4.1: Behavioral Presentation of Human Sexuality
Sexual Health
Sexual health is a state of physical, emotional, mental and social well-being in relation to sexuality; it is not merely the absence of disease, dysfunction or infirmity.
 Love, affection and sexual intimacy all play a role in healthy relationships. They also contribute to your sense of well-being. A number of disorders can affect the ability to have or enjoy sex in both men and women. Concerns about infertility or fear of unplanned pregnancy can also come into play. In addition, a number of diseases and disorders affect sexual health.  In men, treatment of prostate cancer can cause erectile dysfunction.
Public Health Approach to Sexual Health
Sexuality affects individuals and society across a broad spectrum of activities: through health, but also through factors at multiple levels, such as gender relations, reproduction, and economics. Physiologic, behavioral, and affective measurement of sexuality and sexual behavior is complicated by cultural values and norms, but is essential to individual health (including happiness) as well as public health.
 In addition, survey data reveal several individual and relationship factors that are important to sexual health at all levels, with overall health noted as the greatest predictor of sexual satisfaction.
Infidelity
Sexuality is important to society not only because of health implications; it also affects gender and property relations, reproduction potential, and economics.  For instance, infidelity between couples often leads to hurt and divorce, and in Western countries, between 25% and 50% of divorcees cite a spouse’s infidelity as the primary cause of divorce.
Sexual Dysfunction
Sexual dysfunction can pose public health problems, as it is related to public health issues and affects people’s happiness and general well-being.
According to the National Health and Social Life Survey,
- The prevalence of sexual dysfunction was found to be higher among women than men.

- Sexual problems increase with age, but sex-related personal distress decreases.
Sexual Frequency
Sexual frequency is important for sexual relationship satisfaction. Sexual intercourse frequency is noted as being the most important factor when predicting sexual satisfaction. Satisfaction declines with age but not as steeply as sexual frequency declines. However, although satisfaction is lower in women, satisfaction levels do not change over time among women, compared with men.
Sex and Health
Duration and age matter, but health matters most of all. Health proves to be a critical predictor of sexual satisfaction. Among those indicating their health is at least “very good,” more than half say they are satisfied with their sex lives. The majority of older Americans do not practice safe sex, even if they have multiple partners. It was reported that only 1 in 5 sexually active, dating singles use condoms regularly. Many older Americans report dating more than one person at a time and being sexually active with more than one sex partner (6% of men and 1% of women).
 Despite having a lower overall reported health rating, Hispanics report being happier with their sex lives compared with the general population. Sexuality was found to be a higher priority for older Hispanics, who report higher levels of sexual activity and satisfaction. Having a partner matters.
Sexual Happiness
The most important indicator of the sexual happiness of older Americans is having a steady sex partner. That indicator is less important than the frequency of sexual intercourse, good health, low levels of stress, and the absence of financial worries. There are still behavioral differences between older men and women, and older men and women continue to rank the importance of sex and the enjoyment of sex differently—even as they age. Older men continue to have more sex and think about sex more than older women; they see it as more important to their quality of life. Older men report having more frequent orgasms than women (2 out of 3 men, compared with 1 in 3 women), but their frequency of orgasm drops with age. Older men are twice as likely (21% compared with 11%) to admit sexual activity outside their relationship than women.
 Enactment of the Trafficking Victims Protection Act of 2000 (TVPA) made sex trafficking a serious violation of Federal law. The TVPA also recognizes labor trafficking, which is discussed in a separate fact sheet.
As defined by the TVPA, the term ‘commercial sex act’ means any sex act on account of which anything of value is given to or received by any person.
The TVPA recognizes that traffickers use psychological and well as physical coercion and bondage, and it defines coercion to include: threats of serious harm to or physical restraint against any person; any scheme, plan, or pattern intended to cause a person to believe that failure to perform an act would result in serious harm to or physical restraint against any person; or the abuse or threatened abuse of the legal process.
 There are a number of common patterns for luring victims into situations of sex trafficking, including:
- A promise of a good job in another country
- A false marriage proposal turned into a bondage situation
- Being sold into the sex trade by parents, husbands, boyfriends
- Being kidnapped by traffickers
Sex traffickers frequently subject their victims to debt-bondage, an illegal practice in which the traffickers tell their victims that they owe money (often relating to the victims’ living expenses and transport into the country) and that they must pledge their personal services to repay the debt.
Sex traffickers use a variety of methods to “condition” their victims including starvation, confinement, beatings, physical abuse, rape, gang rape, threats of violence to the victims and the victims’ families, forced drug use and the threat of shaming their victims by revealing their activities to their family and their families’ friends.
Victims face numerous health risks. Physical risks include drug and alcohol addiction; physical injuries (broken bones, concussions, burns, vaginal/anal tearings); traumatic brain injury (TBI) resulting in memory loss, dizziness, headaches, numbness; sexually transmitted diseases (e.g., HIV/AIDS, gonorrhea, syphilis, UTIs, pubic lice); sterility, miscarriages, menstrual problems; other diseases (e.g., TB, hepatitis, malaria, pneumonia); and forced or coerced abortions.
 Victims are at risk for Posttraumatic Stress Disorder (PTSD)—acute anxiety, depression, insomnia, physical hyper-alertness, self-loathing that is long-lasting and resistant to change (complex-PTSD).
Victims may also suffer from traumatic bonding—a form of coercive control in which the perpetrator instills in the victim fear as well as gratitude for being allowed to live.
 Sex trafficking operations can be found in highly-visible venues such as street prostitution, as well as more underground systems such as closed-brothels that operate out of residential homes. Sex trafficking also takes place in a variety of public and private locations such as massage parlors, spas, strip clubs and other fronts for prostitution. Victims may start off dancing or stripping in clubs and then be coerced into situations of prostitution and pornography.
Find out about how to identify and interact with victims of human trafficking.
- What would you do if you think you identified someone who might be a victim of human trafficking?
Sources
Sexual Health: Sexual Health , CDC, http://www.cdc.gov/sexualhealth /
Sexuality: Sexuality , NLM, NIH, http://www.nlm.nih.gov/medlineplus/sexualhealth.html
Public Health Approach to Sexual Health: Public Health Approaches to Sexual Health , Centers for Disease Control and Prevention. A Public Health Approach for Advancing Sexual Health in the United States: Rationale and Options for Implementation, Meeting Report of an External Consultation. Atlanta, Georgia: Centers for Disease Control and Prevention; December, 2010. http://www.cdc.gov/sexualhealth/docs/SexualHealthReport-2011-508.pdf
Sex Trafficking: Sex Trafficking Fact Sheet, Administration for Children and Families, USDHHS, www.acf.hhs.gov/trafficking/about/fact_sex.html


3.3: Cloud Computing
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- Ly-Huong T. Pham, Tejal Desai-Naik, Laurie Hammond, & Wael Abdeljabbar
- ASCCC Open Educational Resources Initiative (OERI)
Historically, for software to run on a computer, an individual copy of the software had to be installed on the computer, either from a disk or, more recently, after being downloaded from the Internet. The concept of “cloud” computing changes this model.
“The cloud” refers to applications, services, and data stored in data centers, server farms, and storage servers and accessed by users via the Internet. In most cases, the users don’t know where their data is actually stored. Individuals and organizations use cloud computing.
You probably already use cloud computing in some forms. For example, if you access your email via your web browser, you are using a form of cloud computing. If you use Google Drive’s applications, you are using cloud computing. Simultaneously, these are free versions of cloud computing, big business in providing applications and data storage over the web. Commercial and large applications can also exist on the cloud, such as the entire suite of CRM from Salesforce is offered via the cloud. Cloud computing is not limited to web applications: it can also be used for phone or video streaming services.
Advantages of Cloud Computing
- No software to install or upgrades to maintain.
- Available from any computer that has access to the Internet.
- Can scale to a large number of users easily.
- New applications can be up and running very quickly.
- Services can be leased for a limited time on an as-needed basis.
- Your information is not lost if your hard disk crashes or your laptop is stolen.
- You are not limited by the available memory or disk space on your computer.
Disadvantages of Cloud Computing
- Your information is stored on someone else’s computer
- You must have Internet access to use it. If you do not have access, you’re out of luck.
- You are relying on a third party to provide these services.
- You don’t know how your data is protected from theft or sold by your own cloud service provider.
Cloud computing can greatly impact how organizations manage technology. For example, why is an IT department needed to purchase, configure, and manage personal computers and software when all that is really needed is an Internet connection?
Using a Private Cloud
Many organizations are understandably nervous about giving up control of their data and applications using cloud computing. But they also see the value in reducing the need for installing software and adding disk storage to local computers. A solution to this problem lies in the concept of a private cloud. While there are various private cloud models, the basic idea is for the cloud service provider to rent a specific portion of their server space exclusive to a specific organization. The organization has full control over that server space while still gaining some of the benefits of cloud computing.
Virtualization
One technology that is utilized extensively as part of cloud computing is “virtualization.” Virtualization is using software to create a virtual machine that simulates a computer with an operating system. For example, using virtualization, a single computer that runs Microsoft Windows can host a virtual machine that looks like a computer with a specific Linux-based OS. This ability maximizes the use of available resources on a single machine. Companies such as EMC provide virtualization software that allows cloud service providers to provision web servers to their clients quickly and efficiently. Organizations are also implementing virtualization to reduce the number of servers needed to provide the necessary services. For more detail on how virtualization works, see this informational page from VMWare .


3.9: Summary
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The discipline of anthropology is centered on the concept of culture. What we make, what we do, and what we think constitute the basic elements of culture. These elements combine in aggregates such as symbols, rituals, and social structures. Since the 19th century, anthropologists have developed various modes of analysis for understanding culture, some examining change over time and others considering the functions of culture at one particular point in time. While it is an incredibly useful tool for understanding human social life, the concept of culture is riddled with paradox. Though durable and integrated, culture is subject to constant change, mobility, contest, and variability.


4.3: Fitting Linear Models to Data
Learning Objectives
- Draw and interpret scatter diagrams.
- Use a graphing utility to find the line of best fit.
- Distinguish between linear and nonlinear relations.
- Fit a regression line to a set of data and use the linear model to make predictions.
A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot.
Drawing and Interpreting Scatter Plots
A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model , or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure \(\PageIndex{1}\) shows a sample scatter plot.
Notice this scatter plot does not indicate a linear relationship . The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam.
Example \(\PageIndex{1}\): Using a Scatter Plot to Investigate Cricket Chirps
Table shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[1]. Plot this data, and determine whether the data appears to be linearly related.
| Chirps | 44 | 35 | 20.4 | 33 | 31 | 35 | 18.5 | 37 | 26 |
| Temperature | 80.5 | 70.5 | 57 | 66 | 68 | 72 | 52 | 73.5 | 53 |
Solution
Plotting this data, as depicted in Figure \(\PageIndex{2}\) suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so.
Finding the Line of Best Fit
Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-intercept. We can approximate the slope of the line by extending it until we can estimate the \(\frac{\text{rise}}{\text{run}}\).
Example \(\PageIndex{2}\): Finding a Line of Best Fit
Find a linear function that fits the data in Table \(\PageIndex{1}\) by “eyeballing” a line that seems to fit.
Solution
On a graph, we could try sketching a line.
Using the starting and ending points of our hand drawn line, points \((0, 30)\) and \((50, 90)\), this graph has a slope of
\[m=\dfrac{60}{50}=1.2\]
and a y-intercept at 30. This gives an equation of
\[T(c)=1.2c+30\]
where \(c\) is the number of chirps in 15 seconds, and \(T(c)\) is the temperature in degrees Fahrenheit. The resulting equation is represented in Figure \(\PageIndex{3}\).
Analysis
This linear equation can then be used to approximate answers to various questions we might ask about the trend.
Recognizing Interpolation or Extrapolation
While the data for most examples does not fall perfectly on the line, the equation is our best guess as to how the relationship will behave outside of the values for which we have data. We use a process known as interpolation when we predict a value inside the domain and range of the data. The process of extrapolation is used when we predict a value outside the domain and range of the data.
Figure \(\PageIndex{4}\) compares the two processes for the cricket-chirp data addressed in Example \(\PageIndex{2}\). We can see that interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44. Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.
There is a difference between making predictions inside the domain and range of values for which we have data and outside that domain and range. Predicting a value outside of the domain and range has its limitations. When our model no longer applies after a certain point, it is sometimes called model breakdown . For example, predicting a cost function for a period of two years may involve examining the data where the input is the time in years and the output is the cost. But if we try to extrapolate a cost when \(x=50\), that is in 50 years, the model would not apply because we could not account for factors fifty years in the future.
Interpolation and Extrapolation
Different methods of making predictions are used to analyze data.
- The method of extrapolation involves predicting a value outside the domain and/or range of the data.
- Model breakdown occurs at the point when the model no longer applies.
Example \(\PageIndex{3}\): Understanding Interpolation and Extrapolation
Use the cricket data from Table \(\PageIndex{1}\) to answer the following questions:
- Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
- Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss whether it is reasonable.
Solution
a. The number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model:
\[\begin{align} T(30)&=30+1.2(30) \\ &=66 \text{ degrees} \end{align}\]
Based on the data we have, this value seems reasonable.
b. The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation because 40 is outside the range of our data. Using our model:
\[\begin{align} 40&=30+1.2c \\ 10&=1.2c \\ c&\approx8.33 \end{align}\]
We can compare the regions of interpolation and extrapolation using Figure \(\PageIndex{5}\).
Analysis
Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.
Exercise \(\PageIndex{1}\)
According to the data from Table \(\PageIndex{1}\), what temperature can we predict it is if we counted 20 chirps in 15 seconds?
Solution
54°F
Finding the Line of Best Fit Using a Graphing Utility
While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values[2]. One such technique is called least squares regression and can be computed by many graphing calculators, spreadsheet software, statistical software, and many web-based calculators[3]. Least squares regression is one means to determine the line that best fits the data, and here we will refer to this method as linear regression.
Given data of input and corresponding outputs from a linear function, find the best fit line using linear regression.
- Enter the input in List 1 (L1).
- Enter the output in List 2 (L2).
- On a graphing utility, select Linear Regression (LinReg).
Example \(\PageIndex{4}\): Finding a Least Squares Regression Line
Find the least squares regression line using the cricket-chirp data in Table \(\PageIndex{1}\).
Solution
Enter the input (chirps) in List 1 (L1).
Enter the output (temperature) in List 2 (L2). See Table \(\PageIndex{2}\).
| L1 | 44 | 35 | 20.4 | 33 | 31 | 35 | 18.5 | 37 | 26 |
| L2 | 80.5 | 70.5 | 57 | 66 | 68 | 72 | 52 | 73.5 | 53 |
On a graphing utility, select Linear Regression (LinReg). Using the cricket chirp data from earlier, with technology we obtain the equation:
\[T(c)=30.281+1.143c\]
Analysis
Notice that this line is quite similar to the equation we “eyeballed” but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:
\[\begin{align} T(30)&=30.281+1.143(30) \\ &=64.571 \\ &\approx 64.6 \text{ degrees} \end{align}\]
The graph of the scatter plot with the least squares regression line is shown in Figure \(\PageIndex{6}\).
Will there ever be a case where two different lines will serve as the best fit for the data?
No. There is only one best fit line.
Distinguishing Between Linear and Non-Linear Models
As we saw above with the cricket-chirp model, some data exhibit strong linear trends, but other data, like the final exam scores plotted by age, are clearly nonlinear. Most calculators and computer software can also provide us with the correlation coefficient , which is a measure of how closely the line fits the data. Many graphing calculators require the user to turn a ”diagnostic on” selection to find the correlation coefficient, which mathematicians label as \(r\). The correlation coefficient provides an easy way to get an idea of how close to a line the data falls.
We should compute the correlation coefficient only for data that follows a linear pattern or to determine the degree to which a data set is linear. If the data exhibits a nonlinear pattern, the correlation coefficient for a linear regression is meaningless. To get a sense for the relationship between the value of \(r\) and the graph of the data, Figure \(\PageIndex{7}\) shows some large data sets with their correlation coefficients. Remember, for all plots, the horizontal axis shows the input and the vertical axis shows the output.
Correlation Coefficient
The correlation coefficient is a value, \(r\), between –1 and 1.
- \(r>0\) suggests a positive (increasing) relationship
- \(r<0\) suggests a negative (decreasing) relationship
- The closer the value is to 0, the more scattered the data.
- The closer the value is to 1 or –1, the less scattered the data is.
Example \(\PageIndex{5}\): Finding a Correlation Coefficient
Calculate the correlation coefficient for cricket-chirp data in Table \(\PageIndex{1}\).
Solution
Because the data appear to follow a linear pattern, we can use technology to calculate \(r\). Enter the inputs and corresponding outputs and select the Linear Regression. The calculator will also provide you with the correlation coefficient, \(r=0.9509\). This value is very close to 1, which suggests a strong increasing linear relationship.
Note: For some calculators, the Diagnostics must be turned "on" in order to get the correlation coefficient when linear regression is performed: [2nd]>[0]>[alpha][x–1], then scroll to DIAGNOSTICSON.
Predicting with a Regression Line
Once we determine that a set of data is linear using the correlation coefficient, we can use the regression line to make predictions. As we learned above, a regression line is a line that is closest to the data in the scatter plot, which means that only one such line is a best fit for the data.
Example \(\PageIndex{6}\): Using a Regression Line to Make Predictions
Gasoline consumption in the United States has been steadily increasing. Consumption data from 1994 to 2004 is shown in Table \(\PageIndex{3}\). Determine whether the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.
| Year | '94 | '95 | '96 | '97 | '98 | '99 | '00 | '01 | '02 | '03 | '04 |
| Consumption (billions of gallons) | 113 | 116 | 118 | 119 | 123 | 125 | 126 | 128 | 131 | 133 | 136 |
The scatter plot of the data, including the least squares regression line, is shown in Figure \(\PageIndex{8}\).
We can introduce new input variable, \(t\),representing years since 1994.
The least squares regression equation is:
\[C(t)=113.318+2.209t\]
Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend.
Using this to predict consumption in 2008 \((t=14)\),
\[\begin{align} C(14)&=113.318+2.209(14) \\ &=144.244 \end{align}\]
The model predicts 144.244 billion gallons of gasoline consumption in 2008.
Exercise \(\PageIndex{1}\)
Use the model we created using technology in Example \(\PageIndex{6}\) to predict the gas consumption in 2011. Is this an interpolation or an extrapolation?
- Answer
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150.871 billion gallons; extrapolation
Key Concepts
- Scatter plots show the relationship between two sets of data.
- Scatter plots may represent linear or non-linear models.
- The line of best fit may be estimated or calculated, using a calculator or statistical software.
- Interpolation can be used to predict values inside the domain and range of the data, whereas extrapolation can be used to predict values outside the domain and range of the data.
- The correlation coefficient, \(r\), indicates the degree of linear relationship between data.
- A regression line best fits the data.
- The least squares regression line is found by minimizing the squares of the distances of points from a line passing through the data and may be used to make predictions regarding either of the variables.


10.3: Acid Rain
Acid rain is a term referring to a mixture of wet and dry deposition (deposited material) from the atmosphere containing higher than normal amounts of nitric and sulfuric acids. The precursors, or chemical forerunners, of acid rain formation result from both natural sources, such as volcanoes and decaying vegetation, and man-made sources, primarily emissions of sulfur dioxide (SO 2 ) and nitrogen oxides (NO x ) resulting from fossil fuel combustion. Acid rain occurs when these gases react in the atmosphere with water, oxygen, and other chemicals to form various acidic compounds. The result is a mild solution of sulfuric acid and nitric acid. When sulfur dioxide and nitrogen oxides are released from power plants and other sources, prevailing winds blow these compounds across state and national borders, sometimes over hundreds of miles.
Acid rain is measured using a scale called “pH.” The lower a substance’s pH, the more acidic it is. Pure water has a pH of 7.0. However, normal rain is slightly acidic because carbon dioxide (CO 2 ) dissolves into it forming weak carbonic acid, giving the resulting mixture a pH of approximately 5.6 at typical atmospheric concentrations of CO 2 . As of 2000, the most acidic rain falling in the U.S. has a pH of about 4.3.
Effects of Acid Rain
Acid rain causes acidification of lakes and streams and contributes to the damage of trees at high elevations (for example, red spruce trees above 2,000 feet) and many sensitive forest soils. In addition, acid rain accelerates the decay of building materials and paints, including irreplaceable buildings, statues, and sculptures that are part of our nation’s cultural heritage. Prior to falling to the earth, sulfur dioxide (SO 2 ) and nitrogen oxide (NO x ) gases and their particulate matter derivatives—sulfates and nitrates—contribute to visibility degradation and harm public health.
The ecological effects of acid rain are most clearly seen in the aquatic, or water, environments, such as streams, lakes, and marshes. Most lakes and streams have a pH between 6 and 8, although some lakes are naturally acidic even without the effects of acid rain. Acid rain primarily affects sensitive bodies of water, which are located in watersheds whose soils have a limited ability to neutralize acidic compounds (called “buffering capacity”). Lakes and streams become acidic (i.e., the pH value goes down) when the water itself and its surrounding soil cannot buffer the acid rain enough to neutralize it. In areas where buffering capacity is low, acid rain releases aluminum from soils into lakes and streams; aluminum is highly toxic to many species of aquatic organisms. Acid rain causes slower growth, injury, or death of forests . Of course, acid rain is not the only cause of such conditions. Other factors contribute to the overall stress of these areas, including air pollutants, insects, disease, drought, or very cold weather. In most cases, in fact, the impacts of acid rain on trees are due to the combined effects of acid rain and these other environmental stressors.
Acid rain and the dry deposition of acidic particles contribute to the corrosion of metals (such as bronze) and the deterioration of paint and stone (such as marble and limestone). These effects significantly reduce the societal value of buildings, bridges, cultural objects (such as statues, monuments, and tombstones), and cars (Figure \(\PageIndex{2}\)).
Sulfates and nitrates that form in the atmosphere from sulfur dioxide (SO 2 ) and nitrogen oxides (NO x ) emissions contribute to visibility impairment, meaning we cannot see as far or as clearly through the air. The pollutants that cause acid rain—sulfur dioxide (SO 2 ) and nitrogen oxides (NO x )—damage human health . These gases interact in the atmosphere to form fine sulfate and nitrate particles that can be transported long distances by winds and inhaled deep into people’s lungs. Fine particles can also penetrate indoors. Many scientific studies have identified a relationship between elevated levels of fine particles and increased illness and premature death from heart and lung disorders, such as asthma and bronchitis.
Contributors and Attributions
- Essentials of Environmental Science by Kamala Doršner is licensed under CC BY 4.0 . Modified from the original by Matthew R. Fisher.


12.5.3: Broadband
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The Growth of Broadband
In the early days of the Internet, most access was done via a modem over an analog telephone line. A modem (short for “modulator-demodulator”) was connected to the incoming phone line and a computer in order to connect you to a network. Speeds were measured in bits-per-second (bps), with speeds growing from 1200 bps to 56,000 bps over the years. Connection to the Internet via these modems is called dial-up access. Dial-up was very inconvenient because it tied up the phone line. As the web became more and more interactive, dial-up also hindered usage, as users wanted to transfer more and more data. As a point of reference, downloading a typical 3.5 mb song would take 24 minutes at 1200 bps and 2 minutes at 28,800 bps.
A broadband connection is defined as one that has speeds of at least 256,000 bps, though most connections today are much faster, measured in millions of bits per second (megabits or mbps) or even billions (gigabits). For the home user, a broadband connection is usually accomplished via the cable television lines or phone lines (DSL). Both cable and DSL have similar prices and speeds, though each individual may find that one is better than the other for their specific area. Speeds for cable and DSL can vary during different times of the day or week, depending upon how much data traffic is being used. In more remote areas, where cable and phone companies do not provide access, home Internet connections can be made via satellite. The average home broadband speed is anywhere between 3 mbps and 30 mbps. At 10 mbps, downloading a typical 3.5 mb song would take less than a second. For businesses who require more bandwidth and reliability, telecommunications companies can provide other options, such as T1 and T3 lines.
Broadband access is important because it impacts how the Internet is used. When a community has access to broadband, it allows them to interact more online and increases the usage of digital tools overall. Access to broadband is now considered a basic human right by the United Nations, as declared in their 2011 statement :
“Broadband technologies are fundamentally transforming the way we live,” the Broadband Commission for Digital Development, set up last year by the UN Educational Scientific and Cultural Organization ( UNESCO ) and the UN International Telecommunications Union ( ITU ), said in issuing “ The Broadband Challenge ” at a leadership summit in Geneva.
“It is vital that no one be excluded from the new global knowledge societies we are building. We believe that communication is not just a human need – it is a right.” [1]


5.10F: Anammox
- Describe the overall process of ANaerobic AMMonium OXidation (Anammox) and its purpose
Anammox, an abbreviation for ANaerobic AMMonium OXidation, is a globally significant microbial process of the nitrogen cycle. The bacteria mediating this process were identified in 1999, and at the time were a great surprise to the scientific community. Anammox takes place in many natural environments, contributing up to 50% of the dinitrogen gas produced in the oceans. In this biological process, nitrite and ammonium are converted directly into dinitrogen gas. The overall catabolic reaction is:
NH 4 + + NO 2 − → N 2 + 2H 2 O.
This form of metabolism involves the coupling of ammonia oxidation to nitrite reduction. Since oxygen is not required for the process, these organisms are strict anaerobes. Amazingly, hydrazine (N 2 H 4 — rocket fuel) is produced as an intermediate during anammox metabolism. To deal with the high toxicity of hydrazine, anammox bacteria have a hydrazine-containing intracellular organelle called the anammoxasome (a compartment inside the cytoplasm which is the locus of anammox catabolism), which is surrounded by an unusual and highly compact ladderane lipid membrane. Further, the membranes of these bacteria mainly consist of ladderane lipids so far unique in biology. Of special interest is the conversion to hydrazine (normally used as a high-energy rocket fuel, and poisonous to most living organisms) as an intermediate. A final striking feature of the organism is the extremely slow growth rate. The doubling time is nearly two weeks. The anammox process was originally found to occur only from 20°C to 43°C but more recently, anammox has been observed at temperatures from 36°C to 52°C in hot springs and 60°C to 85°C at hydrothermal vents located along the Mid-Atlantic Ridge.
Anammox organisms are autotrophs although the mechanism for carbon dioxide fixation is still unclear. Because of this property, these organisms could be used industrially to remove nitrogen in wastewater treatment processes. The bacteria that perform the anammox process belong to the bacterial phylum Planctomycetes (e.g. Candidatus Brocadia anammoxidans), of which Planctomyces and Pirellula are the best known genera. Currently five genera of anammox bacteria have been (provisionally) defined: Brocadia, Kuenenia, Anammoxoglobus, Jettenia (all fresh water species), and Scalindua (marine species).
Key Points
- The bacteria mediating this process were identified in 1999, and at the time were a great surprise for the scientific community.
- This form of metabolism occurs in members of the Planctomycetes (e.g. Candidatus Brocadia anammoxidans) and involves the coupling of ammonia oxidation to nitrite reduction.
- To deal with the high toxicity of hydrazine, anammox bacteria contain a hydrazine-containing intracellular organelle called the anammoxasome, surrounded by highly compact ladderane lipid membrane. These lipids are unique in nature, as is the use of hydrazine as a metabolic intermediate.
Key Terms
- Anammox : An abbreviation for ANaerobic AMMonium OXidation, a globally significant microbial process of the nitrogen cycle.
- anaerobes : Organisms that do not require oxygen for growth.
- ladderane : Any of a class of polycyclic hydrocarbons, consisting of repeating cyclobutane moieties, that resemble ladders


13.4: Problems on Transform Methods
Exercise \(\PageIndex{1}\)
Calculate directly the generating function \(g_X(s)\) for the geometric \((p)\) distribution.
- Answer
-
\(g_X (s) = E[s^2] = \sum_{k = 0}^{\infty} p_k s^k = p \sum_{k = 0}^{\infty} q^k s^k = \dfrac{p}{1 - qs}\) (geometric series)
Exercise \(\PageIndex{2}\)
Calculate directly the generating function \(g_X(s)\) for the Poisson \((\mu)\) distribution.
- Answer
-
\(g_X (s) = E[s^X] = \sum_{k = 0}^{\infty} p_k s^k = e^{-\mu} \sum_{k = 0}^{\infty} \dfrac{\mu^k s^k}{k!} = e^{-\mu} e^{\mu s} = e^{\mu (s - 1)}\)
Exercise \(\PageIndex{3}\)
A projection bulb has life (in hours) represented by \(X\) ~ exponential (1/50). The unit will be replaced immediately upon failure or at 60 hours, whichever comes first. Determine the moment generating function for the time \(Y\) to replacement.
- Answer
-
\(Y = I_{[0, a]} (X) X + I_{(a, \infty)} (X) a\) \(e^{sY} = I_{[0, a)} (X) e^{sX} + I_{(a, \infty) (X) e^{as}\)
\(M_Y (s) = \int_{0}^{a} e^{st} \lambda e^{-\lambda t}\ dt + s^{sa} \int_{a}^{\infty} \lambda e^{-\lambda t}\ dt\)
\(= \dfrac{\lambda}{\lambda - s} [1 - e^{(\lambda - s) a}] + e^{-(\lambda - s) a}\)
Exercise \(\PageIndex{4}\)
Simple random variable \(X\) has distribution
\(X =\) [-3 -2 0 1 4] \(PX =\) [0.15 0.20 0.30 0.25 0.10]
a. Determine the moment generating function for \(X\)
b. Show by direct calculation the \(M_X' (0) = E[X]\) and \(M_X'' (0) = E[X^2]\).
- Answer
-
\(M_X (s) = 0.15 e^{-3s} + 0.20 e^{-2s} + 0.30 + 0.25 e^s + 0.10 e^{4s}\)
\(M_X' (s) = -3 \cdot 0.15 e^{-3s} - 2 \cdot 0.20 e^{-2s} + 0 + 0.25 e^{s} + 4 \cdot 0.10 e^{4s}\)
\(M_X''(s) = (-3)^2 \cdot 0.15 e^{-3s} + (-2)^2 \cdot 0.20 e^{-2s} + 0 + 0.25 e^{s} + 4^2 \cdot 0.10 e^{4s}\)
Setting \(s = 0\) and using \(e^0 = 1\) give the desired results.
Exercise \(\PageIndex{5}\)
Use the moment generating function to obtain the variances for the following distributions
EXponential \((\lambda)\) Gamma (\(\alpha, \lambda\)) Normal (\(\mu, \sigma^2\))
- Answer
-
a. Exponential:
\(M_X(s) = \dfrac{\lambda}{\lambda - s}\) \(M_X'(s) = \dfrac{\lambda}{(\lambda - s)^2}\) \(M_X''(s) = \dfrac{2\lambda}{(\lambda - s)^3}\)
\(E[X] = \dfrac{\lambda}{\lambda^2} = \dfrac{1}{\lambda}\) \(E[X^2] = \dfrac{2\lambda}{\lambda^3} = \dfrac{2}{\lambda^2}\) \(\text{Var}[X] = \dfrac{2}{\lambda^2} - (\dfrac{1}{\lambda})^2= \dfrac{1}{\lambda^2}\)
b. Gamma (\(\alpha, \lambda\)):
\(M_X (s) = (\dfrac{\lambda}{\lambda - s})^{\alpha}\) \(M_X' (s) = \alpha (\dfrac{\lambda}{\lambda - s})^{\alpha - 1}\) \(\dfrac{\lambda}{(\lambda - s)^2} = \alpha (\dfrac{\lambda}{\lambda - s})^{\alpha} \dfrac{1}{\lambda - s}\)
\(M_X'' (s) = \alpha^2 (\dfrac{\lambda}{\lambda - s})^{\alpha}\dfrac{1}{\lambda - s} \dfrac{1}{\lambda - s} + \alpha (\dfrac{\lambda}{\lambda - s})^{\alpha} \dfrac{1}{(\lambda - s)^2}\)
\(E[X] =\dfrac{\alpha}{\lambda}\) \(E[X^2] =\dfrac{\alpha^2 + \alpha}{\lambda^2}\) \(\text{Var} [X] = \dfrac{\alpha}{\lambda^2}\)
c. Normal(\(\mu, \sigma\)):
\(M_X (s) = \text{exp} (\dfrac{\sigma^2 s^2}{2} + \mu s)\) \(M_X'(s) = M_X (s) \cdot (\sigma^2 s + \mu)\)
\(M_X''(s) = M_X (s) \cdot (\sigma^2 s + \mu)^2 + M_X (s) \sigma^2\)
\(E[X] = \mu\) \(E[X^2] = \mu^2 + \sigma^2\) \(\text{Var} [X] = \sigma^2\)
Exercise \(\PageIndex{6}\)
The pair \(\{X, Y\}\) is iid with common moment generating function \(\dfrac{\lambda^3}{(\lambda - s)^3}\). Determine the moment generating function for \(Z = 2X - 4Y + 3\).
- Answer
-
\(M_Z(s) = e^{3s} (\dfrac{\lambda}{\lambda - 2s})^3 (\dfrac{\lambda}{\lambda + 4s})^3\)
Exercise \(\PageIndex{7}\)
The pair \(\{X, Y\}\) is iid with common moment generating function \(M_X (s) = (0.6 + 0.4e^s)\). Determine the moment generating function for \(Z = 5X + 2Y\).
- Answer
-
\(M_Z (s) = (0.6 + 0.4e^{5s})(0.6 + 0.4e^{2s})
Exercise \(\PageIndex{8}\)
Use the moment generating function for the symmetric triangular distribution on \((-c, c)\) as derived in the section "Three Basic Transforms".
- Obtain an expression for the symmetric triangular distribution on \((a, b)\) for any \(a < b\).
- Use the result of part (a) to show that the sum of two independent random variables uniform on \((a, b)\) has symmetric triangular distribution on \((2a, 2b)\).
- Answer
-
Let \(m = (a + b)/2\) and \(c = (b - a)/2\). If \(Y\) ~ symetric triangular on \((-c, c)\), then \(X = Y + m\) is symmetric triangular on \((m - c, m + c) = (a, b)\) and
\(M_X (s) = e^{ms} M_Y (s) = \dfrac{e^{cs} + e^{-cs} - 2}{c^2s^2} e^{ms} = \dfrac{e^{(m + c)s} + e^{(m - c)s} - 2e^{ms}}{c^2s^2} = \dfrac{e^{hs} + e^{as} - 2e^{\dfrac{a+b}{2}s}}{(\dfrac{b - a}{2})^2s^2}\)
\(M_{X + Y} (s) = [\dfrac{e^{sb} - e^{sa}}{s(b - a)}]^2 = \dfrac{e^{s2b}+ e^{s2a} - 2e^{s(b + a)}}{s^2 (b - a)^2}\)
Exercise \(\PageIndex{9}\)
Random variable \(X\) has moment generating function \(\dfrac{p^2}{(1 - qe^s)^2}\).
a. Use derivatives to determine \(E[X]\) and \(\text{Var} [X]\).
b. Recognize the distribution from the form and compare \(E[X]\) and \(\text{Var} [X]\) with the result of part (a).
- Answer
-
\([p^2 (1 - qe^s)^{-2}]' = \dfrac{2p^2qe^s}{(1 - qe^s)^3}\) so that \(E[X] = 2q/p\)
\([p^2 (1 - qe^s)^{-2}]'' = \dfrac{6p^2 q^2 e^s}{(1 - qe^s)^4} + \dfrac{2p^2qe^s}{(1 - qe^s)^3}\) so that \(E[X^2] = \dfrac{6q^2}{p^2} + \dfrac{2q}{p}\)
\(\text{Var} [X] = \dfrac{2q^2}{p^2} + \dfrac{2q}{p} = \dfrac{2(q^2 + pq)}{p^2} = \dfrac{2q}{p^2}\)
\(X\) ~ negative binomial \((2, p)\), which has \(E[X] = 2q/p\) and \(\text{Var} [X] = 2q/p^2\).
Exercise \(\PageIndex{10}\)
The pair \(\{X, Y\}\) is independent. \(X\) ~ Poisson (4) and \(Y\) ~ geometric (0, 3). Determine the generating function \(g_Z\) for \(Z = 3X + 2Y\).
- Answer
-
\(g_Z (s) = g_X (s^3) g_Y (s^2) = e^{4(s^3-1)} \cdot \dfrac{0.3}{1 - qs^2}\)
Exercise \(\PageIndex{11}\)
Random variable \(X\) has moment generating function
\(M_X (s) = \dfrac{1}{1 - 3s} \cdot \text{exp} (16s^2/2 + 3s)\)
By recognizing forms and using rules of combinations, determine \(E[X]\) and \(\text{Var} [X]\).
- Answer
-
\(X = X_1 + X_2\) with \(X_1\) ~ exponential (1/3) \(X_2\) ~ \(N\)(3, 16)
\(E[X] = 3 + 3 = 6\) \(\text{Var} [X] = 9 + 16 = 25\)
Exercise \(\PageIndex{12}\)
Random variable \(X\) has moment generating function
\(M_X (s) = \dfrac{\text{exp} (3(e^s - 1))}{1 - 5s} \cdot \text{exp} (16s^2/2 + 3s)\)
By recognizing forms and using rules of combinations, determine \(E[X]\) and \(\text{Var} [X]\).
- Answer
-
\(X = X_1 + X_2 + X_3\), with \(X_1\) ~ Poisson (3), \(X_2\) ~ exponential (1/5), \(X_3\0 ~ \(N\) (3, 16)
\(E[X] = 3 + 5 + 3 = 11\) \(\text{Var} [X] = 3 + 25 + 16 = 44\)
Exercise \(\PageIndex{13}\)
Suppose the class \(\{A, B, C\}\) of events is independent, with respective probabilities 0.3, 0.5, 0.2. Consider
\(X = -3I_A + 2I_B + 4I_C\)
a. Determine the moment generating functions for and use properties of moment generating functions to determine the moment generating function for \(X\).
b. Use the moment generating function to determine the distribution for \(X\).
c. Use canonic to determine the distribution. Compare with result (b).
d. Use distributions for the separate terms; determine the distribution for the sum with mgsum3. Compare with result (b).
- Answer
-
\(M_X (s) = (0.7 + 0.3 e^{-3s})(0.5 + 0.5 e^{2s}) (0.8 + 0.2 e^{4s}) =\)
\(0.12 e^{-3s} + 0.12 e^{-s} + 0.28 + 0.03 e^{s} + 0.28 e^{2s} + 0.03 e^{3s} + 0.07 e^{4s} + 0.07 e^{6s}\)
The distribution is
\(X = \) [-3 -1 0 1 2 3 4 6] \(PX =\) [0.12 0.12 0.28 0.03 0.28 0.03 0.07 0.07]
c = [-3 2 4 0]; P = 0.1*[3 5 2]; canonic Enter row vector of coefficients c Enter row vector of minterm probabilities minprob(P) Use row matrices X and PX for calculations Call for XDBN to view the distribution P1 = [0.7 0.3]; P2 = [0.5 0.5]; P3 = [0.8 0.2]; X1 = [0 -3]; X2 = [0 2]; X3 = [0 4]; [x,px] = mgsum3(X1,X2,X3,P1,P2,P3); disp([X;PX;x;px]') -3.0000 0.1200 -3.0000 0.1200 -1.0000 0.1200 -1.0000 0.1200 0 0.2800 0 0.2800 1.0000 0.0300 1.0000 0.0300 2.0000 0.2800 2.0000 0.2800 3.0000 0.0300 3.0000 0.0300 4.0000 0.0700 4.0000 0.0700 6.0000 0.0700 6.0000 0.0700
Exercise \(\PageIndex{14}\)
Suppose the pair \(\{X, Y\}\) is independent, with both \(X\) and \(Y\) binomial. Use generating functions to show under what condition, if any, \(X + Y\) is binomial.
- Answer
-
Binomial iff both have same \(p\), as shown below.
\(g_{X + Y} (s) = (q_1 + p_1 s)^n (q_2 + p_2s)^m = (q + ps)^{n + m}\) iff \(p_1 = p_2\)
Exercise \(\PageIndex{15}\)
Suppose the pair \(\{X, Y\}\) is independent, with both \(X\) and \(Y\) Poisson.
a. Use generating functions to show under what condition \(X + Y\) is Poisson.
b. What about \(X - Y\)? Justify your answer.
- Answer
-
Always Poisson, as the argument below shows.
\(g_{X + Y} (s) = e^{\mu(s - 1)} e^{v(s - 1)} = e^{(\mu + v) (s - 1)}\)
However, \(Y\) ~ \(X\) could have negative values.
Exercise \(\PageIndex{16}\)
Suppose the pair \(\{X, Y\}\) is independent, \(Y\) is nonnegative integer-valued, \(X\) is Poisson and \(X + Y\) is Poisson. Use the generating functions to show that \(Y\) is Poisson.
- Answer
-
\(E[X+Y] = \mu + v\), where \(v = E[Y] > 0\), \(g_X (s) = e^{\mu(s - 1)}\) and \(g_{X + Y} (s) = g_X (s) g_Y (s) = e^{(\mu + s) (s - 1)\). Division by \(g_X (s)\) gives \(g_Y (s) = e^{v(s - 1)}\).
Exercise \(\PageIndex{17}\)
Suppose the pair \(\{X, Y\}\) is iid, binomial (6, 0.51). By the result of Exercise 13.4.14
\(X + Y\) is binomial. Use mgsum to obtain the distribution for \(Z = 2X + 4Y\). Does \(Z\) have the binomial distribution? Is the result surprising? Examine the first few possible values for \(Z\). Write the generating function for \(Z\); does it have the form for the binomial distribution?

Let \(G = g(X) = 3X^2 - 2X\) and \(H = h(Y) = 2Y^2 + Y + 3\).
a. Use the mgsum to obtain the distribution for \(G + H\).
b. Use icalc and csort to obtain the distribution for \(G + H\) and compare with the result of part (a).
- Answer
-
X = 0:5; Y = 0:7; PX = ibinom(5,0.33,X); PY = ibinom(7,0.47,Y); G = 3*X.^2 - 2*X; H = 2*Y.^2 + Y + 3; [Z,PZ] = mgsum(G,H,PX,PY); icalc Enter row matrix of X-values X Enter row matrix of Y-values Y Enter X probabilities PX Enter Y probabilities PY Use array operations on matrices X, Y, PX, PY, t, u, and P M = 3*t.^2 - 2*t + 2*u.^2 + u + 3; [z,pz] = csort(M,P); e = max(abs(pz - PZ)) % Comparison of p values e = 0
Exercise \(\PageIndex{19}\)
Suppose the pair \(\{X, Y\}\) is independent, with \(X\) ~ binomial (8, 0.39) and \(Y\) ~ uniform on {-1.3, -0.5, 1.3, 2.2, 3.5}. Let
\(U = 3X^2 - 2X + 1\) and \(V = Y^3 + 2Y - 3\)
a. Use mgsum to obtain the distribution for \(U + V\).
b. Use icalc and csort to obtain the distribution for \(U + V\) and compare with the result of part (a).
- Answer
-
X = 0:8; Y = [-1.3 -0.5 1.3 2.2 3.5]; PX = ibinom(8,0.39,X); PY = (1/5)*ones(1,5); U = 3*X.^2 - 2*X + 1; V = Y.^3 + 2*Y - 3; [Z,PZ] = mgsum(U,V,PX,PY); icalc Enter row matrix of X-values X Enter row matrix of Y-values Y Enter X probabilities PX Enter Y probabilities PY Use array operations on matrices X, Y, PX, PY, t, u, and P M = 3*t.^2 - 2*t + 1 + u.^3 + 2*u - 3; [z,pz] = csort(M,P); e = max(abs(pz - PZ)) e = 0
Exercise \(\PageIndex{20}\)
If \(X\) is a nonnegative integer-valued random variable, express the generating function as a power series.
a. Show that the \(k\)th derivative at \(s = 1\) is
\(g_X^{(k)} (1) = E[X(X - 1)(X - 2) \cdot \cdot \cdot (X - k + 1)]\)
b. Use this to show the \(\text{Var} [X] = g_X''(1) + g_X'(1) - [g_X'(1)]^2\).
- Answer
-
Since power series may be differentiated term by term
\(g_X^{(n)} (s) = \sum_{k = 0}^{\infty} k (k - 1) \cdot (k - n + 1) p_k s^{k - n}\) so that
\(g_X^{(n)} (1) = \sum_{k = 0}^{\infty} k(k - 1) \cdot (k - n + 1) p_k = E[X(X - 1) \cdot\cdot\cdot (X - n + 1)]\)
\(\text{Var} [X] = E[X^2] - E^2[X] = E[X(X - 1)] + E[X] - E^2[X] = g_X''(1) + g_X' (1) - [g_X'(1)]^2\)
Exercise \(\PageIndex{21}\)
Let \(M_X (\cdot)\) be the moment generating function for \(X\).
a. Show that \(\text{Var}[X]\) is the second derivative of \(e^{-s\mu} M_X(s)\) evaluated at \(s = 0\).
b. Use this fact to show that \(X\) ~ \(N(\mu, \sigma^2)\), then \(\text{Var} [X] = \sigma^2\).
- Answer
-
\(f(s) = e^{-s \mu} M_X (s)\) \(f''(s) = e^{-s\mu} [-\mu M_X' (s) + \mu^2 M_X (s) + M_X''(s) - \mu M_X'(s)]\)
Setting \(s = 0\) and using the result on moments gives
\(f''(0) = -\mu^2 + \mu^2 + E[X^2] - \mu^2 = \text{Var} [X]\)
Exercise \(\PageIndex{22}\)
Use derivatives of \(M_{M_m} (s)\) to obtain the mean and variance of the negative binomial (\(m, p\)) distribution.
- Answer
-
To simplify writing use \(f(s)\) for \(M_X (S)\).
\(f(s) = \dfrac{p^m}{(1 - qe^s)^m}\) \(f'(s) = \dfrac{mp^mqe^s}{(1 - qe^s)^{m + 1}}\) \(f''(s) = \dfrac{mp^m qe^s}{1 - qe^s)^{m + 1}} + \dfrac{m(m+1) p^m q^2 e^{2s}}{1 - qe^s)^{m + 2}}\)
\(E[X] = \dfrac{mp^m q}{(1 - q)^{m + 1}} = \dfrac{mq}{p}\) \(E[X^2] = \dfrac{mq}{p} + \dfrac{m(m+1)p^mq^2}{(1-q)^{m + 2}}\)
\(\text{Var} [X] = \dfrac{mq}{p} + \dfrac{m(m + 1) q^2}{p^2} - \dfrac{m^2 q^2}{p^2} = \dfrac{mq}{p^2}\)
Exercise \(\PageIndex{23}\)
Use moment generating functions to show that variances add for the sum or difference of independent random variables.
- Answer
-
To simplify writing, set \(f(s) = M_X (s)\), \(g(s) = M_Y (s)\), and \(h(s) = M_X (s) M_Y(s)\)
\(h'(s) = f'(s) g(s) + f(s) g'(s)\) \(h''(s) = f''(s) g(s) + f'(s) g'(s) + f'(s) g'(s) + f(s) g''(s)\)
Setting \(s = 0\) yields
\(E[X + Y] = E[X] + E[Y]\) \(E[(X + Y)^2] = E[X^2] + 2E[X]E[Y] + E[Y^2]\) \(E^2 [X + Y] = E^2[X] + 2E[X] E[Y] + E^2[Y]\)
Taking the difference gives \(\text{Var}[X + Y] = \text{Var} [X] + \text{Var} [Y]\). A similar treatment with \(g(s)\) replaced by \(g(-s)\) shows \(\text{Var} [X - Y] = \text{Var} [X] + \text{Var} [Y]\).
Exercise \(\PageIndex{24}\)
The pair \(\{X, Y\}\) is iid \(N\)(3,5). Use the moment generating function to show that \(Z = 2X - 2Y + 3\) is normal (see Example 3 from " Transform Methods " for general result).
- Answer
-
\(M_{3X} (s) = M_X (3s) = \text{exp} (\dfrac{9 \cdot 5s^2}{2} + 3 \cdot 3s)\) \(M_{-2Y} (s) = M_Y(-2s) = \text{exp} (\dfrac{4 \cdot 5s^2}{2} - 2 \cdot 3s)\)
\(M_Z (s) = e^{3s} \text{exp} (\dfrac{(45 + 20)s^2}{2} + (9 - 6) s) = \text{exp} (\dfrac{65s^2}{2} + 6s)\)
Exercise \(\PageIndex{25}\)
Use the central limit theorem to show that for large enough sample size (usually 20 or more), the sample average
\(A_n = \dfrac{1}{n} \sum_{i = 1}^{n} X_i\)
is approximately \(N(\mu, \sigma^2/n)\) for any reasonable population distribution having mean value \(\mu\) and variance \(\sigma^2\).
- Answer
-
\(E[A_n] = \dfrac{1}{n} \sum_{i = 1}^{n} \mu = \mu\) \(\text{Var} [A_n] = \dfrac{1}{n^2} \sum_{i = 1}^{n} \sigma^2 = \dfrac{\sigma^2}{n}\)
By the central limit theorem, \(A_n\) is approximately normal, with the mean and variance above.
Exercise \(\PageIndex{26}\)
A population has standard deviation approximately three. It is desired to determine the sample size n needed to ensure that with probability 0.95 the sample average will be within 0.5 of the mean value.
- Use the Chebyshev inequality to estimate the needed sample size.
- Use the normal approximation to estimate \(n\) (see Example 1 from " Simple Random Samples and Statistics ").
- Answer
-
Chevyshev inequality:
\(P(\dfrac{|A_n - \mu|}{\sigma/\sqrt{n}} \ge \dfrac{0.5 \sqrt{n}}{3}) \le \dfrac{3^2}{0.5^2 n} \le 0.05\) implies \(n \ge 720\)
Normal approximation: Use of the table in Example 1 from " Simple Random Samples and Statistics " shows
\(n \ge (3/0.5)^2 3.84 = 128\)


Unfortunately, sensationalized media exposure of violent youth has led to exaggerated public fear of juvenile crime, get tough legislation, and a perceived need to “do something” about juvenile crime.
This punitive position is nothing new. Before the inception of the juvenile justice system a mere 100 years ago, youth were treated the same as adults. They were considered culpable for their actions and housed alongside adult offenders in jails and prisons. Recent research has utilized neuroscience to support the need to treat juveniles differently because they are different. The sections of the brain that govern characteristics associated with moral culpability do not stop maturing until the early 20s. Therefore, it is assumed that someone under age 20, such as a juvenile delinquent, has an underdeveloped brain.
Ted Talks
: Stephen Case The youth crime ‘problem’ is examined as a social construction and moral panic created by institutions in Western societies. The talk traces the evolution of youth crime into a phenomenon persistently misrepresented as an escalating social epidemic. The developmental life stages of ‘childhood’ and ‘adolescence’ as inventions are explored, highlighting differences between young people and adults. In this way, ‘youth crime’ can be identified as a social problem requiring distinct responses. A running theme is a child as a source of adult anxiety and fear, motivating societies to create structures, processes, theories, and images of youth crime that punish lawbreakers. The ‘solution’ is the ‘positive youth justice’ model. Children should not be punished as if they are adults but their criminal behavior should be seen as a normal part of growing up. Instead, they should be worked with to meet their needs, to embrace their human rights and to promote their life chances.
https://www.youtube.com/watch?v=QYWPyiZIpV8


14.6: The Emergence of Drug Resistance
- Describe how the Kirby-Bauer disk diffusion test determines the susceptibility of a microbe to an antibacterial drug.
- Explain the significance of the minimal inhibitory concentration and the minimal bactericidal concentration relative to the effectiveness of an antimicrobial drug.
Testing the effectiveness of antimicrobial drugs against specific organisms is important in identifying their spectrum of activity and the therapeutic dosage. This type of test, generally described as antimicrobial susceptibility testing (AST), is commonly performed in a clinical laboratory. In this section, we will discuss common methods of testing the effectiveness of antimicrobials.
The Kirby-Bauer Disk Diffusion Test
The Kirby-Bauer disk diffusion test has long been used as a starting point for determining the susceptibility of specific microbes to various antimicrobial drugs. The Kirby-Bauer assay starts with a Mueller-Hinton agar plate on which a confluent lawn is inoculated with a patient’s isolated bacterial pathogen. Filter paper disks impregnated with known amounts of antibacterial drugs to be tested are then placed on the agar plate. As the bacterial inoculum grows, antibiotic diffuses from the circular disk into the agar and interacts with the growing bacteria. Antibacterial activity is observed as a clear circular zone of inhibition around the drug-impregnated disk, similar to the disk-diffusion assay. The diameter of the zone of inhibition, measured in millimeters and compared to a standardized chart, determines the susceptibility or resistance of the bacterial pathogen to the drug.
There are multiple factors that determine the size of a zone of inhibition in this assay, including drug solubility, rate of drug diffusion through agar, the thickness of the agar medium, and the drug concentration impregnated into the disk. Due to a lack of standardization of these factors, interpretation of the Kirby-Bauer disk diffusion assay provides only limited information on susceptibility and resistance to the drugs tested. The assay cannot distinguish between bacteriostatic and bactericidal activities, and differences in zone sizes cannot be used to compare drug potencies or efficacies. Comparison of zone sizes to a standardized chart will only provide information on the antibacterials to which a bacterial pathogen is susceptible or resistant.
How does one use the information from a Kirby-Bauer assay to predict the therapeutic effectiveness of an antimicrobial drug in a patient?
Unfortunately, infectious diseases don’t take a time-out for lab work. As a result, physicians rarely have the luxury of conducting susceptibility testing before they write a prescription. Instead, they rely primarily on the empirical evidence (i.e., the signs and symptoms of disease) and their professional experience to make an educated guess as to the diagnosis, causative agent(s), and drug most likely to be effective. This approach allows treatment to begin sooner so the patient does not have to wait for lab test results. In many cases, the prescription is effective; however, in an age of increased antimicrobial resistance, it is becoming increasingly more difficult to select the most appropriate empiric therapy. Selecting an inappropriate empiric therapy not only puts the patient at risk but may promote greater resistance to the drug prescribed.
Recently, studies have shown that antibiograms are useful tools in the decision-making process of selecting appropriate empiric therapy. An antibiogram is a compilation of local antibiotic susceptibility data broken down by bacterial pathogen. In a November 2014 study published in the journal Infection Control and Hospital Epidemiology , researchers determined that 85% of the prescriptions ordered in skilled nursing facilities were decided upon empirically, but only 35% of those prescriptions were deemed appropriate when compared with the eventual pathogen identification and susceptibility profile obtained from the clinical laboratory. However, in one nursing facility where use of antibiograms was implemented to direct selection of empiric therapy, appropriateness of empiric therapy increased from 32% before antibiogram implementation to 45% after implementation of antibiograms. 1 Although these data are preliminary, they do suggest that health-care facilities can reduce the number of inappropriate prescriptions by using antibiograms to select empiric therapy, thus benefiting patients and minimizing opportunities for antimicrobial resistance to develop.
Dilution Tests
As discussed, the limitations of the Kirby-Bauer disk diffusion test do not allow for a direct comparison of antibacterial potencies to guide selection of the best therapeutic choice. However, antibacterial dilution tests can be used to determine a particular drug’s minimal inhibitory concentration (MIC), the lowest concentration of drug that inhibits visible bacterial growth, and minimal bactericidal concentration (MBC), the lowest drug concentration that kills ≥99.9% of the starting inoculum. Determining these concentrations helps identify the correct drug for a particular pathogen. For the macrobroth dilution assay, a dilution series of the drug in broth is made in test tubes and the same number of cells of a test bacterial strain is added to each tube (Figure \(\PageIndex{1}\)). The MIC is determined by examining the tubes to find the lowest drug concentration that inhibits visible growth; this is observed as turbidity (cloudiness) in the broth. Tubes with no visible growth are then inoculated onto agar media without antibiotic to determine the MBC. Generally, serum levels of an antibacterial should be at least three to five times above the MIC for treatment of an infection.
The MIC assay can also be performed using 96-well microdilution trays, which allow for the use of small volumes and automated dispensing devices, as well as the testing of multiple antimicrobials and/or microorganisms in one tray (Figure \(\PageIndex{2}\)). MICs are interpreted as the lowest concentration that inhibits visible growth, the same as for the macrobroth dilution in test tubes. Growth may also be interpreted visually or by using a spectrophotometer or similar device to detect turbidity or a color change if an appropriate biochemical substrate that changes color in the presence of bacterial growth is also included in each well.
The Etest is an alternative method used to determine MIC, and is a combination of the Kirby-Bauer disk diffusion testand dilution methods. Similar to the Kirby-Bauer assay, a confluent lawn of a bacterial isolate is inoculated onto the surface of an agar plate. Rather than using circular disks impregnated with one concentration of drug, however, commercially available plastic strips that contain a gradient of an antibacterial are placed on the surface of the inoculated agar plate (Figure \(\PageIndex{3}\)). As the bacterial inoculum grows, antibiotic diffuses from the plastic strips into the agar and interacts with the bacterial cells. Because the rate of drug diffusion is directly related to concentration, an elliptical zone of inhibition is observed with the Etest drug gradient, rather than a circular zone of inhibition observed with the Kirby-Bauer assay. To interpret the results, the intersection of the elliptical zone with the gradient on the drug-containing strip indicates the MIC. Because multiple strips containing different antimicrobials can be placed on the same plate, the MIC of multiple antimicrobials can be determined concurrently and directly compared. However, unlike the macrobroth and microbroth dilution methods, the MBC cannot be determined with the Etest.
Compare and contrast MIC and MBC.
Marisa’s UTI was likely caused by the catheterizations she had in Vietnam. Most bacteria that cause UTIs are members of the normal gut microbiota, but they can cause infections when introduced to the urinary tract, as might have occurred when the catheter was inserted. Alternatively, if the catheter itself was not sterile, bacteria on its surface could have been introduced into Marisa’s body. The antimicrobial therapy Marisa received in Cambodia may also have been a complicating factor because it may have selected for antimicrobial-resistant strains already present in her body. These bacteria would have already contained genes for antimicrobial resistance, either acquired by spontaneous mutation or through horizontal gene transfer, and, therefore, had the best evolutionary advantage for adaptation and growth in the presence of the antimicrobial therapy. As a result, one of these resistant strains may have been subsequently introduced into her urinary tract.
Laboratory testing at the CDC confirmed that the strain of Klebsiella pneumoniae from Marisa’s urine sample was positive for the presence of NDM, a very active carbapenemasethat is beginning to emerge as a new problem in antimicrobial resistance. While NDM-positive strains are resistant to a wide range of antimicrobials, they have shown susceptibility to tigecycline (structurally related to tetracycline) and the polymyxins B and E (colistin).
To prevent her infection from spreading, Marisa was isolated from the other patients in a separate room. All hospital staff interacting with her were advised to follow strict protocols to prevent surface and equipment contamination. This would include especially stringent hand hygiene practices and careful disinfection of all items coming into contact with her.
Marisa’s infection finally responded to tigecycline and eventually cleared. She was discharged a few weeks after admission, and a follow-up stool sample showed her stool to be free of NDM-containing K. pneumoniae , meaning that she was no longer harboring the highly resistant bacterium.
Key Concepts and Summary
- The Kirby-Bauer disk diffusion test helps determine the susceptibility of a microorganism to various antimicrobial drugs. However, the zones of inhibition measured must be correlated to known standards to determine susceptibility and resistance, and do not provide information on bactericidal versus bacteriostatic activity, or allow for direct comparison of drug potencies.
- Antibiograms are useful for monitoring local trends in antimicrobial resistance/susceptibility and for directing appropriate selection of empiric antibacterial therapy.
- There are several laboratory methods available for determining the minimum inhibitory concentration (MIC) of an antimicrobial drug against a specific microbe. The minimal bactericidal concentration (MBC) can also be determined, typically as a follow-up experiment to MIC determination using the tube dilution method.
Footnotes
- 1 J.P. Furuno et al. “Using Antibiograms to Improve Antibiotic Prescribing in Skilled Nursing Facilities.” Infection Control and Hospital Epidemiology 35 no. Suppl S3 (2014):S56–61.
Visit this website to view an interactive antibiogram provided by Stanford University.


35.3: Urinary Analgesics
By the end of this section, you should be able to:
- 35.3.1 Identify the characteristics of urinary analgesic drugs used for urinary and bladder disorders.
- 35.3.2 Explain the indications, actions, adverse reactions, contraindications, and interactions of urinary analgesic drugs used for urinary and bladder disorders.
- 35.3.3 Describe nursing implications of urinary analgesic drugs used for urinary and bladder disorders.
- 35.3.4 Explain the client education related to urinary analgesic drugs used for urinary and bladder disorders.
Urinary analgesics are medications used to relieve pain and discomfort associated with the urinary tract. They are primarily used to treat conditions such as UTIs and interstitial cystitis (also known as bladder pain syndrome ). Urinary analgesics work by numbing or reducing the sensitivity of the urinary tract, thus alleviating pain during urination.
It is important to note that urinary analgesics do not treat the underlying cause of the pain (such as bacterial infection in the case of UTIs); rather, they provide symptomatic relief. Therefore, it is crucial for the client to consult with a health care provider to determine the cause of urinary pain and to receive appropriate treatment. Additionally, urinary analgesics should be used only as directed and for a limited time because prolonged use without addressing the underlying cause may mask important signs and symptoms and delay appropriate diagnosis and treatment.
Phenazopyridine Hydrochloride
Phenazopyridine hydrochloride , an azo dye, is indicated for the symptomatic relief of pain, burning, urgency, frequency, and other discomfort arising from irritation of the lower urinary tract mucosa caused by infection, trauma, surgery, endoscopic procedures, or the passage of catheters (Eastham & Patel, 2022). This drug should not be used for more than 2 consecutive days to avoid potentially masking a bacterial infection requiring antibiotics.
Clinical Tip
Phenazopyridine Hydrochloride and Urine Color
Phenazopyridine hydrochloride may turn the urine, saliva, or tears a reddish-orange color. This is an expected effect of the medication and does not harm the client.
Adverse Effects and Contraindications
Adverse effects include headache; rash; pruritus; gastrointestinal disturbance; reddish-orange color of saliva, tears, or urine; hemolytic anemia; and renal and hepatic toxicity. Phenazopyridine hydrochloride is contraindicated in clients with hypersensitivity and in those with renal insufficiency.
Table 35.5 is a drug prototype table for phenazopyridine hydrochloride. It lists drug class, mechanism of action, adult dosage, indications, therapeutic effects, drug and food interactions, adverse effects, and contraindications.
|
Drug Class
Urinary analgesic Mechanism of Action Exerts an analgesic action locally on the urinary tract mucosa |
Drug Dosage
200 mg orally 3 times a day after meals. Use should not exceed 2 days. |
|
Indications
Symptomatic relief of pain and burning arising from irritation of the lower urinary tract mucosa Therapeutic Effects Relief of urinary frequency and urgency and of pain and burning of the urinary tract mucosa |
Drug Interactions
Ketoconazole Lidocaine Procaine Tetracaine Food Interactions No significant interactions |
|
Adverse Effects
Headache Rash Pruritus Nausea, vomiting Abdominal discomfort Reddish-orange color of saliva, tears, and urine Hemolytic anemia Renal toxicity Hepatic toxicity Yellowish tinge to skin or sclera, which may indicate accumulation due to impaired renal excretion |
Contraindications
Hypersensitivity Renal insufficiency |
Nursing Implications
The nurse should do the following for clients who are taking urinary analgesics:
- Before administering the drug, check the client’s medical history, current drug list (including over-the-counter medications), and allergies.
- Before administering the drug, assess the client’s renal and liver function because these drugs may cause renal or hepatic toxicity in clients who have insufficiency in these areas.
- Notify the health care provider if the client continues to have urinary issues.
- Provide client teaching regarding the drug and when to call the health care provider. See below for additional client teaching guidelines.
Client Teaching Guidelines
The client taking a urinary analgesic should:
- Take the medication as directed by their health care provider.
- Report any use of over-the-counter medications that may contain phenazopyridine hydrochloride to the health care provider because concomitant use may cause renal toxicity.
- Report any urinary tract symptoms that do not resolve or worsen to the health care provider because treatment may need to be modified.
The client taking a urinary analgesic should not:
- Take extra doses of this medication because doing so may cause serious adverse effects such as hemolytic anemia or renal or hepatic toxicity.
- Wear contact lenses while taking phenazopyridine hydrochloride because it may stain them a reddish-orange.
- Drink alcohol while taking urinary analgesics because their interaction may cause serious effects such as confusion, disorientation, and cardiac arrhythmias.
- Take this medication longer than 2 days because doing so could mask symptoms of a bacterial infection that is not resolving.


2.5: Resistors
What is a Resistor?
Special components called resistors are made for the express purpose of creating a precise quantity of resistance for insertion into a circuit. They are typically constructed of metal wire or carbon, and engineered to maintain a stable resistance value over a wide range of environmental conditions. Unlike lamps, they do not produce light, but they do produce heat as electric power is dissipated by them in a working circuit. Typically, though, the purpose of a resistor is not to produce usable heat, but simply to provide a precise quantity of electrical resistance.
Resistor Schematic Symbols
The most common schematic symbol for a resistor is a zig-zag line:
Resistor values in ohms are usually shown as an adjacent number, and if several resistors are present in a circuit, they will be labeled with a unique identifier number such as R 1 , R 2 , R 3 , etc. As you can see, resistor symbols can be shown either horizontally or vertically:
Real resistors look nothing like the zig-zag symbol. Instead, they look like small tubes or cylinders with two wires protruding for connection to a circuit. Here is a sampling of different kinds and sizes of resistors:
In keeping more with their physical appearance, an alternative schematic symbol for a resistor looks like a small, rectangular box:
Resistors can also be shown to have varying rather than fixed resistances. This might be for the purpose of describing an actual physical device designed for the purpose of providing an adjustable resistance, or it could be to show some component that just happens to have an unstable resistance:
In fact, any time you see a component symbol drawn with a diagonal arrow through it, that component has a variable rather than a fixed value. This symbol “modifier” (the diagonal arrow) is standard electronic symbol convention.
Variable Resistors
Variable resistors must have some physical means of adjustment, either a rotating shaft or lever that can be moved to vary the amount of electrical resistance. Here is a photograph showing some devices called potentiometers , which can be used as variable resistors:
Power Rating of Resistors
Because resistors dissipate heat energy as the electric currents through them overcome the “friction” of their resistance, resistors are also rated in terms of how much heat energy they can dissipate without overheating and sustaining damage. Naturally, this power rating is specified in the physical unit of “watts.” Most resistors found in small electronic devices such as portable radios are rated at 1/4 (0.25) watt or less. The power rating of any resistor is roughly proportional to its physical size. Note in the first resistor photograph how the power ratings relate with size: the bigger the resistor, the higher its power dissipation rating. Also, note how resistances (in ohms) have nothing to do with size!
Although it may seem pointless now to have a device doing nothing but resisting electric current, resistors are extremely useful devices in circuits. Because they are simple and so commonly used throughout the world of electricity and electronics, we’ll spend a considerable amount of time analyzing circuits composed of nothing but resistors and batteries.
How are Resistors Useful?
For a practical illustration of resistors’ usefulness, examine the photograph below. It is a picture of a printed circuit board , or PCB : an assembly made of sandwiched layers of insulating phenolic fiber-board and conductive copper strips, into which components may be inserted and secured by a low-temperature welding process called “soldering.” The various components on this circuit board are identified by printed labels. Resistors are denoted by any label beginning with the letter “R”.
This particular circuit board is a computer accessory called a “modem,” which allows digital information transfer over telephone lines. There are at least a dozen resistors (all rated at 1/4 watt power dissipation) that can be seen on this modem’s board. Every one of the black rectangles (called “integrated circuits” or “chips”) contain their own array of resistors for their internal functions, as well.
Another circuit board example shows resistors packaged in even smaller units, called “surface mount devices.” This particular circuit board is the underside of a personal computer hard disk drive, and once again the resistors soldered onto it are designated with labels beginning with the letter “R”:
There are over one hundred surface-mount resistors on this circuit board, and this count, of course, does not include the number of resistors internal to the black “chips.” These two photographs should convince anyone that resistors—devices that “merely” oppose the flow of electrons—are very important components in the realm of electronics!
“Load” on Schematic Diagrams
In schematic diagrams, resistor symbols are sometimes used to illustrate any general type of device in a circuit doing something useful with electrical energy. Any non-specific electrical device is generally called a load , so if you see a schematic diagram showing a resistor symbol labeled “load,” especially in a tutorial circuit diagram explaining some concept unrelated to the actual use of electrical power, that symbol may just be a kind of shorthand representation of something else more practical than a resistor.
Analyzing Resistor Circuits
To summarize what we’ve learned in this lesson, let’s analyze the following circuit, determining all that we can from the information given:
All we’ve been given here to start with is the battery voltage (10 volts) and the circuit current (2 amps). We don’t know the resistor’s resistance in ohms or the power dissipated by it in watts. Surveying our array of Ohm’s Law equations, we find two equations that give us answers from known quantities of voltage and current:
Inserting the known quantities of voltage (E) and current (I) into these two equations, we can determine circuit resistance (R) and power dissipation (P):
For the circuit conditions of 10 volts and 2 amps, the resistor’s resistance must be 5 Ω. If we were designing a circuit to operate at these values, we would have to specify a resistor with a minimum power rating of 20 watts, or else it would overheat and fail.
Resistor Materials
Resistors can be found in a variety of different materials, each one with its own properties and specific areas of use. Most electrical engineers use the types found below:
Wirewound (WW)
Wire Wound Resistors are manufactured by winding resistance wire around a non-conductive core in a spiral. They are typically produced for high precision and power applications. The core is usually made of ceramic or fiberglass and the resistance wire is made of nickel-chromium alloy and are not suitable for applications with frequencies higher than 50kHz. Low noise and stability with respect to temperature variations are standard characteristics of Wire Wound Resistors. Resistance values are available from 0.1 up to 100 kW, with accuracies between 0.1% and 20%.
Metal Film
Nichrome or tantalum nitride are typically used for metal film resistors. A combination of a ceramic material and a metal typically make up the resistive material. The resistance value is changed by cutting a spiral pattern in them film, much like carbon film with a laser or abrasive. Metal film resistors are usually less stable over temperature than wire wound resistors, but handle higher frequencies better.
Metal Oxide Film
Metal oxide resistors use metal oxides such as tin oxide, making them slightly different from metal film resistors. These resistors are reliable and stable and operate at higher temperatures than metal film resistors. Because of this, metal oxide film resistors are used in applications that require high endurance.
Foil
Developed in the 1960’s, the foil resistor is still one of the most accurate and stable types of resistor that you’ll find and are used for applications with high precision requirements. A ceramic substrate that has a thin bulk metal foil cemented to it makes up the resistive element. Foil Resistors feature a very low temperature coefficient of resistance.
Carbon Composition (CCR)
Until the 1960s Carbon Composition Resistors were the standard for most applications. They are reliable, but not very accurate (their tolerance cannot be better than about 5%). A mixture of fine carbon particles and non-conductive ceramic material are used for the resistive element of CCR Resistors. The substance is molded into the shape of a cylinder and baked. The dimensions of the body and the ratio of carbon to ceramic material determine the resistance value. More carbon used in the process means there will be a lower resistance. CCR resistors are still useful for certain applications because of their ability to withstand high energy pulses, a good example application would be in a power supply.
Carbon Film
Carbon film resistors have a thin carbon film (with a spiral cut in the film to increase the resistive path) on an insulating cylindrical core. This allows for the resistance value to be more accurate and also increases the resistance value. Carbon film resistors are much more accurate than carbon composition resistors. Special carbon film resistors are used in applications that require high pulse stability.
Key Performance Indicators (KPIs)
The KPIs for each resistor material can be found below:
Review
Devices called resistors are built to provide precise amounts of resistance in electric circuits. Resistors are rated both in terms of their resistance (ohms) and their ability to dissipate heat energy (watts).
- Resistor resistance ratings cannot be determined from the physical size of the resistor(s) in question, although approximate power ratings can. The larger the resistor is, the more power it can safely dissipate without suffering damage.
- Any device that performs some useful task with electric power is generally known as a load . Sometimes resistor symbols are used in schematic diagrams to designate a non-specific load, rather than an actual resistor.


7.5: Apply the SQ3R method
Now that you are familiar with the steps of the SQ3R Method, you may want to apply them to a text you are reading this week. To see how the steps are applied to an actual reading activity, watch the video below. At several points in the video, you will have the opportunity to pause and try the steps in the method. When you are finished the video or reading, go to the next chapter to move on in the workshop.
Apply it!
Commit to trying the SQ3R method once this week as you complete your course readings. As you do, consider the following questions:
- How does the SQ3R method change how you approach your reading?
- How will you adapt and personalize this process to your own learning strengths and the specific requirements of your courses?
The first place I will begin is the Learning Objectives. I notice that in this textbook, they are located at the beginning of the chapter. I read these carefully to discover the main concepts that I will learn by reading. The next part of the chapter I’ll review is the key takeaways at the end of the chapter. Remember – there’s no rule that says that I need to read each page in order. By reading the key takeaways, I gain a sense of the most important information in the chapter. This will help me to focus my reading later.
Now, I’ll go back to the beginning of the chapter, and briefly skim the contents. I’ll pay particular attention to the headings and to any key diagrams. I’m noticing a key diagram for both Maslow’s hierarchy of needs and the ERG theory. I also notice two other key headings as I skim: I now know I will read about two factor theory, and acquired needs theory. From the information I’ve gained in the survey step, I’ve determined that my goals for reading are:
- To be able to describe the four theories of motivation.
- To identify how these theories are similar and different.
- And to understand how each theory explains employee behavior.
My next step is to begin questioning and reading. I’ll base my questions on key headings I notice. The first heading I read is Maslow’s Hierarchy of Needs . What questions can I ask about this? You may want to pause this video here, and try to create 3-4 questions you might want to ask. Then, resume the video to see how the questioning process works.
Here are the questions I’ve developed:
- What is Maslow’s hierarchy of needs?
- What are the levels in Maslow’s hierarchy? (I remember that there are levels from my survey step)
- Why are there different levels in the hierarchy?
- How does Maslow’s theory explain employee behavior?
I’ve added my questions to my notetaking page. I begin reading looking for the answer to my first question. I find the answer here, in the first paragraph. The theory is based on a simple premise: Human beings have needs that are hierarchically ranked. There are some needs that are basic to all human beings, and in their absence nothing else matters. As we satisfy these basic needs, we start looking to satisfy higher order needs.
Now, I want to add this information to my notes. To get the most benefit of this step, I will recite the information in my own words, then write it down. The step of putting information into my own words ensures that I understand it clearly.
I pause and think about how I can express what I’ve read in my own words. I can say it like this: Maslow’s theory states that everyone has levels (a hierarchy) of needs. When our basic needs are met, we move to fulfill our higher levels of need. I’ll now add this information to my notes.
You will notice that I have left a wide margin on my notetaking page. This space allows me to add additional thoughts, images, and questions about the material later on. I may want to add additional information I learn in class.
I’ll move through the same steps to answer my other three questions. You may want to pause this video here, and try these steps out for yourself.
As I’m reading, I will also take note of key terms in bold letters. For example, I see that physiological needs is a key term in this chapter. These are words that I want to be able to define, as they are important to my understanding of the course material.
I will work through the chapter, following the same steps for each main chapter section: create questions, read to find the answers, recite my answer, and write it in my notes in my own words.
Now that you have seen how the SQ3R method might be applied to a textbook chapter, try it! Notice how this changes your reading process? How do you want to use this information to read in the future?


20: Organic Chemistry
Organic chemistry involving the scientific study of the structure, properties, and reactions of organic compounds and organic materials, i.e., matter in its various forms that contain carbon atoms. Study of structure includes many physical and chemical methods to determine the chemical composition and the chemical constitution of organic compounds and materials. Study of properties includes both physical properties and chemical properties, and uses similar methods as well as methods to evaluate chemical reactivity, with the aim to understand the behavior of the organic matter.
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- 20.1: Prelude to Organic Chemistry
- Early chemists regarded substances isolated from organisms as a different type of matter that could not be synthesized artificially, and these substances were thus known as organic compounds. The widespread belief called vitalism held that organic compounds were formed by a vital force present only in living organisms. Thr defining trait of organic molecules is the presence of carbon as the principal element, bonded to hydrogen and other carbon atoms.
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- 20.2: Hydrocarbons
- Strong, stable bonds between carbon atoms produce complex molecules containing chains, branches, and rings. Hydrocarbons are organic compounds composed of only carbon and hydrogen. The alkanes are saturated hydrocarbons—that is, hydrocarbons that contain only single bonds. Alkenes contain one or more carbon-carbon double bonds. Alkynes contain one or more carbon-carbon triple bonds. Aromatic hydrocarbons contain ring of delocalized π electrons.
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- 20.3: Alcohols and Ethers
- Many organic compounds that are not hydrocarbons can be thought of as derivatives of hydrocarbons. A hydrocarbon derivative can be formed by replacing one or more hydrogen atoms of a hydrocarbon by a functional group, which contains at least one atom of an element other than carbon or hydrogen. The properties of hydrocarbon derivatives are determined largely by the functional group. The –OH group is the functional group of an alcohol. The –R–O–R– group is the functional group of an ether.
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- 20.4: Aldehydes, Ketones, Carboxylic Acids, and Esters
- The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom.
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- 20.5: Amines and Amides
- The addition of nitrogen into an organic framework leads to two families of molecules. Compounds containing a nitrogen atom bonded in a hydrocarbon framework are classified as amines. Compounds that have a nitrogen atom bonded to one side of a carbonyl group are classified as amides.  Amines and carboxylic acids can combine in a condensation reaction to form amides.
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- 20.E: Organic Chemistry (Exercises)
- These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.