
A space was used in formulas that reference multiple ranges; a comma separates range references NUM! A formula has invalid numeric data for the type of operation REF! The wrong type of operand or function argument is used. | 2,909 | 14,765 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2021-25 | latest | en | 0.843736 | # Common excel math functions Note that further math-related Excel functions are also provided in the Excel Statistical Functions and Excel Engineering Functions categories. The tables below list all the current built-in Excel math functions, grouped by category, to help you to find the function you need. Selecting a function link will take you to a full description of the function, with examples of use and common errors. Note that some of the Excel math functions listed below were introduced in recent versions of Excel, and so are not available in earlier versions. Excel Functions. Basic Numeric Information. Courseworks completed paper date Performs a specified calculation e. Rounding Functions. Rounds a number away from zero i. Rounds a number upregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number upregardless of the sign of the number, to a multiple of significance. New in Excel Rounds a number up to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number towards zeroi. Rounds a number downregardless of the sign of the number, to a multiple of significance New in Excel Rounds a number down, to the nearest integer or to the nearest multiple of significance New in Excel Rounds a number up or downto the nearest multiple of significance. Truncates a number towards zero i. Returns the unit matrix for a specified dimension New in Excel Conditional Sums. Adds the cells in a supplied range, that satisfy multiple criteria New in Excel Returns the sum of the products of corresponding values in two or more supplied arrays. Returns the sum of the difference of squares of corresponding values in two supplied arrays. Returns the sum of the sum of squares of corresponding values in two supplied arrays. Returns the sum of squares of differences of corresponding values in two supplied arrays. Returns the secant of an angle New in Excel Returns the hyperbolic secant of an angle New in Excel Returns the cosecant of an angle New in Excel Cheat Sheet of Excel formulas and function is always a customized worksheet where we can have all those function details, shortcut keys to execute any function or formulas, custom way to use 2 or more function together and guideline to use them. Also, we can choose those formulas which are complicated to apply for users in a cheat sheet. Start Your Free Excel Course. As we can see, here several string functions are listed. Please refer the below screenshot where I have applied the functions on strings and shown their workings. Please refer the below screenshot where I have applied the functions on number values and shown their workings. Please refer the below screenshot where I have applied the functions on values and shown their workings. This has been a guide to Excel Formulas Cheat Sheet. You can also go through our other suggested articles —. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy. Forgot Password? Call Our Course Advisors. Cheat Sheet of Excel Formulas. Popular Course in this category. Course Price View Course. Free Excel Course. Login details for this Free course will be emailed to you. Email ID. Contact No.So, the total production quantity is Similarly, apply the same logic to get the total salary amount. Now we know what overall sum values are. Out of these overall total of employees, we need to find the average salary per employee. Select the range of cells for which we are finding the average value, so our range of cells will be from D2 to D We know the average salary per person; for further drill-down, we want to know what is the average salary based on gender. Marketing plan bibliography design dates So totally there are 10 employees on the list. After counting the total number of employees, we may need to count how many male and female employees are there. MOD function will return the remainder when one number is divided by another number. For example, when you divide number 11 by 2, we will get the remainder as 1 because only till 10 number 2 can divide. When we have fraction or decimal values, we may need to round those decimal values to the nearest integer number. For example, we need to round the number 3. As you can see above, B2 cell value Like this, we can use various mathematical functions in excel to do mathematical operations in excel quickly and easily. This has been a guide to Mathematical Function in Excel. Here we discuss how to calculate mathematical function in excel using Sum, Average, Averageif, Counta, Countif, Mod, and Round formulas along with practical examples. You may learn more about excel from the following articles —. Free Excel Course. Login details for this Free course will be emailed to you. This website or its third-party tools use cookies, which are necessary to its functioning and required to achieve the purposes illustrated in the cookie policy. By closing this banner, scrolling this page, clicking a link or continuing to browse otherwise, you agree to our Privacy Policy.The tutorial provides a list of Excel basic formulas and functions with examples and links to related in-depth tutorials. Being primarily designed as a spreadsheet program, Microsoft Excel is extremely powerful and versatile when it comes to calculating numbers or solving math and engineering problems. It enables you to total or average a column of numbers in the blink of an eye. Apart from that, you can compute a compound interest and weighted average, get the optimal budget for your advertising campaign, minimize the shipment costs or make the optimal work schedule for your employees. All this is done by entering formulas in cells. This tutorial aims to teach you the essentials of Excel functions and show how to use basic formulas in Excel. Before providing the basic Excel formulas list, let's define the key terms just to make sure we are on the same page. So, what do we call an Excel formula and Excel function? Function is a predefined formula already available in Excel. Functions perform specific calculations in a particular order based on the specified values, called arguments, or parameters. You can find all available Excel functions in the Function Library on the Formulas tab:. Of course, it's next to impossible to memorize all of them, and you actually don't need to. The Function Wizard will help you find the function best suited for a particular task, while the Excel Formula Intellisense will prompt the function's syntax and arguments as soon as you type the function's name preceded by an equal sign in a cell:. Clicking the function's name will turn it into a blue hyperlink, which will open the Help topic for that function. What follows below is a list of 10 simple yet really helpful functions that are a necessary skill for everyone who wishes to turn from an Excel novice to an Excel professional. The first Excel function you should be familiar with is the one that performs the basic arithmetic operation of addition:. In the syntax of all Excel functions, an argument enclosed in [square brackets] is optional, other arguments are required. Meaning, your Sum formula should include at least 1 number, reference to a cell or a range of cells. For example:. If necessary, you can perform other calculations within a single formula, for example, add up values in cells B2 through B6, and then divide the sum by To sum with conditions, use the SUMIF function: in the 1st argument, you enter the range of cells to be tested against the criteria A2:A6in the 2nd argument - the criteria itself D2and in the last argument - the cells to sum B2:B6 :. In your Excel worksheets, the formulas may look something similar to this:. Popular book review editing websites for college Its syntax is similar to SUM's:. Sums values in cells B2 through B6, and then divides the result by 5. ## Mathematical Function in Excel And what do you call adding up a group of numbers and then dividing the sum by the count of those numbers? Yep, an average!The basic functions covered below are among the most popular formulas in Excel—the ones everyone should know. To help you learn, we've also provided a spreadsheet with all the formula examples we cover below. After you enter one of these functions in A1, you can then reformat the Date and Time or use the system default. It also subtracts, multiplies, divides, and uses any of the comparison operators to return a result of 1 true or 0 false. Excel frames the column of numbers in green borders and displays the formula in the current cell. Remember your high school math? If the numbers inside the formula are not grouped properly, the answer will be wrong. Notice the screenshot below figure 2. For this exercise, you can enter the same values in H, I, and J, with or without the blank rows in between again, added for easier viewing. Note that as we build each formula, we are combining the steps, eventually, into a single formula. Athletics activities department We start out with three separate formulas. The formula in K3 is wrong. It requires grouping the numbers according to the order of calculation using commas or parentheses. Check your numbers again with your calculator and note that this formula is correct. Note that the syntax the structure or layout of the formula is correct in cells N7 and N8, but incorrect in N6. By combining these formulas into one, you can eliminate columns K and L. The RAND function is really simple and traditionally used for statistical analysis, cryptography, gaming, gambling, and probability theory, among dozens of other things. Note; however, that every time you enter new data and press the Enter key, the list of random numbers you just created changes. If you need to maintain your random numbers lists, you must format the cells as values. Now the list contains values instead of functions, so it will not change. Notice in the formula bar that the random numbers have 15 digits after the decimal Excel defaults to 9which you can change, if necessary as displayed in cell F3. Just click the Increase Decimal button in the Number group under the Home tab. Again, you must copy the list and Paste as Values to maintain a static list. McGregor needs to order for his shop.Excel is a great way to organize and keep track of your data. There are more than functions in Excel. If you would like to know more about a function, simply follow the links we added for each of them. Got a different version? No problem, you can still follow the exact same steps. ### Excel Formula Symbols Cheat Sheet (13 Cool Tips) By definition, a function is a predefined formula in Excel which does calculations in the order specified by its parameters. Because learning how to add numbers in Excel is one of the most fundamental skills you need to learn. As you know, addition is an integral part of almost any calculation and task in Excel. As their name implies, they add the values in a specified range only when the criteria are met. The differences between the two are in the number of criteria you can specify. These functions are useful when dealing with large data sets and manual calculations are inefficient and impractical. This function calculates the arithmetic mean of a set of numbers or the sum of the values divided by the number of values. People use the averages every day, from school grades to statistics. These functions shine when you need to get averages from specifics sets in a range. It may seem rather rudimentary. But in actuality, this function is used in a lot of computations and scenarios. This function is used in many things like counting how many items there are in a listcounting specific casesand others. But what if you only need to count a specific subset of cells? These functions are useful in tasks like project managementsales inventoryorder fulfillmentand others. This function returns the sum of the product of two or more arrays. This is an important Excel function since this is used to calculate weighted averages as well as simplify a lot of tasks like sales inventory. Ever had the need to come up with random values between a specified minimum and maximum values?Use this handy Cheat Sheet to discover great functions and tips to help you get the most out of Excel. Some Excel functions apply to specific subject areas, but others are general and apply to all needs. The following list shows an array of Excel functions used by one and all. Check here for a quickie reference to the purpose of each Excel function. Here is list of Excel functions associated with text, along with a description of what each function does:. Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. The following is the order in which mathematical operators and syntax are applied both in Excel and in general mathematics. In Excel formulas, you can refer to other cells either relatively or absolutely. When you copy and paste a formula in Excel, how you create the references within the formula | tells Excel what to change in the formula it pastes. You can also mix relative and absolute references so that, when you move or copy a formula, the row changes but the column does not, or vice versa. If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A handful of errors can appear in a cell when a formula or function in Excel cannot be resolved. Knowing their meaning helps correct the problem. He has written numerous articles and books on a variety of technical topics. Microsoft Excel Tutorial - Beginners Level 1 His latest projects include large-scale cloud-based applications and mobile app development. Cheat Sheet. Excel Order of Operations to Keep in Mind Mathematics dictates a protocol of how formulas are interpreted, and Excel follows that protocol. Excel Cell References Worth Remembering In Excel formulas, you can refer to other cells either relatively or absolutely. Excel Error Messages to Get to Know If you create a formula in Excel that contains an error or circular reference, Excel lets you know about it with an error message. A formula or a function inside a formula cannot find the referenced data NAME? Text in the formula is not recognized NULL! A space was used in formulas that reference multiple ranges; a comma separates range references NUM! A formula has invalid numeric data for the type of operation REF! The wrong type of operand or function argument is used. |
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