9.3: Solve Money Applications
- Solve coin word problems
- Solve ticket and stamp word problems
Before you get started, take this readiness quiz.
- Multiply: 14(0.25). If you missed this problem, review Example 5.3.5 .
- Simplify: 100(0.2 + 0.05n). If you missed this problem, review Example 7.4.6 .
- Solve: 0.25x + 0.10(x + 4) = 2.5 If you missed this problem, review Example 8.6.8 .
Solve Coin Word Problems
Imagine taking a handful of coins from your pocket or purse and placing them on your desk. How would you determine the value of that pile of coins?
If you can form a step-by-step plan for finding the total value of the coins, it will help you as you begin solving coin word problems.
One way to bring some order to the mess of coins would be to separate the coins into stacks according to their value. Quarters would go with quarters, dimes with dimes, nickels with nickels, and so on. To get the total value of all the coins, you would add the total value of each pile.
Figure \(\PageIndex{1}\) - To determine the total value of a stack of nickels, multiply the number of nickels times the value of one nickel.(Credit: Darren Hester via ppdigital)
How would you determine the value of each pile? Think about the dime pile—how much is it worth? If you count the number of dimes, you'll know how many you have—the number of dimes.
But this does not tell you the value of all the dimes. Say you counted 17 dimes, how much are they worth? Each dime is worth $0.10 —that is the value of one dime. To find the total value of the pile of 17 dimes, multiply 17 by $0.10 to get $1.70. This is the total value of all 17 dimes.
\[\begin{split} 17 \cdot \$0.10 &= \$ 1.70 \\ number\; \cdot value &= total\; value \end{split}\]
For coins of the same type, the total value can be found as follows:
\[number\; \cdot value = total\; value\]
where number is the number of coins, value is the value of each coin, and total value is the total value of all the coins.
You could continue this process for each type of coin, and then you would know the total value of each type of coin. To get the total value of all the coins, add the total value of each type of coin.
Let's look at a specific case. Suppose there are 14 quarters, 17 dimes, 21 nickels, and 39 pennies. We'll make a table to organize the information – the type of coin, the number of each, and the value.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | 14 | 0.25 | 3.50 |
| Dimes | 17 | 0.10 | 1.70 |
| Nickels | 21 | 0.05 | 1.05 |
| Pennies | 39 | 0.01 | 0.39 |
| 6.64 |
The total value of all the coins is $6.64. Notice how Table \(\PageIndex{1}\) helped us organize all the information. Let's see how this method is used to solve a coin word problem.
Adalberto has $2.25 in dimes and nickels in his pocket. He has nine more nickels than dimes. How many of each type of coin does he have?
Solution
Step 1. Read the problem. Make sure you understand all the words and ideas.
- Determine the types of coins involved.
Think about the strategy we used to find the value of the handful of coins. The first thing you need is to notice what types of coins are involved. Adalberto has dimes and nickels.
-
Create a table to organize the information.
- Label the columns ‘type’, ‘number’, ‘value’, ‘total value’.
- List the types of coins.
- Write in the value of each type of coin.
- Write in the total value of all the coins.
We can work this problem all in cents or in dollars. Here we will do it in dollars and put in the dollar sign ($) in the table as a reminder.
The value of a dime is $0.10 and the value of a nickel is $0.05. The total value of all the coins is $2.25.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Dimes | 0.10 | ||
| Nickels | 0.05 | ||
| 2.25 |
Step 2. Identify what you are looking for.
- We are asked to find the number of dimes and nickels Adalberto has.
Step 3. Name what you are looking for.
- Use variable expressions to represent the number of each type of coin.
-
Multiply the number times the value to get the total value of each type of coin. In this problem you cannot count each type of coin—that is what you are looking for—but you have a clue. There are nine more nickels than dimes. The number of nickels is nine more than the number of dimes.
- Let d = number of dimes.
- d + 9 = number of nickels
- Fill in the “number” column to help get everything organized.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Dimes | d | 0.10 | |
| Nickels | d + 9 | 0.05 | |
| 2.25 |
Now we have all the information we need from the problem!
You multiply the number times the value to get the total value of each type of coin. While you do not know the actual number, you do have an expression to represent it.
And so now multiply number • value and write the results in the Total Value column.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Dimes | d | 0.10 | 0.10d |
| Nickels | d + 9 | 0.05 | 0.05(d + 9) |
| 2.25 |
Step 4. Translate into an equation. Restate the problem in one sentence. Then translate into an equation.
Step 5. Solve the equation using good algebra techniques.
| Write the equation. | $$0.10d + 0.05(d + 9) = 2.25$$ |
| Distribute. | $$0.10d + 0.05d + 0.45 = 2.25$$ |
| Combine like terms. | $$0.15d + 0.45 = 2.25$$ |
| Subtract 0.45 from each side. | $$0.15d = 1.80$$ |
| Divide to find the number of dimes. | $$d = 12$$ |
| The number of nickels is d + 9. | $$\begin{split} d + 9& \\ \textcolor{red}{12} + 9& \\ 21& \end{split}$$ |
Step 6. Check .
\[\begin{split} 12\; dimes:\; 12(0.10) &= 1.20 \\ 21\; nickels:\; 21(0.05) &= 1.05 \\ \hline &\quad \$ 2.25\; \checkmark \end{split}\]
Step 7. Answer the question.
Adalberto has twelve dimes and twenty-one nickels.
If this were a homework exercise, our work might look like this:
Check:
\[\begin{split} 12\; dimes \quad 12(0.10) &= 1.20 \\ 21\; nickels \quad 21(0.05) &= 1.05 \\ \hline &\quad \$ 2.25 \end{split}\]
Michaela has $2.05 in dimes and nickels in her change purse. She has seven more dimes than nickels. How many coins of each type does she have?
- Answer
-
9 nickels, 16 dimes
Liliana has $2.10 in nickels and quarters in her backpack. She has 12 more nickels than quarters. How many coins of each type does she have?
- Answer
-
17 nickels, 5 quarters
Step 1. Read the problem. Make sure you understand all the words and ideas, and create a table to organize the information.
Step 2. Identify what you are looking for.
Step 3. Name what you are looking for. Choose a variable to represent that quantity.
- Use variable expressions to represent the number of each type of coin and write them in the table.
- Multiply the number times the value to get the total value of each type of coin.
Step 4. Translate into an equation. Write the equation by adding the total values of all the types of coins.
Step 5. Solve the equation using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.
You may find it helpful to put all the numbers into the table to make sure they check.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
Maria has $2.43 in quarters and pennies in her wallet. She has twice as many pennies as quarters. How many coins of each type does she have?
Solution
Step 1. Read the problem.
- Determine the types of coins involved. We know that Maria has quarters and pennies.
-
Create a table to organize the information.
- Label the columns type, number, value, total value.
- List the types of coins.
- Write in the value of each type of coin.
- Write in the total value of all the coins.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | 0.25 | ||
| Pennies | 0.01 | ||
| 2.43 |
Step 2. Identify what you are looking for.
- We are looking for the number of quarters and pennies.
Step 3. Name: Represent the number of quarters and pennies using variables.
-
We know Maria has twice as many pennies as quarters. The number of pennies is defined in terms of quarters.
- Let q represent the number of quarters. Then the number of pennies is 2q.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | q | 0.25 | |
| Pennies | 2q | 0.01 | |
| 2.43 |
Multiply the ‘number’ and the ‘value’ to get the ‘total value’ of each type of coin.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Quarters | q | 0.25 | 0.25q |
| Pennies | 2q | 0.01 | 0.01(2q) |
| 2.43 |
Step 4. Translate . Write the equation by adding the 'total value’ of all the types of coins.
Step 5. Solve the equation.
| Write the equation. | $$0.25q + 0.01(2q) = 2.43$$ |
| Multiply. | $$0.25q + 0.02q = 2.43$$ |
| Combine like terms. | $$0.27q = 2.43$$ |
| Divide by 0.27. | $$q = 9\; quarters$$ |
| The number of pennies is 2q. | $$\begin{split} &2q \\ &2\; \cdot\; \textcolor{red}{9} \\ &18\; pennies \end{split}$$ |
Step 6. Check the answer in the problem. Maria has 9 quarters and 18 pennies. Does this make $2.43?
\[\begin{split} 9\; quarters \quad 9(0.25) &= 2.25 \\ 18\; pennies \quad 18(0.01) &= 0.18 \\ \hline Total \qquad \qquad \qquad \quad &\quad \$ 2.43\; \checkmark \end{split}\]
Step 7. Answer the question. Maria has nine quarters and eighteen pennies.
Sumanta has $4.20 in nickels and dimes in her desk drawer. She has twice as many nickels as dimes. How many coins of each type does she have?
- Answer
-
42 nickels, 21 dimes
Alison has three times as many dimes as quarters in her purse. She has $9.35 altogether. How many coins of each type does she have?
- Answer
-
51 dimes, 17 quarters
In the next example, we'll show only the completed table—make sure you understand how to fill it in step by step.
Danny has $2.14 worth of pennies and nickels in his piggy bank. The number of nickels is two more than ten times the number of pennies. How many nickels and how many pennies does Danny have?
Solution
| Step 1: Read the problem. | |
| Determine the types of coins involved. Create a table. | Pennies and nickels |
| Write in the value of each type of coin. |
Pennies are worth $0.01. Nickels are worth $0.05. |
| Step 2: Identify what you are looking for. | the number of pennies and nickels |
| Step 3: Name . Represent the number of each type of coin using variables. The number of nickels is defined in terms of the number of pennies, so start with pennies. | Let p = number of pennies |
| The number of nickels is two more than then times the number of pennies. | 10p + 2 = number of nickels |
Multiply the number and the value to get the total value of each type of coin.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| pennies | p | 0.01 | 0.01p |
| nickels | 10p + 2 | 0.05 | 0.05(10p + 2) |
| $2.14 |
Step 4. Translate : Write the equation by adding the total value of all the types of coins.
Step 5. Solve the equation.
| $$0.01p + 0.50p + 0.10 = 2.14$$ | |
| $$0.51p + 0.10 = 2.14$$ | |
| $$0.51p = 2.04$$ | |
| $$p = 4\; pennies$$ | |
| How many nickels? | $$10p + 2$$ |
| $$10(\textcolor{red}{4}) + 2$$ | |
| $$42\; nickels$$ |
Step 6. Check . Is the total value of 4 pennies and 42 nickels equal to $2.14?
\[\begin{split} 4(0.01) + 42(0.05) &\stackrel{?}{=} 2.14 \\ 2.14 &= 2.14\; \checkmark \end{split}\]
Step 7. Answer the question. Danny has 4 pennies and 42 nickels.
Jesse has $6.55 worth of quarters and nickels in his pocket. The number of nickels is five more than two times the number of quarters. How many nickels and how many quarters does Jesse have?
- Answer
-
41 nickels, 18 quarters
Elaine has $7.00 in dimes and nickels in her coin jar. The number of dimes that Elaine has is seven less than three times the number of nickels. How many of each coin does Elaine have?
- Answer
-
22 nickels, 59 dimes
Solve Ticket and Stamp Word Problems
The strategies we used for coin problems can be easily applied to some other kinds of problems too. Problems involving tickets or stamps are very similar to coin problems, for example. Like coins, tickets and stamps have different values; so we can organize the information in tables much like we did for coin problems.
At a school concert, the total value of tickets sold was $1,506. Student tickets sold for $6 each and adult tickets sold for $9 each. The number of adult tickets sold was 5 less than three times the number of student tickets sold. How many student tickets and how many adult tickets were sold?
Solution
Step 1: Read the problem.
- Determine the types of tickets involved. There are student tickets and adult tickets.
- Create a table to organize the information.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Student | 6 | ||
| Adult | 9 | ||
| 1,506 |
Step 2. Identify what you are looking for. We are looking for the number of student and adult tickets.
Step 3. Name . Represent the number of each type of ticket using variables.
- We know the number of adult tickets sold was 5 less than three times the number of student tickets sold. Let s be the number of student tickets.
- Then 3s − 5 is the number of adult tickets.
- Multiply the number times the value to get the total value of each type of ticket.
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| Student | s | 6 | 6s |
| Adult | 3s - 5 | 9 | 9(9s - 5) |
| 1,506 |
Step 4. Translate : Write the equation by adding the total values of each type of ticket.
\[6s + 9(3s − 5) = 1506\]
Step 5. Solve the equation.
\[\begin{split} 6s + 27s − 45 &= 1506 \\ 33s − 45 &= 1506 \\ 33s &= 1551 \\ s &= 47\; students \end{split}\]
Substitute to find the number of adults.
\[\begin{split} 3s - 5 &= number\; of\; adults \\ 3(\textcolor{red}{47}) - 5 &= 136\; adults \end{split}\]
Step 6. Check . There were 47 student tickets at $6 each and 136 adult tickets at $9 each. Is the total value $1506? We find the total value of each type of ticket by multiplying the number of tickets times its value; we then add to get the total value of all the tickets sold.
\[\begin{split} 47 \cdot 6 &= 282 \\ 136 \cdot 9 &= 1224 \\ \hline &\quad 1506 \end{split}\]
Step 7. Answer the question. They sold 47 student tickets and 136 adult tickets.
The first day of a water polo tournament, the total value of tickets sold was $17,610. One-day passes sold for $20 and tournament passes sold for $30. The number of tournament passes sold was 37 more than the number of day passes sold. How many day passes and how many tournament passes were sold?
- Answer
-
330 day passes, 367 tournament passes
At the movie theater, the total value of tickets sold was $2,612.50. Adult tickets sold for $10 each and senior/ child tickets sold for $7.50 each. The number of senior/child tickets sold was 25 less than twice the number of adult tickets sold. How many senior/child tickets and how many adult tickets were sold?
- Answer
-
112 adult tickets, 199 senior/child tickets
Now we'll do one where we fill in the table all at once.
Monica paid $10.44 for stamps she needed to mail the invitations to her sister's baby shower. The number of 49-cent stamps was four more than twice the number of 8-cent stamps. How many 49-cent stamps and how many 8-cent stamps did Monica buy?
Solution
The type of stamps are 49-cent stamps and 8-cent stamps. Their names also give the value. “The number of 49 cent stamps was four more than twice the number of 8 cent stamps.”
Let x = number of 8-cent stamps
2x + 4 = number of 49-cent stamps
| Type | Number | Value ($) | Total Value ($) |
|---|---|---|---|
| 49-cent stamps | 2x + 4 | 0.49 | 0.49(2x + 4) |
| 8-cent stamps | x | 0.08 | 0.08x |
| 10.44 |
| Write the equation from the total values. | $$0.49(2x + 4) + 0.08x = 10.44$$ |
| Solve the equation. | $$\begin{split} 0.98x + 1.96 + 0.08x &= 10.44 \\ 1.06x + 1.96 &= 10.44 \\ 1.06x &= 8.48 \\ x &= 8 \end{split}$$ |
| Monica bought 8 eight-cent stamps. | |
| Find the number of 49-cent stamps she bought by evaluating. | 2x + 4 for x = 8.$$\begin{split} &2x + 4 \\ &2 \cdot 8 + 4 \\ &16 + 4 \\ &20 \end{split}$$ |
| Check. | $$\begin{split} 8(0.08) + 20(0.49) &\stackrel{?}{=} 10.44 \\ 0.64 + 9.80 &\stackrel{?}{=} 10.44 \\ 10.44 &= 10.44\; \checkmark \end{split}$$ |
Monica bought eight 8-cent stamps and twenty 49-cent stamps.
Eric paid $16.64 for stamps so he could mail thank you notes for his wedding gifts. The number of 49-cent stamps was eight more than twice the number of 8-cent stamps. How many 49-cent stamps and how many 8-cent stamps did Eric buy?
- Answer
-
32 at 49 cents, 12 at 8 cents
Kailee paid $14.84 for stamps. The number of 49-cent stamps was four less than three times the number of 21-cent stamps. How many 49-cent stamps and how many 21-cent stamps did Kailee buy?
- Answer
-
26 at 49 cents, 10 at 21 cents
Practice Makes Perfect
Solve Coin Word Problems
In the following exercises, solve the coin word problems.
- Jaime has $2.60 in dimes and nickels. The number of dimes is 14 more than the number of nickels. How many of each coin does he have?
- Lee has $1.75 in dimes and nickels. The number of nickels is 11 more than the number of dimes. How many of each coin does he have?
- Ngo has a collection of dimes and quarters with a total value of $3.50. The number of dimes is 7 more than the number of quarters. How many of each coin does he have?
- Connor has a collection of dimes and quarters with a total value of $6.30. The number of dimes is 14 more than the number of quarters. How many of each coin does he have?
- Carolyn has $2.55 in her purse in nickels and dimes. The number of nickels is 9 less than three times the number of dimes. Find the number of each type of coin.
- Julio has $2.75 in his pocket in nickels and dimes. The number of dimes is 10 less than twice the number of nickels. Find the number of each type of coin.
- Chi has $11.30 in dimes and quarters. The number of dimes is 3 more than three times the number of quarters. How many dimes and nickels does Chi have?
- Tyler has $9.70 in dimes and quarters. The number of quarters is 8 more than four times the number of dimes. How many of each coin does he have?
- A cash box of $1 and $5 bills is worth $45. The number of $1 bills is 3 more than the number of $5 bills. How many of each bill does it contain?
- Joe's wallet contains $1 and $5 bills worth $47. The number of $1 bills is 5 more than the number of $5 bills. How many of each bill does he have?
- In a cash drawer there is $125 in $5 and $10 bills. The number of $10 bills is twice the number of $5 bills. How many of each are in the drawer?
- John has $175 in $5 and $10 bills in his drawer. The number of $5 bills is three times the number of $10 bills. How many of each are in the drawer?
- Mukul has $3.75 in quarters, dimes and nickels in his pocket. He has five more dimes than quarters and nine more nickels than quarters. How many of each coin are in his pocket?
- Vina has $4.70 in quarters, dimes and nickels in her purse. She has eight more dimes than quarters and six more nickels than quarters. How many of each coin are in her purse?
Solve Ticket and Stamp Word Problems
In the following exercises, solve the ticket and stamp word problems.
- The play took in $550 one night. The number of $8 adult tickets was 10 less than twice the number of $5 child tickets. How many of each ticket were sold?
- If the number of $8 child tickets is seventeen less than three times the number of $12 adult tickets and the theater took in $584, how many of each ticket were sold?
- The movie theater took in $1,220 one Monday night. The number of $7 child tickets was ten more than twice the number of $9 adult tickets. How many of each were sold?
- The ball game took in $1,340 one Saturday. The number of $12 adult tickets was 15 more than twice the number of $5 child tickets. How many of each were sold?
- Julie went to the post office and bought both $0.49 stamps and $0.34 postcards for her office's bills She spent $62.60. The number of stamps was 20 more than twice the number of postcards. How many of each did she buy?
- Before he left for college out of state, Jason went to the post office and bought both $0.49 stamps and $0.34 postcards and spent $12.52. The number of stamps was 4 more than twice the number of postcards. How many of each did he buy?
- Maria spent $16.80 at the post office. She bought three times as many $0.49 stamps as $0.21 stamps. How many of each did she buy?
- Hector spent $43.40 at the post office. He bought four times as many $0.49 stamps as $0.21 stamps. How many of each did he buy?
- Hilda has $210 worth of $10 and $12 stock shares. The numbers of $10 shares is 5 more than twice the number of $12 shares. How many of each does she have?
- Mario invested $475 in $45 and $25 stock shares. The number of $25 shares was 5 less than three times the number of $45 shares. How many of each type of share did he buy?
Everyday Math
- Parent Volunteer As the treasurer of her daughter's Girl Scout troop, Laney collected money for some girls and adults to go to a 3-day camp. Each girl paid $75 and each adult paid $30. The total amount of money collected for camp was $765. If the number of girls is three times the number of adults, how many girls and how many adults paid for camp?
- Parent Volunteer Laurie was completing the treasurer's report for her son's Boy Scout troop at the end of the school year. She didn't remember how many boys had paid the $24 full-year registration fee and how many had paid a $16 partial-year fee. She knew that the number of boys who paid for a full-year was ten more than the number who paid for a partial-year. If $400 was collected for all the registrations, how many boys had paid the full-year fee and how many had paid the partial-year fee?
Writing Exercises
- Suppose you have 6 quarters, 9 dimes, and 4 pennies. Explain how you find the total value of all the coins.
- Do you find it helpful to use a table when solving coin problems? Why or why not?
- In the table used to solve coin problems, one column is labeled “number” and another column is labeled ‘“value.” What is the difference between the number and the value?
- What similarities and differences did you see between solving the coin problems and the ticket and stamp problems?
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) After reviewing this checklist, what will you do to become confident for all objectives?
Contributors and Attributions
-
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


2.14: What’s Wrong with Approval Voting?
Approval voting can very easily violate the Majority Criterion.
Consider the voting schedule:
\(\begin{array}{|l|l|l|l|}
\hline & 80 & 15 & 5 \\
\hline 1^{\text {st }} \text { choice } & \text { A } & \text { B } & \text { C } \\
\hline 2^{\text {nd }} \text { choice } & \text { B } & \text { C } & \text { B } \\
\hline 3^{\text {rd }} \text { choice } & \text { C } & \text { A } & \text { A } \\
\hline
\end{array}\)
Solution
Clearly A is the majority winner. Now suppose that this election was held using Approval Voting, and every voter marked approval of their top two candidates.
- A would receive approval from 80 voters
- B would receive approval from 100 voters
- C would receive approval from 20 voters
B would be the winner. Some argue that Approval Voting tends to vote the least disliked choice, rather than the most liked candidate.
Additionally, Approval Voting is susceptible to strategic insincere voting, in which a voter does not vote their true preference to try to increase the chances of their choice winning. For example, in the movie example above, suppose Bob and Alice would much rather watch Scream. They remove The Matrix from their approval list, resulting in a different result.
\(\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|}
\hline & \text { Bob } & \text { Ann } & \text { Marv } & \text { Alice } & \text { Eve } & \text { Omar } & \text { Lupe } & \text { Dave } & \text { Tish } & \text { Jim } \\
\hline \text { Titanic } & & \mathrm{X} & \mathrm{X} & & & \mathrm{X} & & \mathrm{X} & & \mathrm{X} \\
\hline \text { Scream } & \mathrm{X} & & \mathrm{X} & \mathrm{X} & & \mathrm{X} & \mathrm{X} & & \mathrm{X} & \\
\hline \text { The Matrix } & & \mathrm{X} & \mathrm{X} & & \mathrm{X} & & \mathrm{X} & & & \mathrm{X} \\
\hline
\end{array}\)
Totaling the results, we find Titanic received 5 approvals, Scream received 6 approvals, and The Matrix received 5 approvals. By voting insincerely, Bob and Alice were able to sway the result in favor of their preference.


21.1: Introduction
Chapter Objectives
After studying this chapter, you will be able to:
- Identify the components and anatomy of the lymphatic system
- Discuss the role of the innate immune response against pathogens
- Describe the power of the adaptive immune response to cure disease
- Explain immunological deficiencies and over-reactions of the immune system
- Discuss the role of the immune response in transplantation and cancer
- Describe the interaction of the immune and lymphatic systems with other body systems
In June 1981, the Centers for Disease Control and Prevention (CDC), in Atlanta, Georgia, published a report of an unusual cluster of five patients in Los Angeles, California. All five were diagnosed with a rare pneumonia caused by a fungus called Pneumocystis jirovecii (formerly known as Pneumocystis carinii ).
Why was this unusual? Although commonly found in the lungs of healthy individuals, this fungus is an opportunistic pathogen that causes disease in individuals with suppressed or underdeveloped immune systems. The very young, whose immune systems have yet to mature, and the elderly, whose immune systems have declined with age, are particularly susceptible. The five patients from LA, though, were between 29 and 36 years of age and should have been in the prime of their lives, immunologically speaking. What could be going on?
A few days later, a cluster of eight cases was reported in New York City, also involving young patients, this time exhibiting a rare form of skin cancer known as Kaposi’s sarcoma. This cancer of the cells that line the blood and lymphatic vessels was previously observed as a relatively innocuous disease of the elderly. The disease that doctors saw in 1981 was frighteningly more severe, with multiple, fast-growing lesions that spread to all parts of the body, including the trunk and face. Could the immune systems of these young patients have been compromised in some way? Indeed, when they were tested, they exhibited extremely low numbers of a specific type of white blood cell in their bloodstreams, indicating that they had somehow lost a major part of the immune system.
Acquired immune deficiency syndrome, or AIDS, turned out to be a new disease caused by the previously unknown human immunodeficiency virus (HIV). Although nearly 100 percent fatal in those with active HIV infections in the early years, the development of anti-HIV drugs has transformed HIV infection into a chronic, manageable disease and not the certain death sentence it once was. One positive outcome resulting from the emergence of HIV disease was that the public’s attention became focused as never before on the importance of having a functional and healthy immune system.


8.10: Practice
Harriet is part of a group of five children who share four pies. Jeff is part of a group of seven children who share four pies. Jean is part of a group of seven children who share six pies.
- Who gets more pie, Harriet or Jeff? Justify your answer!
- Who gets more pie, Jeff or Jean? Justify your answer!
- Who gets more pie, Harriet or Jean? Justify your answer!
Yesterday was Zoe’s birthday, and she had a big rectangular cake. Today, \(\frac{2}{5}\) of the cake is left. The leftover cake is shown here.
Draw a picture of the original (whole) cake and explain your work.
Use benchmarks and intuitive methods to arrange the fractions below in ascending order. Explain how you decided. (The point of this problem is to think more and compute less!): \[\frac{2}{5}, \qquad \frac{1}{3}, \qquad \frac{5}{8}, \qquad \frac{1}{4}, \qquad \frac{2}{3}, \qquad \frac{3}{4}, \qquad \frac{4}{7} \ldotp \nonumber \]
Which of these fractions has the larger value? Justify your choice. \[\frac{10001}{10002} \qquad or \qquad \frac{10000001}{10000002} \ldotp \nonumber \]
Solve each division problem. Look for a shortcut, and explain your work.
\[\frac{251+251+251+251}{4} \nonumber \]
\[\frac{377+377+377+377+377}{5} \nonumber \]
\[\frac{123123+123123+123123+123123+123123+123123}{3} \nonumber \]
Yoko says
\[\frac{16}{64} = \frac{1}{4} \nonumber \]
because she cancels the sixes:
\[\frac{16}{64} = \frac{1}{4} \ldotp \nonumber \]
But note:
\[\frac{16}{64} = \frac{1 \cdot 16}{4 \cdot 16} = \frac{1}{4} \nonumber \]
So is Yoko right? Does her cancelation rule always work? If it does not always work, can you find any other example where it works? Can you find every example where it works?
Jimmy says that a fraction does not change in value if you add the same amount to the numerator and the denominator. Is he right? If you were Jimmy’s teacher, how would you respond?
- Shelly says that if \(ab < cd\) then \(\frac{a}{b} < \frac{c}{d}\). Is Shelly’s claim always true, sometimes true, or never true? If you were Shelly’s teacher, what would you say to her?
- Rob says that if \(ad < bc\) then \(\frac{a}{b} < \frac{c}{d}\). Is Rob’s claim always true, sometimes true, or never true? If you were Rob’s teacher, what would you say to him?
Jill, her brother, and another partner own a pizza restaurant. If Jill owns \(\frac{1}{3}\) of the restaurant and her brother owns \(\frac{1}{4}\) of the restaurant, what fraction does the third partner own?
John spent a quarter of his life as a boy growing up, one-sixth of his life in college, and one-half of his life as a teacher. He spent his last six years in retirement. How old was he when he died?
Nana was planning to make a red, white, and blue quilt. One-third was to be red and two-fifths was to be white. If the area of the quilt was to be 30 square feet, how many square feet would be blue? [1]
Ku’u Hae Aloha (My Beloved Flag), Hawaiian cotton quilt from Waimea, before 1918, Honolulu Academy of Arts.
Rafael ate one-fourth of a pizza and Rocco ate one-third of it. What fraction of the pizza did they eat?
Tangrams [2] are a seven-piece puzzle, and the seven pieces can be assembled into a big square.
- If the large square shown above is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
- The tangram puzzle contains a small square. If the small square (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
- The tangram set contains two large triangles. If a large triangle (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
- The tangram set contains one medium triangle. If the medium triangle (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers.
- The tangram set contains two small triangles. If a small triangle (the single tangram piece) is one whole, assign a fraction value to each of the seven tangram pieces. Justify your answers
Mikiko said her family made two square pizzas at home. One of the pizzas was 8 inches on each side, and the other was 12 inches on each side. Mikiko ate \(\frac{1}{4}\) of the small pizza and \(\frac{1}{12}\) of the large pizza. So she said that she ate
\[\frac{1}{4} + \frac{1}{12} = \frac{3}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3} \nonumber \]
of a pizza. Do you agree with Mikiko’s calculation? Did she eat \(\frac{1}{3}\) of a whole pizza? Carefully justify your answer. (This question is tricky. It’s probably a good idea to draw a picture!)
Look at the triangle of numbers. There are lots of patterns here! Find as many as you can. In particular, try to answer these questions:
- What pattern describes the first number in each row?
- How is each fraction related to the two fractions below it?
- Can you write down the next two rows of the triangle?
Marie made a sheet cake at home, but she saved some to bring to work and share with her co-workers the next day. Answer these questions about Marie’s cake. (Draw a picture!)
- Suppose Marie saved \(\frac{1}{2}\) of the cake for her co-workers and the co-workers ate \(\frac{3}{4}\) of this. What fraction of the entire cake did they eat?
- What if Marie saved \(\frac{1}{6}\) instead, and they ate \(\frac{2}{3}\) of this?
- What if she saved \(\frac{5}{7}\) of the cake and they ate \(\frac{1}{2}\) of this?
An elementary school held a “Family Math Night” event, and 405 students showed up. Two-thirds of the students who showed up won a door prize. How many students won prizes?
For each picture shown:
- What multiplication problem is represented?
- What is the product?
For each problem, use only the digits 0, 1, 2,. . . , 9 at most once each in place of the variables. Find the value closest to 1. Note that can be a different value in each of the three problems. Justify your answer: How do you know it is the closest to 1?
- $\frac{a}{b} \ldotp$
- $\frac{a}{b} \cdot \frac{c}{d} \ldotp$
- $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f} \ldotp$
A town plans to build a community garden that will cover \(\frac{2}{3}\) of a square mile on an old farm. One side of the garden area will be along an existing fence that is \(\frac{3}{4}\) of a mile long. If the garden is a rectangle, how long is the other side?
Nate used \(90 \frac{1}{2}\) pounds of seed to plant \(\frac{1}{4}\) acres of wheat. How many pounds of seed did he use per acre?
The family-sized box of laundry detergent contains 35 cups of detergent. Your family’s machine requires \(1 \frac{1}{4}\) cup per load. How many loads of laundry can your family do with one box of detergent?
Jessica bikes to campus every day. When she is one-third of the way between her home and campus, she passes a grocery store. When she is halfway to school, she passes a Subway sandwich shop. This morning, Jessica passed the grocery store at 8:30am, and she passed Subway at 8:35am. What time did she get to campus?
If you place a full container of flour on a balance scale and place on the other side a \(\frac{1}{3}\) pound weight plus a container of flour (the same size) that is \(\frac{3}{4}\) full, then the scale balances. How much does the full container of flour weigh?
Geoff spent \(\frac{3}{4}\) of his allowance on a movie. He spent \(\frac{11}{18}\) of what was left on snacks at school. He also spent $3 on a magazine, and that left him with \(\frac{1}{24}\) of his total allowance, which he put into his savings account. How much money did Geoff save that week?
Lily was flying to San Francisco from Honolulu. Halfway there, she fell asleep. When she woke up, the distance remaining was half the distance traveled while she slept. For what fraction of the trip was Lily asleep?
- Image used under Creative Commons CC0 1.0 Universal Public Domain Dedication. ↵
- Tangram image from Wikimedia Commons, public domain. ↵


8.2: Common Resources
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The Tragedy of the Commons
The tragedy of the commons is the overexploitation of a common good by individual, rational actors.
learning Objectives
- Describe the tragedy of the commons
Common Goods
Common goods are goods that are rivalrous and non-excludable. This means that anyone has access to the good, but that the use of the good by one person reduces the ability of someone else to use it. A classic example of a common good are fish stocks in international waters; no one is excluded from fishing, but as people withdraw fish without limits being imposed, the stocks for later fishermen are potentially depleted.
Tragedy of Commons
The tragedy of the commons is the depletion of a common good by individuals who are acting independently and rationally according to each one’s self-interest. Consider, the example of fish in international waters. Each individual fisherman, acting independently, will rationally choose to catch some of the fish to sell. This makes sense: there is a resource that the fisherman is able to use to generate a profit. However, when a lot of fishermen, all thinking this way, catch the fish, the total stock of fish may be depleted. When the stock of fish is depleted, none of the fishermen are able to continue fishing, even though, in the long run, each fisherman would have preferred that the fish not be depleted. The tragedy of the commons describes such situations in which people withdraw resources to secure short-term gains without regard for the long-term consequences.
Not all common goods, however, suffer from the tragedy of the commons. If individuals have enlightened self-interest, they will realize the negative long-term effects of their short-term decisions. This would be the same as the fishermen realizing that they should limit their fishing to preserve the stock of fish in the long-term.
In the absence of enlightened self-interest, the government may step in and impose regulations or taxes to discourage the behavior that leads to the tragedy of the commons. This would be like the government imposing limits on the amount of fish that can be caught.
Bluefin Tuna Caught in Net : Fish populations are at risk of becoming fully extinct due to overfishing. The Food and Agriculture Association estimated 70% of the world’s fish species are either fully exploited or depleted.
The Free-Rider Problem
The free-rider problem is when individuals benefit from a public good without paying their share of the cost.
learning objectives
- Describe the Free-Rider Problem
It is easy to think about public goods as free. In your everyday life, you benefit from public goods such as roads and bridges even though no transaction occurs when you use them. However, even public goods need to be paid for. In the case of roads and bridges, everyone pays taxes to the government, who then uses the taxes to pay for public goods.
Roads : Free riders are able to use roads without paying their taxes because roads are a non-excludable public good.
Public goods, as you may recall, are both non-rivalrous and non-excludable. It is the second trait- the non-excludability- that leads to what is called the free-rider problem. The free-rider problem is that some people may benefit from a public good without paying their share of the cost.
Since public goods are non-excludable, free-riders not only can’t be prevented from using the good, but actually have an incentive to continue to free-ride. If they will be able to use the public good whether they pay their share of the costs, they might as well not pay.
Take the military, for example. National security is a public good: it is both non-rivalrous and non-excludable. In order to have such a public good, everyone pays taxes which are then used by the government to finance the military. However, there are undoubtedly people who have not paid their taxes. These people, without having paid their share of the cost of having a military, still benefit from the protection the military provides. They are free-riders.
Of course, there are commonly regulations that attempt to discourage free-riding. For government-provided public goods, the government makes sure that everyone pays their share of the costs by enforcing tax laws. The threat of fines or jail time are enough of a threat that most people find it more appealing (in the US, at least) to pay their share of public goods via taxes than to free-ride.
Key Points
- Common goods are non-excludable and rivalrous.
- When individuals act independently and rationally, they may collectively trade long-term benefit for short-term gain.
- Enlightened self-interest and government intervention are two ways that the tragedy of the commons may be avoided.
- Public goods are non-excludable, but have a cost, so those who don’t pay their share of the cost can still easily benefit from the good.
- Free-riders have an incentive to free ride because they can benefit from a good at a reduced personal cost.
- The providers of public goods often create enforcement mechanisms to mitigate the free-rider problem.
Key Terms
- Common good : Goods which are rivalrous and non-excludable.
- Enlightened Self-Interest : The ability for individuals to realize when their actions, collectively, will trade long-term benefit for short-term gain.
- public good : A good that is non-rivalrous and non-excludable.
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5.38: The Weibull Distribution
In this section, we will study a two-parameter family of distributions that has special importance in reliability.
The Basic Weibull Distribution
Distribution Functions
The basic Weibull distribution with shape parameter \( k \in (0, \infty) \) is a continuous distribution on \( [0, \infty) \) with distribution function \( G \) given by \[ G(t) = 1 - \exp\left(-t^k\right), \quad t \in [0, \infty) \] The special case \( k = 1 \) gives the standard Weibull distribution .
Proof
Clearly \( G \) is continuous and increasing on \( [0, \infty) \) with \( G(0) = 0 \) and \( G(t) \to 1 \) as \( t \to \infty \).
The Weibull distribution is named for Waloddi Weibull. Weibull was not the first person to use the distribution, but was the first to study it extensively and recognize its wide use in applications. The standard Weibull distribution is the same as the standard exponential distribution. But as we will see, every Weibull random variable can be obtained from a standard Weibull variable by a simple deterministic transformation, so the terminology is justified.
The probability density function \( g \) is given by \[ g(t) = k t^{k - 1} \exp\left(-t^k\right), \quad t \in (0, \infty) \]
- If \(0 \lt k \lt 1\), \(g\) is decreasing and concave upward with \( g(t) \to \infty \) as \( t \downarrow 0 \).
- If \( k = 1 \), \( g \) is decreasing and concave upward with mode \( t = 0 \).
- If \(k \gt 1\), \(g\) increases and then decreases, with mode \(t = \left( \frac{k - 1}{k} \right)^{1/k}\).
- If \( 1 \lt k \le 2 \), \( g \) is concave downward and then upward, with inflection point at \( t = \left[\frac{3 (k - 1) + \sqrt{(5 k - 1)(k - 1)}}{2 k}\right]^{1/k} \)
- If \( k \gt 2 \), \( g \) is concave upward, then downward, then upward again, with inflection points at \( t = \left[\frac{3 (k - 1) \pm \sqrt{(5 k - 1)(k - 1)}}{2 k}\right]^{1/k} \)
Proof
These results follow from basic calculus. The PDF is \( g = G^\prime \) where \( G \) is the CDF above. The first order properties come from \[ g^\prime(t) = k t^{k-2} \exp\left(-t^k\right)\left[-k t^k + (k - 1)\right] \] The second order properties come from \[ g^{\prime\prime}(t) = k t^{k-3} \exp\left(-t^k\right)\left[k^2 t^{2 k} - 3 k (k - 1) t^k + (k - 1)(k - 2)\right] \]
So the Weibull density function has a rich variety of shapes, depending on the shape parameter, and has the classic unimodal shape when \( k \gt 1 \). If \( k \ge 1 \), \( g \) is defined at 0 also.
In the special distribution simulator, select the Weibull distribution. Vary the shape parameter and note the shape of the probability density function. For selected values of the shape parameter, run the simulation 1000 times and compare the empirical density function to the probability density function.
The quantile function \( G^{-1} \) is given by \[ G^{-1}(p) = [-\ln(1 - p)]^{1/k}, \quad p \in [0, 1) \]
- The first quartile is \( q_1 = (\ln 4 - \ln 3)^{1/k} \).
- The median is \( q_2 = (\ln 2)^{1/k} \).
- The third quartile is \( q_3 = (\ln 4)^{1/k} \).
Proof
The formula for \( G^{-1}(p) \) comes from solving \( G(t) = p \) for \( t \) in terms of \( p \).
Open the special distribution calculator and select the Weibull distribution. Vary the shape parameter and note the shape of the distribution and probability density functions. For selected values of the parameter, compute the median and the first and third quartiles.
The reliability function \( G^c \) is given by \[ G^c(t) = \exp(-t^k), \quad t \in [0, \infty) \]
Proof
This follows trivially from the CDF above, since \( G^c = 1 - G \).
The failure rate function \( r \) is given by \[ r(t) = k t^{k-1}, \quad t \in (0, \infty) \]
- If \( 0 \lt k \lt 1 \), \(r\) is decreasing with \( r(t) \to \infty \) as \( t \downarrow 0 \) and \( r(t) \to 0 \) as \( t \to \infty \).
- If \( k = 1 \), \(r\) is constant 1.
- If \( k \gt 1 \), \(r\) is increasing with \( r(0) = 0 \) and \( r(t) \to \infty \) as \( t \to \infty \).
Proof
The formula for \( r \) follows immediately from the PDF \( g \) and the reliability function \( G^c \) given above, since \( r = g \big/ G^c \).
Thus, the Weibull distribution can be used to model devices with decreasing failure rate, constant failure rate, or increasing failure rate. This versatility is one reason for the wide use of the Weibull distribution in reliability. If \( k \ge 1 \), \( r \) is defined at 0 also.
Moments
Suppose that \(Z\) has the basic Weibull distribution with shape parameter \(k \in (0, \infty)\). The moments of \(Z\), and hence the mean and variance of \(Z\) can be expressed in terms of the gamma function \( \Gamma \)
\(\E(Z^n) = \Gamma\left(1 + \frac{n}{k}\right)\) for \(n \ge 0\).
Proof
For \( n \ge 0 \), \[ \E(Z^n) = \int_0^\infty t^n k t^{k-1} \exp(-t^k) \, dt \] Substituting \(u = t^k\) gives \[ \E(Z^n) = \int_0^\infty u^{n/k} e^{-u} du = \Gamma\left(1 + \frac{n}{k}\right) \]
So the Weibull distribution has moments of all orders. The moment generating function, however, does not have a simple, closed expression in terms of the usual elementary functions.
In particular, the mean and variance of \(Z\) are
- \(\E(Z) = \Gamma\left(1 + \frac{1}{k}\right)\)
- \(\var(Z) = \Gamma\left(1 + \frac{2}{k}\right) - \Gamma^2\left(1 + \frac{1}{k}\right)\)
Note that \( \E(Z) \to 1 \) and \( \var(Z) \to 0 \) as \( k \to \infty \). We will learn more about the limiting distribution below.
In the special distribution simulator, select the Weibull distribution. Vary the shape parameter and note the size and location of the mean \( \pm \) standard deviation bar. For selected values of the shape parameter, run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation.
The skewness and kurtosis also follow easily from the general moment result above, although the formulas are not particularly helpful.
Skewness and kurtosis
- The skewness of \( Z \) is \[ \skw(Z) = \frac{\Gamma(1 + 3 / k) - 3 \Gamma(1 + 1 / k) \Gamma(1 + 2 / k) + 2 \Gamma^3(1 + 1 / k)}{\left[\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)\right]^{3/2}} \]
- The kurtosis of \( Z \) is \[ \kur(Z) = \frac{\Gamma(1 + 4 / k) - 4 \Gamma(1 + 1 / k) \Gamma(1 + 3 / k) + 6 \Gamma^2(1 + 1 / k) \Gamma(1 + 2 / k) - 3 \Gamma^4(1 + 1 / k)}{\left[\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)\right]^2} \]
Proof
The results follow directly from the general moment result and the computational formulas for skewness and kurtosis.
Related Distributions
As noted above, the standard Weibull distribution (shape parameter 1) is the same as the standard exponential distribution. More generally, any basic Weibull variable can be constructed from a standard exponential variable.
Suppose that \( k \in (0, \infty) \).
- If \( U \) has the standard exponential distribution then \( Z = U^{1/k} \) has the basic Weibull distribution with shape parameter \( k \).
- If \( Z \) has the basic Weibull distribution with shape parameter \( k \) then \( U = Z^k \) has the standard exponential distribution.
Proof
We use distribution functions. The basic Weibull CDF is given above; the standard exponential CDF is \( u \mapsto 1 - e^{-u} \) on \( [0, \infty) \). Note that the inverse transformations \( z = u^k \) and \( u = z^{1/k} \) are strictly increasing and map \( [0, \infty) \) onto \( [0, \infty) \).
- \( \P(Z \le z) = \P\left(U \le z^k\right) = 1 - \exp\left(-z^k\right)\) for \( z \in [0, \infty) \).
- \( \P(U \le u) = \P\left(Z \le u^{1/k}\right) = 1 - \exp\left[-\left(u^{1/k}\right)^k\right] = 1 - e^{-u} \) for \( u \in [0, \infty) \).
The basic Weibull distribution has the usual connections with the standard uniform distribution by means of the distribution function and the quantile function given above.
Suppose that \( k \in (0, \infty) \).
- If \( U \) has the standard uniform distribution then \( Z = (-\ln U)^{1/k} \) has the basic Weibull distribution with shape parameter \( k \).
- If \( Z \) has the basic Weibull distribution with shape parameter \( k \) then \( U = \exp\left(-Z^k\right) \) has the standard uniform distribution.
Proof
Let \( G \) denote the CDF of the basic Weibull distribution with shape parameter \( k \) and \( G^{-1} \) the corresponding quantile function , given above.
- If \( U \) has the standard uniform distribution then so does \( 1 - U \). Hence \(Z = G^{-1}(1 - U) = (-\ln U)^{1/k} \) has the basic Weibull distribution with shape parameter \( k \).
- If \( Z \) has the basic Weibull distribution with shape parameter \( k \) then \( G(Z) \) has the standard uniform distribution. But then so does \( U = 1 - G(Z) = \exp\left(-Z^k\right) \).
Since the quantile function has a simple, closed form, the basic Weibull distribution can be simulated using the random quantile method.
Open the random quantile experiment and select the Weibull distribution. Vary the shape parameter and note again the shape of the distribution and density functions. For selected values of the parameter, run the simulation 1000 times and compare the empirical density, mean, and standard deviation to their distributional counterparts.
The limiting distribution with respect to the shape parameter is concentrated at a single point.
The basic Weibull distribution with shape parameter \( k \in (0, \infty) \) converges to point mass at 1 as \( k \to \infty \).
Proof
Once again, let \( G \) denote the basic Weibull CDF with shape parameter \( k \) given above. Note that \( G(t) \to 0 \) as \( k \to \infty \) for \( 0 \le t \lt 1 \); \(G(1) = 1 - e^{-1}\) for all \( k \); and \( G(t) \to 1 \) as \( k \to \infty \) for \( t \gt 1 \). Except for the point of discontinuity \( t = 1 \), the limits are the CDF of point mass at 1.
The General Weibull Distribution
Like most special continuous distributions on \( [0, \infty) \), the basic Weibull distribution is generalized by the inclusion of a scale parameter. A scale transformation often corresponds in applications to a change of units, and for the Weibull distribution this usually means a change in time units.
Suppose that \(Z\) has the basic Weibull distribution with shape parameter \(k \in (0, \infty)\). For \( b \in (0, \infty) \), random variable \(X = b Z\) has the Weibull distribution with shape parameter \(k\) and scale parameter \(b\).
Generalizations of the results given above follow easily from basic properties of the scale transformation.
Distribution Functions
Suppose that \( X \) has the Weibull distribution with shape parameter \( k \in (0, \infty) \) and scale parameter \( b \in (0, \infty) \).
\( X \) distribution function \( F \) given by \[ F(t) = 1 - \exp\left[-\left(\frac{t}{b}\right)^k\right], \quad t \in [0, \infty) \]
Proof
Recall that \( F(t) = G\left(\frac{t}{b}\right) \) for \( t \in [0, \infty) \) where \( G \) is the CDF of the basic Weibull distribution with shape parameter \( k \), given above.
\( X \) has probability density function \( f \) given by \[ f(t) = \frac{k}{b^k} \, t^{k-1} \, \exp \left[ -\left( \frac{t}{b} \right)^k \right], \quad t \in (0, \infty)\]
- If \(0 \lt k \lt 1\), \(f\) is decreasing and concave upward with \( f(t) \to \infty \) as \( t \downarrow 0 \).
- If \( k = 1 \), \( f \) is decreasing and concave upward with mode \( t = 0 \).
- If \(k \gt 1\), \(f\) increases and then decreases, with mode \(t = b \left( \frac{k - 1}{k} \right)^{1/k}\).
- If \( 1 \lt k \le 2 \), \( f \) is concave downward and then upward, with inflection point at \( t = b \left[\frac{3 (k - 1) + \sqrt{(5 k - 1)(k - 1)}}{2 k}\right]^{1/k} \)
- If \( k \gt 2 \), \( f \) is concave upward, then downward, then upward again, with inflection points at \( t = b \left[\frac{3 (k - 1) \pm \sqrt{(5 k - 1)(k - 1)}}{2 k}\right]^{1/k} \)
Proof
Recall that \( f(t) = \frac{1}{b} g\left(\frac{t}{b}\right) \) for \( t \in (0, \infty) \) where \( g \) is the PDF of the corresponding basic Weibull distribution given above.
Open the special distribution simulator and select the Weibull distribution. Vary the parameters and note the shape of the probability density function. For selected values of the parameters, run the simulation 1000 times and compare the empirical density function to the probability density function.
\( X \) has quantile function \( F^{-1} \) given by \[ F^{-1}(p) = b [-\ln(1 - p)]^{1/k}, \quad p \in [0, 1) \]
- The first quartile is \( q_1 = b (\ln 4 - \ln 3)^{1/k} \).
- The median is \( q_2 = b (\ln 2)^{1/k} \).
- The third quartile is \( q_3 = b (\ln 4)^{1/k} \).
Proof
Recall that \( F^{-1}(p) = b G^{-1}(p) \) for \( p \in [0, 1) \) where \( G^{-1} \) is the quantile function of the corresponding basic Weibull distribution given above.
Open the special distribution calculator and select the Weibull distribution. Vary the parameters and note the shape of the distribution and probability density functions. For selected values of the parameters, compute the median and the first and third quartiles.
\( X \) has reliability function \( F^c \) given by \[ F^c(t) = \exp\left[-\left(\frac{t}{b}\right)^k\right], \quad t \in [0, \infty) \]
Proof
This follows trivially from the CDF \( F \) given above, since \( F^c = 1 - F \).
As before, the Weibull distribution has decreasing, constant, or increasing failure rates, depending only on the shape parameter.
\( X \) has failure rate function \( R \) given by \[ R(t) = \frac{k t^{k-1}}{b^k}, \quad t \in (0, \infty) \]
- If \(0 \lt k \lt 1\), \( R \) is decreasing with \( R(t) \to \infty \) as \( t \downarrow 0 \) and \( R(t) \to 0 \) as \( t \to \infty \).
- If \(k = 1\), \( R \) is constant \( \frac{1}{b} \).
- If \( k \gt 1 \), \(R\) is increasing with \( R(0) = 0 \) and \( R(t) \to \infty \) as \( t \to \infty \).
Moments
Suppose again that \( X \) has the Weibull distribution with shape parameter \( k \in (0, \infty) \) and scale parameter \( b \in (0, \infty) \). Recall that by definition , we can take \( X = b Z \) where \( Z \) has the basic Weibull distribution with shape parameter \( k \).
\(\E(X^n) = b^n \Gamma\left(1 + \frac{n}{k}\right)\) for \(n \ge 0\).
Proof
The result then follows from the moments of \( Z \) above, since \( \E(X^n) = b^n \E(Z^n) \).
In particular, the mean and variance of \(X\) are
- \(\E(X) = b \Gamma\left(1 + \frac{1}{k}\right)\)
- \(\var(X) = b^2 \left[\Gamma\left(1 + \frac{2}{k}\right) - \Gamma^2\left(1 + \frac{1}{k}\right)\right]\)
Note that \( \E(X) \to b \) and \( \var(X) \to 0 \) as \( k \to \infty \).
Open the special distribution simulator and select the Weibull distribution. Vary the parameters and note the size and location of the mean \( \pm \) standard deviation bar. For selected values of the parameters, run the simulation 1000 times and compare the empirical mean and standard deviation to the distribution mean and standard deviation.
Skewness and kurtosis
- The skewness of \( X \) is \[ \skw(X) = \frac{\Gamma(1 + 3 / k) - 3 \Gamma(1 + 1 / k) \Gamma(1 + 2 / k) + 2 \Gamma^3(1 + 1 / k)}{\left[\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)\right]^{3/2}} \]
- The kurtosis of \( X \) is \[ \kur(X) = \frac{\Gamma(1 + 4 / k) - 4 \Gamma(1 + 1 / k) \Gamma(1 + 3 / k) + 6 \Gamma^2(1 + 1 / k) \Gamma(1 + 2 / k) - 3 \Gamma^4(1 + 1 / k)}{\left[\Gamma(1 + 2 / k) - \Gamma^2(1 + 1 / k)\right]^2} \]
Proof
Skewness and kurtosis depend only on the standard score of the random variable, and hence are invariant under scale transformations. So the results are the same as the skewness and kurtosis of \( Z \) .
Related Distributions
Since the Weibull distribution is a scale family for each value of the shape parameter, it is trivially closed under scale transformations.
Suppose that \(X\) has the Weibull distribution with shape parameter \(k \in (0, \infty)\) and scale parameter \(b \in (0, \infty)\). If \(c \in (0, \infty)\) then \(Y = c X\) has the Weibull distribution with shape parameter \(k\) and scale parameter \(b c\).
Proof
By definition , we can take \( X = b Z \) where \( Z \) has the basic Weibull distribution with shape parameter \( k \). But then \( Y = c X = (b c) Z \).
The exponential distribution is a special case of the Weibull distribution, the case corresponding to constant failure rate.
The Weibull distribution with shape parameter 1 and scale parameter \( b \in (0, \infty) \) is the exponential distribution with scale parameter \( b \).
Proof
When \( k = 1 \), the Weibull CDF \( F \) is given by \( F(t) = 1 - e^{-t / b} \) for \( t \in [0, \infty) \). But this is also the CDF of the exponential distribution with scale parameter \( b \).
More generally, any Weibull distributed variable can be constructed from the standard variable. The following result is a simple generalization of the connection between the basic Weibull distribution and the exponential distribution.
Suppose that \(k, \, b \in (0, \infty)\).
- If \(X\) has the standard exponential distribution (parameter 1), then \(Y = b \, X^{1/k}\) has the Weibull distribution with shape parameter \(k\) and scale parameter \(b\).
- If \(Y\) has the Weibull distribution with shape parameter \(k\) and scale parameter \(b\), then \(X = (Y / b)^k\) has the standard exponential distribution.
Proof
The results are a simple consequence of the corresponding result above
- If \( X \) has the standard exponential distribution then \( X^{1/k} \) has the basic Weibull distribution with shape parameter \( k \), and hence \( Y = b X^{1/k} \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b \).
- If \( Y \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b \) then \( Y / b \) has the basic Weibull distribution with shape parameter \( k \), and hence \( X = (Y / b)^k \) has the standard exponential distributioon.
The Rayleigh distribution, named for William Strutt, Lord Rayleigh, is also a special case of the Weibull distribution.
The Rayleigh distribution with scale parameter \( b \in (0, \infty) \) is the Weibull distribution with shape parameter \( 2 \) and scale parameter \( \sqrt{2} b \).
Proof
The Rayleigh distribution with scale parameter \( b \) has CDF \( F \) given by \[ F(x) = 1 - \exp\left(-\frac{x^2}{2 b^2}\right), \quad x \in [0, \infty) \] But this is also the Weibull CDF with shape parameter \( 2 \) and scale parameter \( \sqrt{2} b \).
Recall that the minimum of independent, exponentially distributed variables also has an exponential distribution (and the rate parameter of the minimum is the sum of the rate parameters of the variables). The Weibull distribution has a similar, but more restricted property.
Suppose that \( (X_1, X_2, \ldots, X_n) \) is an independent sequence of variables, each having the Weibull distribution with shape parameter \( k \in (0, \infty) \) and scale parameter \( b \in (0, \infty) \). Then \( U = \min\{X_1, X_2, \ldots, X_n\} \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b / n^{1/k} \).
Proof
Recall that the reliability function of the minimum of independent variables is the product of the reliability functions of the variables. It follows that \( U \) has reliability function given by \[ \P(U \gt t) = \left\{\exp\left[-\left(\frac{t}{b}\right)^k\right]\right\}^n = \exp\left[-n \left(\frac{t}{b}\right)^k\right] = \exp\left[-\left(\frac{t}{b / n^{1/k}}\right)^k\right], \quad t \in [0, \infty) \] and so the result follows.
As before, Weibull distribution has the usual connections with the standard uniform distribution by means of the distribution function and the quantile function given above..
Suppose that \( k, \, b \in (0, \infty) \).
- If \( U \) has the standard uniform distribution then \( X = b (-\ln U )^{1/k} \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b \).
- If \( X \) has the basic Weibull distribution with shape parameter \( k \) then \( U = \exp\left[-(X/b)^k\right] \) has the standard uniform distribution.
Proof
Let \( F \) denote the Weibull CDF with shape parameter \( k \) and scale parameter \( b \) and so that \( F^{-1} \) is the corresponding quantile function .
- If \( U \) has the standard uniform distribution then so does \( 1 - U \). Hence \(X = F^{-1}(1 - U) = b (-\ln U )^{1/k} \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b \).
- If \( X \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b \) then \( F(X) \) has the standard uniform distribution. But then so does \( U = 1 - F(X) = \exp\left[-(X/b)^k\right] \).
Again, since the quantile function has a simple, closed form, the Weibull distribution can be simulated using the random quantile method.
Open the random quantile experiment and select the Weibull distribution. Vary the parameters and note again the shape of the distribution and density functions. For selected values of the parameters, run the simulation 1000 times and compare the empirical density, mean, and standard deviation to their distributional counterparts.
The limiting distribution with respect to the shape parameter is concentrated at a single point.
The Weibull distribution with shape parameter \( k \in (0, \infty) \) and scale parameter \( b \in (0, \infty) \) converges to point mass at \( b \) as \( k \to \infty \).
Proof
If \( X \) has the Weibull distribution with shape parameter \( k \) and scale parameter \( b \), then we can write \(X = b Z \) where \( Z \) has the basic Weibull distribution with shape parameter \( k \). We showed above that the distribution of \( Z \) converges to point mass at 1 , so by the continuity theorem for convergence in distribution, the distribution of \( X \) converges to point mass at \( b \).
Finally, the Weibull distribution is a member of the family of general exponential distributions if the shape parameter is fixed.
Suppose that \( X \) has the Weibull distribution with shape parameter \( k \in (0, \infty) \) and scale parameter \( b \in (0, \infty) \). For fixed \( k \), \( X \) has a general exponential distribution with respect to \( b \), with natural parameter \( k - 1 \) and natural statistics \( \ln X \).
Proof
This follows from the definition of the general exponential distribution, since the Weibull PDF can be written in the form \[ f(t) = \frac{k}{b^k}\exp\left(-t^k\right) \exp[(k - 1) \ln t], \quad t \in (0, \infty) \]
Computational Exercises
The lifetime \(T\) of a device (in hours) has the Weibull distribution with shape parameter \(k = 1.2\) and scale parameter \(b = 1000\).
- Find the probability that the device will last at least 1500 hours.
- Approximate the mean and standard deviation of \(T\).
- Compute the failure rate function.
Answer
- \(\P(T \gt 1500) = 0.1966\)
- \(\E(T) = 940.7\), \(\sd(T) = 787.2\)
- \(h(t) = 0.000301 t^{0.2}\)


5.2: Functions of the Skeletal System
By the end of this section, you will be able to:
- Define bone, cartilage, and the skeletal system
- List and describe the functions of the skeletal system
Bone , or osseous tissue , is a hardened connective tissue that forms most of the adult skeleton, the support structure of the body. In the areas of the skeleton where bones move (for example, the ribcage and joints), cartilage , a semi-rigid form of connective tissue, provides flexibility and smooth surfaces for movement. The skeletal system is the body system composed of bones and cartilage and performs the following critical functions for the human body:
- supports the body
- facilitates movement
- protects internal organs
- produces blood cells
- stores and releases minerals and fat
Support, Movement, and Protection
The most apparent functions of the skeletal system are the gross functions—those visible by observation. Simply by looking at a person, you can see how the bones support, facilitate movement, and protect the human body.
Just as the steel beams of a building provide a scaffold to support its weight, the bones and cartilage of your skeletal system compose the scaffold that supports the rest of your body. Without the skeletal system, you would be a limp mass of organs, muscle, and skin.
Bones facilitate movement by serving as points of attachment for your muscles. While some bones only serve as a support for the muscles, others also transmit the forces produced when your muscles contract. From a mechanical point of view, bones act as levers and articulations (where two bones come together, also known as joints) serve as fulcrums. Unless a muscle spans an articulation and contracts, a bone is not going to move. Information on the interactions of the skeletal and muscular systems, that is the musculoskeletal system, will be covered in the upcoming chapters of this book.
Bones also protect internal organs from injury by covering or surrounding them.
CAREER CONNECTION
Orthopedist
An orthopedist is a doctor who specializes in diagnosing and treating disorders and injuries related to the musculoskeletal system. Some orthopedic problems can be treated with medications, exercises, braces, and other devices, but others may be best treated with surgery (Figure \(\PageIndex{2}\)).
While the origin of the word “orthopedics” (ortho- = “straight”; paed- = “child”), literally means “straightening of the child,” orthopedists can have patients who range from pediatric to geriatric. In recent years, orthopedists have even performed prenatal surgery to correct spina bifida, a congenital defect in which the neural canal in the spine of the fetus fails to close completely during embryologic development.
Orthopedists commonly treat bone and joint injuries but they also treat other bone conditions including curvature of the spine. Lateral curvatures (scoliosis) can be severe enough to slip under the shoulder blade (scapula) forcing it up as a hump. Spinal curvatures can also be excessive dorsoventrally (kyphosis) causing a hunch back and thoracic compression. These curvatures often appear in preteens as the result of poor posture, abnormal growth, or indeterminate causes. Mostly, they are readily treated by orthopedists. As people age, accumulated spinal column injuries and diseases like osteoporosis can also lead to curvatures of the spine, hence the stooping you sometimes see in the elderly.
Some orthopedists sub-specialize in sports medicine, which addresses both simple injuries, such as a sprained ankle, and complex injuries, such as a torn rotator cuff in the shoulder. Treatment can range from exercise to surgery.
Mineral Storage, Energy Storage, and Hematopoiesis
On a metabolic level, bone tissue performs several critical functions. For one, the bone matrix acts as a reservoir for a number of minerals important to the functioning of the body, especially calcium, and potassium. These minerals, incorporated into bone tissue, can be released back into the bloodstream to maintain levels needed to support physiological processes. Calcium ions, for example, are essential for muscle contractions and controlling the flow of other ions involved in the transmission of nerve impulses.
 The softer connective tissue that fills the interior of most bone is referred to as bone marrow (Figure \(\PageIndex{4}\)). There are two types of bone marrow: yellow marrow (deep) and red marrow (superficial). Yellow marrow contains adipose tissue; the triglycerides (lipids) stored in the adipocytes of the tissue can serve as a source of energy. Red marrow is where hematopoiesis —the production of blood cells—takes place. Red blood cells, white blood cells, and platelets are all produced in the red marrow.
Concept Review
The major functions of the bones are body support, facilitation of movement, protection of internal organs, storage of minerals and fat, and hematopoiesis. Together, the muscular system and skeletal system are known as the musculoskeletal system.
Review Questions
Q. Which function of the skeletal system would be especially important if you were in a car accident?
A. storage of minerals
B. protection of internal organs
C. facilitation of movement
D. fat storage
- Answer
-
Answer: B
Q. Bone tissue can be described as ________.
A. dead calcified tissue
B. cartilage
C. the skeletal system
D. hardened connective tissue
- Answer
-
Answer: D
Q. Without red marrow, bones would not be able to ________.
A. store phosphate
B. store calcium
C. facilitate making blood cells
D. move like levers
- Answer
-
Answer: C
Q. Yellow marrow has been identified as ________.
A. an area of fat storage
B. a point of attachment for muscles
C. the hard portion of bone
D. the cause of kyphosis
- Answer
-
Answer: A
Q. Which of the following can be found in areas of movement?
A. hematopoiesis
B. cartilage
C. yellow marrow
D. red marrow
- Answer
-
Answer: B
Q. The skeletal system is primarily made of ________.
A. muscles and tendons
B. bones and cartilage
C. vitreous humor
D. minerals and fat
- Answer
-
Answer: B
Critical Thinking Questions
The skeletal system is composed of bone and cartilage and has many functions. Choose three of its functions and discuss what features of the skeletal system allow it to accomplish these functions.
- Answer
-
Answer:
It supports the body. The rigid, yet flexible skeleton acts as a framework to support the other organs of the body.
It facilitates movement. The skeletal system provides attachment points and moveable joints that allow skeletal muscles to change the shape and positions (move) of the body.
It protects internal organs. Parts of the skeleton enclose or partly enclose various organs of the body including our brain, ears, heart, and lungs. Any trauma to these organs has to be mediated through the skeletal system.
It facilitates blood cell production. The central cavity of long bones is filled with marrow. The red marrow is responsible for forming red and white blood cells.
It stores and releases minerals and fat. The mineral component of bone, in addition to providing hardness to bone, provides a mineral reservoir that can be tapped as needed. Additionally, the yellow marrow, which is found in the central cavity of long bones along with red marrow, serves as a storage site for fat.
Glossary
- bone
- hard, dense connective tissue that forms the structural elements of the skeleton
- cartilage
- semi-rigid connective tissue found on the skeleton in areas where flexibility and smooth surfaces support movement
- hematopoiesis
- production of blood cells, which occurs in the red marrow of the bones
- orthopedist
- doctor who specializes in diagnosing and treating musculoskeletal disorders and injuries
- osseous tissue
- bone tissue; a hard, dense connective tissue that forms the structural elements of the skeleton
- red marrow
- connective tissue in the interior cavity of a bone where hematopoiesis takes place
- skeletal system
- organ system composed of bones and cartilage that provides for movement, support, and protection
- yellow marrow
- connective tissue in the interior cavity of a bone where fat is stored
Contributors and Attributions
-
OpenStax Anatomy & Physiology (CC BY 4.0). Access for free at https://openstax.org/books/anatomy-and-physiology


2.8: Bankruptcy
Many students think that bankruptcy is a death knell. It is not – necessarily. It can actually be a good thing, which ensures the survival of an enterprise that experienced financial stress. There are different forms of bankruptcy. The forms are categorized by “chapters,” under Title 11 of the United States Bankruptcy Code. Let’s briefly see what the principal bankruptcy forms are and how each is different.
Chapter 7 :
This bankruptcy form allows debtors to eliminate most or all of their debts over a short period of time, often just a few months. Only student loans, child support payments and some other debts may survive. Here, a trustee is appointed who then liquidates unsecured debts and makes the proper distributions. Collateral on secured debt may be repossessed. Certain assets will be protected, such as social security insurance. To qualify for Chapter 7, the debtor must satisfy a “means test.” If the test is not satisfied, the debtor may seek relief under Chapter 13.
Chapter 11 :
Here, the debtor retains ownership and control of assets. The debtor is referred to as a “debtor in possession.” The “DIP” runs the day-to-day affairs of the business while the creditors work with the bankruptcy court to work out a plan to be made whole. If the creditors come to an agreement, the business continues operating and certain agreed-upon payments are made. If there is no agreement, the court intercedes; debtors filing a second time are referred to as Chapter 22 filers.
Chapter 13 :
In this form, debtors retain ownership and possession of the assets (in contrast to Chapter 7), and will make payments to a trustee from future earnings , which will then be disbursed to creditors. There is a five-year limit in which this process must be completed . Secured creditors may receive larger payments.
Students’ take-away : Bankruptcy is not always the end of the story.
Advice : One should consult with an attorney to obtain detailed, actionable information regarding these complex laws.
Question : In the Banking Crisis of 2008, the United States government rescued, or bailed out, General Motors; it purchased stock in the company, using taxpayer funds. Was that the right move? Would it have been better for Uncle Sam to have allowed the corporation to fail and file under the Bankruptcy Code?


<IP_ADDRESS>: Classification and Milling of Wheat
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Classification of Wheat
Wheat can be classified in three ways:
- Color (e.g., red, yellow, white)
- Planting season: spring wheat, planted in the spring and harvested in early fall; winter wheat, planted in the fall, harvested the following summer
- Characteristics of the grain: durum, hard bread wheat, and soft wheat
The United States recognizes seven market classes of wheat. The first five, listed below, are the most important:
- Hard red winter (planted in the fall, harvested the following summer)
- Soft red winter
- Hard red spring (planted and harvested the same year)
- Durum
- Red durum
Characteristics of the Major Wheat Groups
The major wheat groups each have differing characteristics. This also determines their use in food production and baking. The following identifies the major characteristics of each of the three major wheat groups.
- Durum wheat: Durum wheat is generally high in gluten-producing proteins. In effect, durum is the “hardest” wheat. Durum wheat is used for making semolina flour, which is made into macaroni and other pasta and is the top choice due to the higher gluten content.
- Hard wheat: Hard wheat includes hard winter wheat and hard spring wheat. Hard Red Winter (HRW) and Hard Red Spring (HRS) are most common in the United States. They contain more gluten-producing proteins than soft wheat, and are used for making bread flours and all-purpose flours.
- Soft wheat: Soft wheat is low in gluten-producing protein. Soft Red Winter (SRW) is the most common in the United States. Soft wheat is milled into cake and pastry flour.
Milling of Wheat
Milling wheat is the process that turns whole grains into flour. The goal of the miller is to produce:
- A consistent product
- A range of flours suitable for a variety of functions
- Flours with predictable performance
The very first mill operation is analyzing the grain, which determines criteria such as the gluten content and amylase (an enzyme that breaks down carbohydrates) activity. It is at this point that decisions about blending are made.
Following analysis, milling may be divided into three stages:
- Cleaning and conditioning – ridding the grain of all impurities and readying it for milling
- Crushing or breaking – breaking down the grain in successive stages to release its component parts
- Reduction – progressive rollings and siftings to refine the flour and separate it into various categories, called streams
Cleaning
Wheat received at the mill contains weeds, seeds, chaff, and other foreign material. Strong drafts of air from the aspirator remove lighter impurities. The disc separator removes barley, oats, and other foreign materials. From there, the wheat goes to the scourers in which it is driven vigorously against perforated steel casings by metal beaters. In this way, much of the dirt lodged in the crease of the wheat berry is removed and carried away by a strong blast of air. Then the magnetic separator removes any iron or steel.
At this point, the wheat is moistened. Machines known as whizzers take off the surface moisture. The wheat is then tempered , or allowed to lie in bins for a short time while still damp, to toughen the bran coat, thus making possible a complete separation of the bran from the flour-producing portion of the wheat berry. After tempering, the wheat is warmed to a uniform temperature before the crushing process starts.
Crushing or Breaking
The objectives at this stage are twofold:
- Separate as much bran and germ as possible from the endosperm
- Maximize the flour from the resulting endosperm
Household grain mills create flour in one step — grain in one end, flour out the other — but the commercial mill breaks the grain down in a succession of very gradual steps, ensuring that little bran and germ are mixed with any endosperm.
Although the process is referred to as crushing , flour mills crack rather than crush the wheat with large steel rollers. The rollers at the beginning of the milling system are corrugated and break the wheat into coarse particles. The grain passes through screens of increasing fineness. Air currents draw off impurities from the middlings . Middlings is the name given to coarse fragments of endosperm, somewhere between the size of semolina and flour. Middlings occur after the “break” of the grain.
Bran and germ are sifted out, and the coarse particles are rolled, sifted, and purified again. This separation of germ and bran from the endosperm is an important goal of the miller. It is done to improve dough-making characteristics and color. As well, the germ contains oil and can affect keeping quality of the flour.
Reduction
In the reduction stage, the coarser particles go through a series of fine rollers and sieves. After the first crushing, the wheat is separated into five or six streams. This is accomplished by means of machines called plansifters that contain sieves, stacked vertically, with meshes of various sizes. The finest mesh is as fine as the finished flour, and some flour is created at an early stage of reduction.
Next, each of the divisions or streams passes through cleaning machines, known as purifiers , a series of sieves arranged horizontally and slightly angled. An upcurrent draught of air assists in eliminating dust. The product is crushed a little more, and each of the resulting streams is again divided into numerous portions by means of sifting. The final crushings are made by perfectly smooth steel rollers that reduce the middlings into flour. The flour is then bleached and put into bulk storage. From bulk storage, the flour is enriched (thiamine, niacin, riboflavin, and iron are added), and either bagged for home and bakery use or made ready for bulk delivery.
Extraction Rates
The extraction rate is a figure representing the percentage of flour produced from a given quantity of grain. For example, if 82 kg of flour is produced from 100 kg of grain, the extraction rate is 82% (82÷100×100). Extraction rates vary depending on the type of flour produced. A whole grain flour, which contains all of the germ, bran, and endosperm, can have an extraction rate of close to 100%, while white all-purpose flours generally have extraction rates of around 70%. Since many of the nutrients are found in the germ and bran, flours with a higher extraction rate have a higher nutritional value.
Whole Grain and Artisan Milling
Whole grain and artisan milling is the type of milling that was practiced before the consumer market demanded smooth white flours that are refined and have chemical additives to expedite the aging of flours. Artisan milling produces flours that are less refined and better suited to traditional bread, but also contain little to no additives and have higher nutritional content. For that reason, demand for these types of flour is on the rise.
Artisan millers (also known as micro millers) process many non-stream grains, including spelt, kamut, buckwheat, and other non-gluten grains and pulses. This offers bakers opportunities to work with different grains and expand their businesses. Artisan flours are readily available directly from millers or through a distributor. Knowing the origin of the grains and the quality of the ingredients in baking is important for artisan bakers.


15.4: Chapter 5
Your Turn
5.1
1. 18 plus 11 OR the sum of 18 and 11
2. 27 times 9 OR the product of 27 and 9
3. 84 divided by 7 OR the quotient of 84 and 7
4. \(p\) minus \(q\) OR the difference of \(p\) and \(q\)
5.2
1. \(43 + 67\)
2. \(\left( {45} \right)\left( 3 \right)\)
3. \(45 \div 3\)
4. \(89-42\)
5.3
1. \(n - 20\)
2. \(\left( {n + 2} \right) \bullet 6\) OR \(6\left( {n + 2} \right)\)
3. \({n^3} - 5\)
4. \(60h + 40\)
5.4
1. \(5y = 50\)
2. \(\frac{1}{2}n = 30\)
3. \(3n - 7 = 2\)
4. \(2x + 7 = 21\)
5.5
1. \(24 - \left( {17 - 6} \right) = 13\)
2. \(\left( {3 \bullet 6} \right) + 13 = 31\)
3. \(\left( {12 - 6} \right) \div \left( {5 - 3} \right) = 3\)
4. \(5 \bullet \left( {{3^2} + 5} \right) = 70\)
5.6
1. \(9\)
2. \(- 6\)
5.7
1. \(5{x^{ 2}}-3x - 7\)
5.8
1. \(- 9x + 4\)
5.9
1. \(2y + 10\)
2. \(- 2a - 2b + 8\)
3. \(112 + 16x\)
4. \(6x + 18\)
5. \(3a + 4\)
5.10
1.
\(2{x^2} - 11x + 12\)
5.11
1.
\(4{x^2} + x - 2\)
5.12
1. \(x = 3\)
5.13
1. \(y = 28\)
5.14
1. Henry has 42 books.
5.15
1. Total cost is $48.85
5.16
1. Answers will vary. For example: You can rent a paddleboard for $25 per hour with a water shoe purchase of $75. If you spent $200, how many hours did you rent the paddle board for?
You rented the paddle board for 5 hours.
You rented the paddle board for 5 hours.
5.17
1. \(12 = - 1\), which is false; therefore, this is a false statement, and the equation has no solution.
5.18
1. \(- 12 = - 12\), which is true; therefore, this is a true statement, and there are infinitely many solutions.
5.19
1. \(t = \frac{I}{{\text{P}}r}\)
5.20
1. \(h = \frac
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5.21
1.
5.22
1. \(0 \leq x \leq 2.5\)
Graph:
Graph:
5.23
1.
5.24
1.
\(p \geq 2\)
\(p \geq 2\)
5.25
1. Taleisha can send/receive 106 or fewer text messages and keep her monthly bill no more than $50.
5.26
1. You could take up to 17 credit hours and stay under $2,000.
5.27
1. Malik must tutor at least 23 hours.
5.28
1. \(a\) = 1 U.S. dollar, and \(b\) = 1.21 Canadian dollars, the ratio is 1 to 1.21; or 1:1.21; or \(\frac{1}1.21\).
5.29
1. With \(a\) = 170 pounds on Earth, and \(b\) = 64 pounds on Mars, the ratio is 170 to 64; or 170:64; or \(\frac17064\).
5.30
1. $219.51
5.31
1. 501.6 pounds
5.32
1. 720 cookies
5.33
1. The constant of proportionality (centimeters divided by inches) is 2.54. This tells you that there are 2.54 centimeters in one inch.
5.34
1. 30.5 hours (or 30½ hours, or 30 hours and 30 minutes)
5.35
1. $207.50
5.36
1. 125.9 miles
5.37
1. The scale is \(1{\text{ inch}} = 91.25{\text{ miles}}\). The other borders would calculate as: eastern and western borders are 273.75 miles, and northern border is 365 miles.
5.38
1. 184 inches, or 15 feet, 4 inches.
5.39
1.
5.40
1.
- Yes
- Yes
2.
- No
- No
3.
- Yes
- Yes
4.
- Yes
- Yes
5.41
1.
5.42
1.
Your friend will pay $1.65.
5.43
1. Yes
2. Yes
3. No
4. No
5. No
5.44
1. \(y \geq - 2x + 3\)
5.45
1.
5.46
1. \(11x + 16.5y \geq 330\)
2.
3. Answers will vary.
5.47
1. \({x^2} + 4x + 3\)
5.48
1. \(2{x^2} - 5x - 3\)
5.49
1. \(\left( x + 2 \right)\left( x + 4 \right)\)
5.50
1. \(\left( {x - 7} \right)\left( {x - 9} \right)\)
5.51
1.
5.52
1. \(x = - 3,\,\,x = 2\)
5.53
1. \(x = 2\) or \(x = - 5\)
5.54
1. \(x = 3\) or \(x = - 5\)
5.55
1. \(x = 5\), \(x = -5\)
5.56
1. \(p = \frac{7}{5}\), \(p = - \frac{7}{5}\)
5.57
1. \(a = 5\), \(a = - 3\)
5.58
1. \(y = 1\), \(y = \frac{2}{3}\)
5.59
1. 16, 15 and –16, –15
5.60
1. \({\text{width}} = 5{\text{ feet}}\), \({\text{length}} = 6{\text{ feet}}\)
5.61
1. 22
2. 6
3. \(3{t^2} - 2t + 1\)
5.62
1. There 205 unread emails after 7 days.
5.63
1. This relation is a function.
2. This relation is not a function.
5.64
1. Both George and Mike have two phone numbers. Each \(x\)-value is not matched with only one \(y\)-value. This relation is not a function.
5.65
1. function
2. not a function
3. function
5.66
1. This graph represents a function.
5.67
1. This graph does not represent a function.
5.68
1. The domain is the set of all \(x\)-values of the relation: \(\left\{ {1,2,3,4,5} \right\}\).
2. The range is the set of all \(y\)-values of the relation: \(\left\{ {1,8,27,64,125} \right\}\).
5.69
1. The ordered pairs of the relation are: \(\left\{ {\left( {-3,\,3} \right),\left( {-2,\,2} \right),\left( {-1,\,0} \right),\left( {0,\,-1} \right),\left( {2,\,-2} \right),\left( {4,\,-4} \right)} \right\}\).
2. The domain is the set of all \(x\)-values of the relation: \(\left\{ {-3,-2,-1,0,2,4} \right\}\).
3. The range is the set of all \(y\)-values of the relation: \(\left\{{-4,-2,-1,0,2,3}\right\}\).
5.70
1. The graph crosses the \(x\)-axis at the point (2, 0). The \(x\)-intercept is (2, 0). The graph crosses the \(y\)-axis at the point (0, −2). The \(y\)-intercept is (0, −2).
5.71
1. The \(x\)-intercept is (4, 0) and the \(y\)-intercept is (0, 12).
5.72
1. \(m = \frac-4{3}\)
5.73
1. \({\text{slope}} = - 1\)
5.74
1. slope \(m = 2\) and \(y{\text{-intercept}} = (0,-1)\)
2. slope \(m = - \frac{1}{4}\) and \(y{\text{-intercept}} = (0,4)\)
5.75
1.
5.76
1.
5.77
1.
5.78
1. (0, 20) is the \(y\text{-intercept}\) and represents that there were 20 teachers at Jones High School in 1990. There is no \(x\text{-intercept}\).
2. In the first 5 years the slope is 2; this means that on average, the school gained 2 teachers every year between 1990 and 1995. Between 1995 and 2000, the slope is 4; on average the school gained 4 teachers every year. Then the slope is 0 between 2000 and 2005 meaning the number of teachers remained the same. There was a decrease in teachers between 2005 and 2010, represented by a slope of –2. Finally, the slope is 4 between 2010 and 2020, which indicates that on average the school gained 4 teachers every year.
3. Answers will vary. Jones High School was founded in 1990 and hired 2 teachers per year until 1995, when they had an increase in students and they hired 4 teachers per year for the next 5 years. Then there was a hiring freeze, and no teachers were hired between 2000 and 2005. After the hiring freeze, the student population decreased, and they lost 2 teachers per year until 2010. Another surge in student population meant Jones High School hired 4 new teachers per year until 2020 when they had 80 teachers at the school.
5.79
1. 50 inches
2. 66 inches
3. The slope, 2, means that the height \(h\) increases 2 inches when the shoe size(s) increases 1 size.
4. The \(h\)-intercept means that when the shoe size is 0, the woman’s height is 50 inches.
5.80
1. $25
2. $85
3. The slope, 4, means that the weekly cost, \(C\), increases by $4 when the number of pizzas sold, \(p\), increases by 1. The \(C\)-intercept means that when the number of pizzas sold is 0, the weekly cost is $25.
4. Graph:
5.81
1. No
2. Yes
5.82
1. No
2. No
5.83
1.
5.84
1. (3, 2)
5.85
1. \((3, 2)\)
5.86
1. There is no solution to this system.
5.87
1. There are infinitely many solutions to this system.
5.88
1. Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.
5.89
1. not a solution
2. a solution
5.90
1. The region containing (0, 0) is the solution to the system of linear inequalities.
5.91
1. The solution is the darkest shaded region.
5.92
1. The solution is the lighter shaded region.
5.93
1. No solution.
5.94
1. The solution is the lighter shaded region.
5.95
1. \(\left\{ \begin{array}{l}240h + 160C \ge 800\\1.40h + 0.50C \le 5\\h \ge 0\\C \ge 0\end{array} \right.\)
2.
3. The point \((3, 2)\) is not in the solution region. Omar would not choose to eat 3 hamburgers and 2 cookies.
4. The point \((2, 4)\) is in the solution region. Omar might choose to eat 2 hamburgers and 4 cookies.
5.96
1. With \(a =\) the number of bags of apples sold, and \(b =\) the number of bunches of bananas sold, the objective function is \(P = 4a + 6b\).
5.97
1. \(T = 20x + 28y\)
5.98
1. The constraints are \(a + b \leq 20\) and \(3a + 5b \leq 70\). The summary is: \(P = 4a + 6b\), \(a + b \leq 20\), and \(3a + 5b \leq 70\).
5.99
1. The constraints are:
\(15 \leq x \leq 22\)
\(13 \leq y \leq 19\)
So the system is:
\(T = 20x + 28y\)
\(15 \leq x \leq 22\)
\(13 \leq y \leq 19\)
\(15 \leq x \leq 22\)
\(13 \leq y \leq 19\)
So the system is:
\(T = 20x + 28y\)
\(15 \leq x \leq 22\)
\(13 \leq y \leq 19\)
5.100
1. The maximum value for the profit \(P\) occurs when \(x = 15\) and \(y = 5\). This means that to maximize their profit, the Robotics Club should sell 15 bags of apples and 5 bunches of bananas every day.
Check Your Understanding
1. \(J = V + 2\), \(V = J - 2\)
2. \(5x+8\), \(2n+3m\)
3. \((8x + 12x) \div \left( {4x - 2x} \right)\)
4. \(- 3\)
5. \({x^2}-2xy + {y^2}\)
6. \(3x\left( {3{x^2} + x-2} \right)\)
7. multiplication
8. division
9. It is a correct solution strategy.
Let
\(\begin{array}{rcl}{x}&{ = }&{38}\\{8\left( {38-2} \right)}&{ \mathop = \limits^? }&{6\left( {38 + 10} \right)}\\{8\left( {36} \right)}&{ \mathop = \limits^? }&{6\left( {48} \right)}\\{288}&{ = }&{288 ✓}\\\end{array}\)
Let
\(\begin{array}{rcl}{x}&{ = }&{38}\\{8\left( {38-2} \right)}&{ \mathop = \limits^? }&{6\left( {38 + 10} \right)}\\{8\left( {36} \right)}&{ \mathop = \limits^? }&{6\left( {48} \right)}\\{288}&{ = }&{288 ✓}\\\end{array}\)
10. It is a correct solution strategy.
Let
\(\begin{array}{rcl}{x}&{ = }&{- \,2}\\{7 + 4\left( {2 + 5\left( { - \,2} \right)} \right)}&{ \mathop = \limits^? }&{3\left( {6\left( { -\,2} \right) + 7} \right)-\left( {13\left( { -\,2} \right) + 36} \right)}\\{7 + 4\left( {2-10} \right)}&{ \mathop = \limits^? }&{3\left( { - 12 + 7} \right)-\left( { -\,26 + 36} \right) }\\{7 + 4\left( { -\,8} \right)}&{ \mathop = \limits^? }&{3\left( { - 5} \right)-\left( {10} \right) }\\{7-32}&{ \mathop = \limits^? }&{ -\,15-10}\\{-\,25}&{ = }&{-\,25✓}\end{array}\)
Let
\(\begin{array}{rcl}{x}&{ = }&{- \,2}\\{7 + 4\left( {2 + 5\left( { - \,2} \right)} \right)}&{ \mathop = \limits^? }&{3\left( {6\left( { -\,2} \right) + 7} \right)-\left( {13\left( { -\,2} \right) + 36} \right)}\\{7 + 4\left( {2-10} \right)}&{ \mathop = \limits^? }&{3\left( { - 12 + 7} \right)-\left( { -\,26 + 36} \right) }\\{7 + 4\left( { -\,8} \right)}&{ \mathop = \limits^? }&{3\left( { - 5} \right)-\left( {10} \right) }\\{7-32}&{ \mathop = \limits^? }&{ -\,15-10}\\{-\,25}&{ = }&{-\,25✓}\end{array}\)
11. This is not a correct solution strategy. The negative sign is not distributed correctly in the second line of the solution strategy. The second line should read \(8x + 7-2x + 9 = 22-4x + 4\).
12. \(
(click for details)
= 1.7x + 3\)
Callstack:
at (Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/15:_Answer_Key/15.04:_Chapter_5), /content/body/div/div/div[101]/div/div[12]/span, line 1, column 1
13. $40.40
14. \(
(click for details)
= 1.6y + 5\)
Callstack:
at (Bookshelves/Applied_Mathematics/Contemporary_Mathematics_(OpenStax)/15:_Answer_Key/15.04:_Chapter_5), /content/body/div/div/div[101]/div/div[14]/span, line 1, column 1
15. $40.20
16. The Enjoyable Cab Company, because the cab fare will be $ 0.20 less than what it would cost to take a taxi from the Nice Cab Company.
17. Luis is; there are infinitely many solutions. If this is solved using the general strategy, it simplifies to \(0 = 0\). This is a true statement, so there are infinitely many solutions.
18.
d
\(F = \frac{9}{5}C + 32\)
19.
b
\(K = C + 273\), although d \(C = K – 273\) is an equivalent formula.
20.
a
\(K = \frac{5}{9}\left( {F-32} \right) + 273\)
21.
b
\(R = \frac{9}{5}C + 492\)
22. d \(\left( { - \infty , - 1} \right)\)
23. b \([ - 5,\infty )\)
24. c \(\left[ {\frac{3}{2},\infty } \right)\)
25. a \(( - 4,3)\)
26.
e
\(6x \geq 24\)
27.
b
\(-6x \gt 18\)
28.
c
\(4x - 3 > - 11\)
29.
c
\(- 3x + 14 > - 13\)
30.
b
\(8x < 764\)
31.
d
\(50x > \text{8,120}\)
32.
a
True
33.
a
True
34.
a
True
35.
b
False
36.
a
True
37. \(12{:}16\)
38. \(16{:}28\)
39. 324.00 British pounds (None of these.)
40. 10 inches
41. Yes he can, but barely. At 37 miles per gallon, Albert can drive \(499.5\) miles. While in theory he can make it, he probably should fill up with gasoline somewhere along the way!
42. \($40.66\)
43.
d
\(y = \frac{5}{3}\)
44.
45.
b
\(y = \frac{1}{2}x + 4\)
46.
47.
 Elimination
68. Substitution
69. Substitution
70. Elimination
71. Substitution
72. Elimination
73. Elimination
74. Substitution
75. c
76. e
77. d
78. b
79. a
80. c
81. a
82. b
83. d
84. a
85. c
86. d
87.
88. b
89. \($140\)