22.1: Activity 1 - Searching Algorithm
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Introduction
This activity involves studying the searching process and how to write algorithms that can find a particular given item from a list of given items. The activity of searching will dedicate itself to providing the answer of a presence or no presence of the searched item. The section will dedicate itself to introducing the sequential and binary types of search.
Searching algorithm
In computer science, a search algorithm is an algorithm for finding an item with specified properties among a collection of items.
Finding a Data Item - The Search Problem
Consider a collection of data items of same type e.g. integers, and imagine we wish to check if a particular data item is in the collection. Call the particular data item of interest to us, the key. The task is then to search among the records in the database to find out which one “matches’’ the key. This can first be done by arranging the data using a structure that can hold a collection of data items of the same type, that is, the arrays or linked lists. The searching can be done using either the sequential or binary search, where the former involves looking at each item in turn while the later involves items that are in order. The example below, is of an array consisting of seven elements long, containing numeric values. That is, the maximum number of comparisons is 7, and occurs when the key we are searching for is in A[6].
Both sequential and binary search algorithms can be used to find the data item:
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For a sequential search the following will do:
for (each item in list) { compare search item to current item if match, save index of matching item break } return index of matching item, or -1 if item not found
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For a binary search, the following algorithm illustrates how it is carried out:
set first = 1, last = N, mid = N /2 while (item not found and first < last) { compare search term to item at mid if match save index break else if search term is less than item at mid, set last = mid-1 else set first = mid+1 set mid = (first+last) /2 } return index of matching item, or -1 if not found
Conclusion
In this activity, you learned two different search algorithms, the sequential and binary searches. Examples illustrating them were also given. Its application to an array was also introduced i.e. how a search can be applied in an array to search for a particular item in an array.
Assessment
Provide a short description of binary search algorithm.


2.5: Conducting
Introduction
Conducting is the art of directing a musical performance, such as a concert, by way of visible gestures with the hands, arms, face and head. The primary duties of the conductor are to unify performers, set the tempo, execute clear preparations and beats (meter), and to listen critically and shape the sound of the ensemble.Communication is non-verbal during a performance, however in rehearsal frequent interruptions allow the conductor to give verbal directions as to how the music should be played or sung.
Conductors act as guides to the orchestras and/or choirs they conduct. They choose the works to be performed and study their scores to which they may make certain adjustments, work out their interpretation, and relay their vision to the performers. They may also attend to organizational matters, such as scheduling rehearsals. Orchestras,choirs, concert bands and other sizable musical ensembles such as big bands are usually led by conductors.
Please watch the following short video about conducting with simple meters. Remember that, in simple meters, each beat can be divided into two equal parts.
The Movements of Conducting
Duple Meter
Examples of duple meter are 2/4, 2/2, or fast 6/8 time. A conductor would sweep out a J-shaped arc to conduct duple meter:
Triple Meter
Examples of triple meter are 3/4 or 3/2. A conductor would sweep out a triangular arc to conduct triple meters.
Quadruple Meter
Example of quadruple meter are 4/4, 4/2, or 4/8. A conductor would sweep out an upside-down T shape to conduct quadruple meter.
Watch the following video. Does Gustavo Dudamel (the conductor) always draw the 4/4 pattern?
Slow, 6/8 Time
When conducting 6/8 time, the conductor will sweep out an upside-down T , bouncing along the bottom to each beat.
Syncopation
In music, syncopation involves a variety of rhythms which are in some way unexpected which make part or all of a tune or piece of music off-beat. More simply, syncopation is a general term for “a disturbance or interruption of the regular flow of rhythm”: a “placement of rhythmic stresses or accents where they wouldn’t normally occur.”
Contributors and Attributions
Revision and adaptation. Provided by : Lumen Learning and Natalia Kuznetsova. License : CC BY-SA: Attribution-ShareAlike
- Provided by : Wikipedia. Located at : https://en.Wikipedia.org/wiki/Conducting . License : CC BY-SA: Attribution-ShareAlike
- Kensington Conductor Mural. Authored by : wiredforlego. Located at : https://www.flickr.com/photos/wiredforsound23/14615664692/ . License : CC BY-NC: Attribution-NonCommercial
- #3-Simple Meters: Conducting Tip with Michelle Willis. Authored by : Michelle Willis. Located at : https://www.youtube.com/watch?v=79Pk-33R2HM . License : All Rights Reserved . License Terms : Standard YouTube License


11.9: Immune System
See Figure 11.13 [1] to review the various structures of the immune system. The immune system provides specific and nonspecific defense mechanisms to protect the body from infection. Examples of a specific defense are white blood cells that circulate in the blood and lymph. The function of these cells is to identify and destroy pathogens such as bacteria and viruses. Examples of nonspecific defenses are the skin and mucous membranes. These parts of the system act as barriers to block entrance of pathogens. For more information, refer to the “ Defenses Against Transmission of Infection ” section in Chapter 4.
Immune system function and decreasing the risk for infection depends on healthy nutrition. In fact, malnutrition is the leading cause of immune system deficiency worldwide. When immune system functions are inadequate, there is a marked increase in the chance of getting an infection. Children in many poor, developing countries who lack good sources of protein and carbohydrates often die from infections that their bodies would have normally fought off with proper nutritional intake. Because their protein and/or energy intake is so low, the immune system cannot perform its functions. [2]
Other nutrients, such as iron, zinc, selenium, copper, folate, and vitamins A, B6, C, D, and E, benefit immune system function. It is best to obtain minerals and vitamins from eating a variety of healthy foods. Deficiencies in these nutrients can cause an increased risk for infection and even death. Zinc deficiency results in suppression of the immune system’s skin barrier functions and is also associated with a decrease in the number of circulating white blood cells. Taking zinc supplements has been found to be therapeutically beneficial for the treatment of leprosy, tuberculosis, pneumonia, and the common cold. [3]
Just as undernutrition compromises immune system health, so can overnutrition that causes obesity. People who are obese are at increased risk for developing diabetes, cardiovascular, and immune system disorders. High intake of saturated and trans fats negatively affects the immune system, whereas increasing intake of omega-3 fatty acids (found in salmon and other oily fish) reduces the risk of developing certain autoimmune disorders, such as rheumatoid arthritis. Rheumatoid arthritis affects the joints but is more severe in terms of pain and swelling than osteoarthritis. It can begin as early as age 18. Nursing assistant interventions for osteoarthritis also can be used to help manage the discomfort of rheumatoid arthritis. [4]
Autoimmune disorders cause the body to view parts of itself as foreign, mistake those parts as pathogens, and launch an attack against those parts. It can affect certain parts of the body like skin, joints, or a specific organ, or it can affect the entire body. In the digestive system, it can cause food intolerances, celiac disease, or irritable bowel syndrome. Refer to the Chapter 6.2 subsection on “ Modified Diets ” for management. See Table 11.9 for common chronic conditions and the interventions nursing assistants can provide to assist with immune disorders. Review section Chapter 4.7, “ Signs and Symptoms of Infection ” for observations related to the immune system that should be reported to the nurse.
Table 11.9 Common Chronic Conditions of the Immune System and Related Interventions
| Diagnosis | Definition | Symptoms to report | Nursing Assistant Interventions |
|---|---|---|---|
| Asthma | Airways become swollen and narrowed in response to an environmental trigger, making it difficult to breathe. Severe asthma attacks can be life-threatening. | Increased shortness of breath during activity or at rest. |
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| Type 1 Diabetes | The immune system attacks the pancreas and reduces the production of insulin. | Excessive thirst, hunger, or urination; confusion; or hot/dry or cold/clammy skin. |
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| Leukemia, Lymphoma, and Other Immune System Cancers | Ineffective white blood cells elevate the risk for life-threatening infection. | Fatigue, weakness, or signs and symptoms of infection. |
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| Human Immunodeficiency Virus (HIV) | The virus attacks white blood cells that identify pathogens in the body, and they are no longer able to fight infection. | Fatigue, weakness, or signs and symptoms of infection. |
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| Thyroid Disorders | As the thyroid is attacked by the immune system, the regulation of hormones that control metabolism, energy levels, body temperature, heart rate, and appetite are affected. | Increased fatigue or cold intolerance or changes in appetite or mood. |
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| Multiple Sclerosis (MS) | The immune system attacks the protective sheath around the spinal cord, causing disruption of signals from the brain to body parts. Functioning can vary from day-to-day. | Decreased mobility or cognition. |
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- “ 2206_The_Location_Structure_and_Histology_of_the_Thymus.jpg ” by OpenStax College is licensed under CC BY 3.0 ↵
- This work is a derivative of Human Nutrition by University of Hawai‘i at Mānoa Food Science and Human Nutrition Program and is licensed under CC BY NC SA 4.0 ↵
- This work is a derivative of Human Nutrition by University of Hawai‘i at Mānoa Food Science and Human Nutrition Program and is licensed under CC BY NC SA 4.0 ↵
- This work is a derivative of Human Nutrition by University of Hawai‘i at Mānoa Food Science and Human Nutrition Program and is licensed under CC BY NC SA 4.0 ↵


11.7: Exercises
Skills
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The table below shows scores on a Math test.
- Complete the frequency table for the Math test scores
- Construct a histogram of the data
- Construct a pie chart of the data
\(\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 80 & 50 & 50 & 90 & 70 & 70 & 100 & 60 & 70 & 80 & 70 & 50 \\
\hline 90 & 100 & 80 & 70 & 30 & 80 & 80 & 70 & 100 & 60 & 60 & 50 \\
\hline
\end{array}\)
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A group of adults where asked how many cars they had in their household
- Complete the frequency table for the car number data
- Construct a histogram of the data
- Construct a pie chart of the data
\(\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 1 & 4 & 2 & 2 & 1 & 2 & 3 & 3 & 1 & 4 & 2 & 2 \\
\hline 1 & 2 & 1 & 3 & 2 & 2 & 1 & 2 & 1 & 1 & 1 & 2 \\
\hline
\end{array}\)
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A group of adults were asked how many children they have in their families. The bar graph to the right shows the number of adults who indicated each number of children.
- How many adults where questioned?
- What percentage of the adults questioned had 0 children?
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Jasmine was interested in how many days it would take an order from Netflix to arrive at her door. The graph below shows the data she collected.
- How many movies did she order?
- What percentage of the movies arrived in one day?
- The bar graph below shows the percentage of students who received each letter grade on their last English paper. The class contains 20 students. What number of students earned an A on their paper?
- Kori categorized her spending for this month into four categories: Rent, Food, Fun, and Other. The percents she spent in each category are pictured here. If she spent a total of $2600 this month, how much did she spend on rent?
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A group of diners were asked how much they would pay for a meal. Their responses were: $7.50, $8.25, $9.00, $8.00, $7.25, $7.50, $8.00, $7.00.
- Find the mean
- Find the median
- Write the 5-number summary for this data
- Find the standard deviation of this data
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You recorded the time in seconds it took for 8 participants to solve a puzzle. The times were: 15.2, 18.8, 19.3, 19.7, 20.2, 21.8, 22.1, 29.4.
- Find the mean
- Find the median
- Write the 5-number summary for this data
- Find the standard deviation of this data
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Refer back to the histogram from question #3.
- Compute the mean number of children for the group surveyed
- Compute the median number of children for the group surveyed
- Write the 5-number summary for this data.
- Create box plot.
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Refer back to the histogram from question #4.
- Compute the mean number of shipping days
- Compute the median number of shipping days
- Write the 5-number summary for this data.
- Create box plot.
Concepts
- The box plot below shows salaries for Actuaries and CPAs. Kendra makes the median salary for an Actuary. Kelsey makes the first quartile salary for a CPA. Who makes more money? How much more?
Exploration
- Studies are often done by pharmaceutical companies to determine the effectiveness of a treatment program. Suppose that a new AIDS antibody drug is currently under study. It is given to patients once the AIDS symptoms have revealed themselves. Of interest is the average length of time in months patients live once starting the treatment. Two researchers each follow a different set of 40 AIDS patients from the start of treatment until their deaths. The following data (in months) are collected.
Researcher 1: 3; 4; 11; 15; 16; 17; 22; 44; 37; 16; 14; 24; 25; 15; 26; 27; 33; 29; 35; 44; 13; 21; 22; 10; 12; 8; 40; 32; 26; 27; 31; 34; 29; 17; 8; 24; 18; 47; 33; 34
Researcher 2: 3; 14; 11; 5; 16; 17; 28; 41; 31; 18; 14; 14; 26; 25; 21; 22; 31; 2; 35; 44; 23; 21; 21; 16; 12; 18; 41; 22; 16; 25; 33; 34; 29; 13; 18; 24; 23; 42; 33; 29
- Create comparative histograms of the data
- Create comparative boxplots of the data
- A graph appears below showing the number of adults and children who prefer each type of soda. There were 130 adults and kids surveyed. Discuss some ways in which the graph below could be improved
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Make up three data sets with 5 numbers each that have:
- the same mean but different standard deviations.
- the same mean but different medians.
- the same median but different means.
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A sample of 30 distance scores measured in yards has a mean of 7, a variance of 16, and a standard deviation of 4.
- You want to convert all your distances from yards to feet, so you multiply each score in the sample by 3. What are the new mean, median, variance, and standard deviation?
- You then decide that you only want to look at the distance past a certain point. Thus, after multiplying the original scores by 3, you decide to subtract 4 feet from each of the scores. Now what are the new mean, median, variance, and standard deviation?
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In your class, design a poll on a topic of interest to you and give it to the class.
- Summarize the data, computing the mean and five-number summary.
- Create a graphical representation of the data.
- Write several sentences about the topic, using your computed statistics as evidence in your writing.


11.1: Sequences and Their Notations
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Learning Objectives
- Write the terms of a sequence defined by an explicit formula.
- Write the terms of a sequence defined by a recursive formula.
- Use factorial notation.
A video game company launches an exciting new advertising campaign. They predict the number of online visits to their website, or hits, will double each day. The model they are using shows \(2\) hits the first day, \(4\) hits the second day, \(8\) hits the third day, and so on (Table \(\PageIndex{1}\)).
| Day | 1 | 2 | 3 | 4 | 5 | … |
|---|---|---|---|---|---|---|
| Hits | 2 | 4 | 8 | 16 | 32 | … |
If their model continues, how many hits will there be at the end of the month? To answer this question, we’ll first need to know how to determine a list of numbers written in a specific order. In this section, we will explore these kinds of ordered lists.
Writing the Terms of a Sequence Defined by an Explicit Formula
One way to describe an ordered list of numbers is as a sequence . A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is
\(\{ 2,4,8,16,32,…\}\)
The ellipsis (…) indicates that the sequence continues indefinitely. Each number in the sequence is called a term. The first five terms of this sequence are \(2\), \(4\), \(8\), \(16\), and \(32\).
Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as \(31\) terms. A more efficient way to determine a specific term is by writing a formula to define the sequence.
One type of formula is an explicit formula , which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence , where nn is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the nth term.
The first term of the sequence is \(2^1=2\), the second term is \(2^2=4\), the third term is \(2^3=8\), and so on. The nth term of the sequence can be found by raising 2 to the nth power. An explicit formula for a sequence is named by a lower case letter \(a\), \(b\), \(c\)... with the subscript \(n\). The explicit formula for this sequence is
\(a_n=2^n\)
Now that we have a formula for the nth term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31 st term of the sequence. We will substitute 31 for \(n\) in the formula.
\[ \begin{align*} a_{31}=2^{31} \\ [5pt] &=2,147,483,648 \end{align*}\]
If the doubling trend continues, the company will get \(2,147,483,648\) hits on the last day of the month. That is over \(2.1\) billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions.
Another way to represent the sequence is by using a table. The first five terms of the sequence and the nth term of the sequence are shown in Table \(\PageIndex{2}\).
| \(n\) | 1 | 2 | 3 | 4 | 5 | \(n\) |
|---|---|---|---|---|---|---|
| \(n^{th}\) term of the sequence, \(a_n\) | 2 | 4 | 8 | 16 | 32 | \(2^n\) |
Graphing provides a visual representation of the sequence as a set of distinct points. We can see from the graph in Figure \(\PageIndex{1}\) that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function.
Figure \(\PageIndex{1}\)
Lastly, we can write this particular sequence as
\(\{2,4,8,16,32,…,2^n,…\}\)
A sequence that continues indefinitely is called an infinite sequence . The domain of an infinite sequence is the set of counting numbers. If we consider only the first \(10\) terms of the sequence, we could write
\(\{2,4,8,16,32,…,2^n,…,1024\}\)
This sequence is called a finite sequence because it does not continue indefinitely.
SEQUENCE
A sequence is a function whose domain is the set of positive integers. A finite sequence is a sequence whose domain consists of only the first \(n\) positive integers. The numbers in a sequence are called terms . The variable \(a\) with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence.
\[a_1,a_2,a_3,…,a_n,…\]
We call \(a_1\) the first term of the sequence, \(a_2\) the second term of the sequence, \(a_3\) the third term of the sequence, and so on. The term \(a_n\) is called the nth term of the sequence, or the general term of the sequence. An explicit formula defines the nth term of a sequence using the position of the term. A sequence that continues indefinitely is an infinite sequence.
Q&A: Does a sequence always have to begin with \(a_1\)?
No. In certain problems, it may be useful to define the initial term as \(a_0\) instead of \(a_1\). In these problems, the domain of the function includes \(0\).
How to: Given an explicit formula, write the first \(n\) terms of a sequence
- Substitute each value of \(n\) into the formula. Begin with \(n=1\) to find the first term, \(a_1\).
- To find the second term, \(a_2\), use \(n=2\).
- Continue in the same manner until you have identified all nn terms.
Example \(\PageIndex{1}\): Writing the Terms of a Sequence Defined by an Explicit Formula
Write the first five terms of the sequence defined by the explicit formula \(a_n=−3n+8\).
Solution
Substitute \(n=1\) into the formula. Repeat with values \(2\) through \(5\) for \(n\).
\(\begin{array}{ll} n=1 & a_1=−3(1)+8=5 \\ n=2 & a_2=−3(2)+8=2 \\ n=3 & a_3=−3(3)+8=−1 \\ n=4 & a_4=−3(4)+8=−4 \\ n=5 & a_5=−3(5)+8=−7 \end{array}\)
The first five terms are \(\{5,2,−1,−4,−7\}\).
Analysis
The sequence values can be listed in a table. A table, such as Table \(\PageIndex{3}\), is a convenient way to input the function into a graphing utility.
| \(n\) | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| \(a_n\) | 5 | 2 | –1 | –4 | –7 |
A graph can be made from this table of values. From the graph in Figure \(\PageIndex{2}\), we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the positive integers only.
Figure \(\PageIndex{2}\)
Exercise \(\PageIndex{1}\)
Write the first five terms of the sequence defined by the explicit formula \(t_n=5n−4\).
- Answer
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The first five terms are \(\{1,6, 11, 16, 21\}\).
Investigating Alternating Sequences
Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as \(n\) increases. Let’s take a look at the following sequence.
\(\{2,−4,6,−8\}\)
Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence.
How to: Given an explicit formula with alternating terms, write the first nn terms of a sequence
- Substitute each value of nn into the formula. Begin with \(n=1\) to find the first term, \(a_1\). The sign of the term is given by the \({(−1)}^n\) in the explicit formula.
- To find the second term, \(a_2\), use \(n=2\).
- Continue in the same manner until you have identified all \(n\) terms.
Example \(\PageIndex{2}\): Writing the Terms of an Alternating Sequence Defined by an Explicit Formula
Write the first five terms of the sequence.
\(a_n=\dfrac{{(−1)}^nn^2}{n+1}\)
Solution
Substitute \(n=1\), \(n=2\), and so on in the formula.
\(\begin{array}{ll} n=1 & a_1=\dfrac{{(−1)}^12^2}{1+1}=−\dfrac{1}{2} \\ n=2 & a_2=\dfrac{{(−1)}^22^2}{2+1}=\dfrac{4}{3} \\ n=3 & a_3=\dfrac{{(−1)}^33^2}{3+1}=−\dfrac{9}{4} \\ n=4 & a_4=\dfrac{{(−1)}^44^2}{4+1}=\dfrac{16}{5} \\ n=5 & a_5=\dfrac{{(−1)}^55^2}{5+1}=−\dfrac{25}{6} \end{array}\)
The first five terms are \(\{−12,43,−94,165,−256\}\).
Analysis
The graph of this function, shown in Figure \(\PageIndex{3}\), looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values.
Figure \(\PageIndex{3}\)
Q&A: In the example above, does the \((–1)\) to the power of \(n\) account for the oscillations of signs?
Yes, the power might be \(n\), \(n+1\), \(n−1\), and so on, but any odd powers will result in a negative term, and any even power will result in a positive term.
Exercise \(\PageIndex{2}\)
Write the first five terms of the sequence:
\(a_n=\dfrac{4n}{{(−2)}^n}\)
- Answer
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The first five terms are \(\{−2, 2, −32, 1, −58\}\).
Investigating Piecewise Explicit Formulas
We’ve learned that sequences are functions whose domain is over the positive integers. This is true for other types of functions, including some piecewise functions. Recall that a piecewise function is a function defined by multiple subsections. A different formula might represent each individual subsection.
How to: Given an explicit formula for a piecewise function, write the first \(n\) terms of a sequence
- Identify the formula to which \(n=1\) applies.
- To find the first term, \(a_1\), use \(n=1\) in the appropriate formula.
- Identify the formula to which \(n=2\) applies.
- To find the second term, \(a_2\), use \(n=2\) in the appropriate formula.
- Continue in the same manner until you have identified all \(n\) terms.
Example \(\PageIndex{3}\): Writing the Terms of a Sequence Defined by a Piecewise Explicit Formula
Write the first six terms of the sequence.
\(a_n=\begin{cases}n^2\space \text{if n is not divisible by } 3\\\dfrac{n}{3} \text{if n is divisible by } 3\end{cases}\)
Solution
Substitute \(n=1\), \(n=2\), and so on in the appropriate formula. Use \(n^2\) when \(n\) is not a multiple of \(3\). Use \(n^3\) when \(n\) is a multiple of \(3\).
\(\begin{array}{ll} a_1=1^2=1 & 1 \text{ is not a multiple of }3. \text{ Use } n^2. \\ a_2=2^2=4 & 2 \text{ is not a multiple of }3 \text{ . Use }n^2. \\ a_3=\dfrac{3}{3}=1 & 3 \text{ is a multiple of }3 \text{. Use }\dfrac{n}{3}. \\ a_4=4^2=16 & 4 \text{ is not a multiple of }3. \text{ Use }n^2. \\ a_5=5^2=25 & 5 \text{ is not a multiple of } 3. \text{Use }n^2. \\ a_6=\dfrac{6}{3}=2 & 6 \text{ is a multiple of }3. \text{Use }\dfrac{n}{3} \end{array}\)
The first six terms are \(\{1, 4, 1, 16, 25, 2\}\).
Analysis
Every third point on the graph shown in Figure \(\PageIndex{4}\) stands out from the two nearby points. This occurs because the sequence was defined by a piecewise function.
Figure \(\PageIndex{4}\)
Exercise \(\PageIndex{3}\)
Write the first six terms of the sequence.
\(a_n=\begin{cases}2n^3 \text{ if } n\text{ is odd} \\ \dfrac{5n}{2} \text{ if } n \text{ is even}\end{cases}\)
- Answer
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The first six terms are \(\{2, 5, 54, 10, 250, 15\}\).
Finding an Explicit Formula
Thus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the nth term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases.
How to: Given the first few terms of a sequence, find an explicit formula for the sequence
- Look for a pattern among the terms.
- If the terms are fractions, look for a separate pattern among the numerators and denominators.
- Look for a pattern among the signs of the terms.
- Write a formula for \(a_n\) in terms of \(n\). Test your formula for \(n=1\), \(n=2\), and \(n=3\).
Example \(\PageIndex{4}\): Writing an Explicit Formula for the n th Term of a Sequence
Write an explicit formula for the nth term of each sequence.
- \(\{−\dfrac{2}{11}, \dfrac{3}{13}, −\dfrac{4}{15}, \dfrac{5}{17}, −\dfrac{6}{19},…\}\)
- \(\{−\dfrac{2}{25}, −\dfrac{2}{125}, −\dfrac{2}{625}, −\dfrac{2}{3,125}, −\dfrac{2}{15,625},…\}\)
- \(\{e^4,e^5,e^6,e^7,e^8,…\}\)
Solution
Look for the pattern in each sequence.
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The terms alternate between positive and negative. We can use \({(−1)}^n\) to make the terms alternate. The numerator can be represented by \(n+1\). The denominator can be represented by \(2n+9\).
\(a_n=\dfrac{{(−1)}^n(n+1)}{2n+9}\)
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The terms are all negative.
So we know that the fraction is negative, the numerator is \(2\), and the denominator can be represented by \(5n+1\).
\(a_n=−\dfrac{2}{5^{n+1}}\)
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The terms are powers of \(e\). For \(n=1\), the first term is \(e^4\) so the exponent must be \(n+3\).
\(a_n=e^{n+3}\)
Exercise \(\PageIndex{4A}\)
Write an explicit formula for the nth term of the sequence.\(\{9,−81,729,−6,561,59,049,…\}\)
- Answer
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\(a_n={(−1)}^{n+1}9^n\)
Exercise \(\PageIndex{4B}\)
Write an explicit formula for the nth term of the sequence.
\(\{−34,−98,−2712,−8116,−24320,...\}\)
- Answer
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\(a_n=−\dfrac{3^n}{4n}\)
Exercise \(\PageIndex{4C}\)
Write an explicit formula for the nth term of the sequence.
\(\left \{ \dfrac{1}{e^2}, \dfrac{1}{e}, 1, e, e^2,... \right \}\)
- Answer
-
\(a_n=e^{n−3}\)
Writing the Terms of a Sequence Defined by a Recursive Formula
Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are \(1, 1, 2, 3, 5, 8, 13, 21, 34,….\) Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with \(13\) petals, and different varieties of daisies that may have \(21\) or \(34\) petals.
Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula , a formula that defines the terms of a sequence using previous terms.
A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining \(a_n\) in terms of preceding terms. For example, suppose we know the following:
\[\begin{align*} a_1 &= 3 \\ a_n &= 2a_{n−1}−1 , \text{ for }n≥2 \end{align*}\]
We can find the subsequent terms of the sequence using the first term.
\[\begin{align*} a_1 &= 3 \\ a_2 &= 2a_1−1=2(3)−1=5 \\ a_3 &= 2a_2−1=2(5)−1=9 \\ a_4 &= 2a_3−1=2(9)−1=17 \end{align*}\]
So the first four terms of the sequence are \(\{3, 5, 9, 17\}\).
The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms.
\[\begin{align*}a_1 &= 1 \\ a_2 &= 1 \\ a_n &= a_{n−1}+a_{n−2} \text{, for }n≥3 \end{align*}\]
To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are \(21\) and \(34\), so
\(a_{10}=a_9+a_8=34+21=55\)
RECURSIVE FORMULA
A recursive formula is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence.
Q&A: Must the first two terms always be given in a recursive formula?
No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined.
How to: Given a recursive formula with only the first term provided, write the first n terms of a sequence
- Identify the initial term, \(a_1\), which is given as part of the formula. This is the first term.
- To find the second term, \(a_2\), substitute the initial term into the formula for \(a_{n−1}\). Solve.
- To find the third term, \(a_3\), substitute the second term into the formula. Solve.
- Repeat until you have solved for the nth term.
Example \(\PageIndex{5}\): Writing the Terms of a Sequence Defined by a Recursive Formula
Write the first five terms of the sequence defined by the recursive formula.
\[\begin{align*}a_1 &= 9 \\ a_n &= 3a_{n−1}−20 , \text{ for }n≥2 \end{align*}\]
Solution
The first term is given in the formula. For each subsequent term, we replace \(a_n−1\) with the value of the preceding term.
\(\begin{array}{ll} n=1 & a_1=9 \\ n=2 & a_2=3a_1−20=3(9)−20=27−20=7 \\ n=3 & a_3=3a_2−20=3(7)−20=21−20=1 \\ n=4 & a_4=3a_3−20=3(1)−20=3−20=−17 \\ n=5 & a_5=3a_4−20=3(−17)−20=−51−20=−71 \end{array}\)
The first five terms are \(\{9, 7, 1, –17, –71\}\). See Figure \(\PageIndex{5}\).
Figure \(\PageIndex{5}\)
Exercise \(\PageIndex{5}\)
Write the first five terms of the sequence defined by the recursive formula.
\[\begin{align*} a_1 &= 2 \\ a_n &= 2a_{n−1}+1 , \text{ for }n≥2 \end{align*}\]
- Answer
-
\(\{2, 5, 11, 23, 47\}\)
How to: Given a recursive formula with two initial terms, write the first \(n\) terms of a sequence
- Identify the initial term, \(a_1\), which is given as part of the formula.
- Identify the second term, \(a_2\), which is given as part of the formula.
- To find the third term, substitute the initial term and the second term into the formula. Evaluate.
- Repeat until you have evaluated the nth term.
Example \(\PageIndex{6}\): Writing the Terms of a Sequence Defined by a Recursive Formula
Write the first six terms of the sequence defined by the recursive formula.
\[\begin{align*} a_1 &=1 \\ a_2 &= 2 \\ a_n &= 3a_{n−1}+4a_{n−2} , \text{ for }n≥3 \end{align*}\]
Solution
The first two terms are given. For each subsequent term, we replace \(a_{n−1}\) and \(a_{n−2}\) with the values of the two preceding terms.
\(\begin{array}{ll} n=3 & a_3=3a_2+4a_1=3(2)+4(1)=10 \\ n=4 & a_4=3a_3+4a_2=3(10)+4(2)=38 \\ n=5 & a_5=3a_4+4a_3=3(38)+4(10)=154 \\ n=6 & a_6=3a_5+4a_4=3(154)+4(38)=614 \end{array}\)
The first six terms are \(\{1,2,10,38,154,614\}\). See Figure \(\PageIndex{6}\).
Figure \(\PageIndex{6}\)
Exercise \(\PageIndex{6}\)
Write the first \(8\) terms of the sequence defined by the recursive formula.
\[\begin{align*}a_1 &= 0 \\ a_2 &= 1 \\ a_3 &= 1 \\ a_n &= \dfrac{a_{n−1}}{a_{n−2}}+a_{n−3} , \text{ for }n≥4 \end{align*}\]
- Answer
-
\(\left \{0, 1, 1, 1, 2, 3, 52, 176 \right \}\).
Using Factorial Notation
The formulas for some sequences include products of consecutive positive integers. \(n\) factorial , written as \(n!\), is the product of the positive integers from \(1\) to \(n\). For example,
\[\begin{align*}4! &= 4⋅3⋅2⋅1=24 \\ 5! &= 5⋅4⋅3⋅2⋅1=120 \end{align*}\]
An example of formula containing a factorial is \(a_n=(n+1)!\). The sixth term of the sequence can be found by substituting \(6\) for \(n\).
\(a_6=(6+1)!=7!=7·6·5·4·3·2·1=5040\)
The factorial of any whole number \(n\) is \(n(n−1)!\) We can therefore also think of \(5!\) as \(5⋅4!\).
FACTORIAL
\(n\) factorial is a mathematical operation that can be defined using a recursive formula. The factorial of \(n\), denoted \(n!\), is defined for a positive integer \(n\) as:
\[\begin{align} 0! &= 1 \\ 1! &= 1 \\ n! &= n(n−1)(n−2)⋯(2)(1), \text{ for }n≥2 \end{align}\]
The special case \(0!\) is defined as \(0!=1\).
Q&A: Can factorials always be found using a calculator?
No. Factorials get large very quickly—faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial.
Example \(\PageIndex{7}\): Writing the Terms of a Sequence Using Factorials
Write the first five terms of the sequence defined by the explicit formula \(a_n=\dfrac{5n}{(n+2)!}\).
Solution
Substitute \(n=1\), \(n=2\), and so on in the formula.
\(\begin{array}{ll} n=1 & a_1=\dfrac{5(1)}{(1+2)!}=\dfrac{5}{3!}=\dfrac{5}{3·2·1}=\dfrac{5}{6} \\ n=2 & a_2=\dfrac{5(2)}{(2+2)!}=\dfrac{10}{4!}=\dfrac{10}{4·3·2·1}=\dfrac{5}{12} \\ n=3 & a_3=\dfrac{5(3)}{(3+2)!}=\dfrac{15}{5!}=\dfrac{15}{5·4·3·2·1}=\dfrac{1}{8} \\ n=4 & a_4=\dfrac{5(4)}{(4+2)!}=\dfrac{20}{6!}=\dfrac{20}{6·5·4·3·2·1}=\dfrac{1}{36} \\ n=5 & a_5=\dfrac{5(5)}{(5+2)!}=\dfrac{25}{7!}=\dfrac{25}{7·6·5·4·3·2·1}=\dfrac{5}{1,008} \end{array}\)
The first five terms are \(\left \{\dfrac{5}{6}, \dfrac{5}{12}, \dfrac{1}{8}, \dfrac{1}{36}, \dfrac{5}{1,008} \right \}\).
Analysis
Figure \(\PageIndex{7}\) shows the graph of the sequence. Notice that, since factorials grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as nn increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero.
Figure \(\PageIndex{7}\)
Exercise \(\PageIndex{7}\)
Write the first five terms of the sequence defined by the explicit formula \(a_n=\dfrac{(n+1)!}{2n}\).
- Answer
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The first five terms are \(\left \{1, \dfrac{3}{2}, 4, 15, 72 \right \}\).
Media
Access this online resource for additional instruction and practice with sequences.
Key Equations
| Formula for a factorial |
\(0!=1\) \(1!=1\) \(n!=n(n−1)(n−2)⋯(2)(1)\), for \(n≥2\) |
Key Concepts
- A sequence is a list of numbers, called terms, written in a specific order.
- Explicit formulas define each term of a sequence using the position of the term. See Example \(\PageIndex{1}\), Example \(\PageIndex{2}\), and Example \(\PageIndex{3}\).
- An explicit formula for thenth nth term of a sequence can be written by analyzing the pattern of several terms. See Example \(\PageIndex{4}\).
- Recursive formulas define each term of a sequence using previous terms.
- Recursive formulas must state the initial term, or terms, of a sequence.
- A set of terms can be written by using a recursive formula. See Example \(\PageIndex{5}\) and Example \(\PageIndex{6}\).
- A factorial is a mathematical operation that can be defined recursively.
- The factorial of \(n\) is the product of all integers from \(1\) to \(n\). See Example \(\PageIndex{7}\).
Contributors and Attributions
-
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .


4.2E: Exercises for Section 12.2
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Taking Derivatives of Vector-Valued Functions
In questions 1 - 10, compute the derivative of each vector-valued function.
1) \(\vecs r(t)=t^3 \,\hat{\mathbf{i}}+3t^2 \,\hat{\mathbf{j}}+\dfrac{t^3}{6}\,\hat{\mathbf{k}}\)
- Answer
- \(\vecs r'(t)= 3t^2 \,\hat{\mathbf{i}}+6t\,\hat{\mathbf{j}}+\frac{1}{2}t^2\,\hat{\mathbf{k}}\)
2) \(\vecs r(t)=\sin(t) \,\hat{\mathbf{i}}+\cos(t) \,\hat{\mathbf{j}}+e^t \,\hat{\mathbf{k}}\)
3) \(\vecs r(t)=e^{−t} \,\hat{\mathbf{i}}+\sin(3t) \,\hat{\mathbf{j}}+10 \sqrt{t} \,\hat{\mathbf{k}}\). A sketch of the graph is shown here. Notice the varying periodic nature of the graph.
- Answer
- \(\vecs r'(t) = −e^{−t}\,\hat{\mathbf{i}}+3\cos (3t)\,\hat{\mathbf{j}}+\dfrac{5}{\sqrt{t}}\,\hat{\mathbf{k}}\)
4) \(\vecs r(t)=e^t \,\hat{\mathbf{i}}+2e^t \,\hat{\mathbf{j}}+\,\hat{\mathbf{k}}\)
5) \(\vecs r(t)=\,\hat{\mathbf{i}}+\,\hat{\mathbf{j}}+\,\hat{\mathbf{k}}\)
- Answer
- \(\vecs r'(t) = ⟨0,0,0⟩ = \vecs 0\)
6) \(\vecs r(t)=te^t \,\hat{\mathbf{i}}+t \ln(t) \,\hat{\mathbf{j}}+\sin(3t)\,\hat{\mathbf{k}}\)
7) \(\vecs r(t)=\left\langle\dfrac{1}{t+1},\,\arctan(t), \,\ln t^3 \right\rangle\)
- Answer
- \(\vecs r'(t) = ⟨\dfrac{−1}{(t+1)^2},\dfrac{1}{1+t^2},\dfrac{3}{t}⟩\)
8) \(\vecs r(t)= \langle\tan 2t, \,\sec 2t, \,\sin ^2 t\rangle\)
9) \(\vecs r(t)=\langle 3,\, 4 \sin (3t),\, t \cos(t)\rangle\)
- Answer
- \(\vecs r'(t) = ⟨0,12 \cos(3t), \cos t−t \sin t⟩\)
10) \(\vecs r(t)=t^2 \,\hat{\mathbf{i}}+te^{−2t} \,\hat{\mathbf{j}}−5e^{−4t} \,\hat{\mathbf{k}}\)
11) a. Describe and sketch the curve represented by the vector-valued function \(\vecs r(t)=⟨6t,6t−t^2⟩\).
b. Locate the highest point on the curve \(\vecs r(t)=⟨6t,6t−t^2⟩\) and give the value of the function at this point.
- Answer
- b. \(\vecs r(t)=⟨18,9⟩\) at \( t=3\)
12) Find the parametric equations of the tangent line to the curve \(\vecs r(t)=⟨t, t^2,t⟩\) at \(t=2\).
13) Find the parametric equations of the tangent line to the curve \(\vecs r(t)=⟨e^t,e^{−t},0⟩\) at \(t=0\).
- Answer
- \(x=1+t,\quad y=1−t,\quad z=0\)
14) Compute the first, second, and third derivatives of \(\vecs r(t)=3t \,\hat{\mathbf{i}}+6\ln(t) \,\hat{\mathbf{j}}+5e^{−3t}\,\hat{\mathbf{k}}\).
Describing Motion with Vector-Valued Functions
In questions 15 - 17, find the velocity and acceleration at the given times, plot the graph of the position function, and draw in the velocity and acceleration vectors at the corresponding locations on the curve.
15) \(\vecs r(t)= t \,\hat{\mathbf{i}}+t^3\,\hat{\mathbf{j}}\), at \(t = 0\) and at \(t = 1\)
- Answer
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\(\begin{array}{lll} \, & t = 0: & t = 1: \\
\vecs r(t) = t \, \mathbf{\hat i} + t^3 \, \mathbf{\hat j}, & \vecs r(0) = \vecs 0, & \vecs r(1) = \mathbf{\hat i} + \mathbf{\hat j} \\
\vecs v(t) = \mathbf{\hat i} + 3t^2 \, \mathbf{\hat j}, & \vecs v(0) = \mathbf{\hat i}, & \vecs v(1) = \mathbf{\hat i} + 3\,\mathbf{\hat j} \\
\vecs a(t) = 6t \, \mathbf{\hat j}, & \vecs a(0) = \vecs 0, & \vecs a(1) = 6\,\mathbf{\hat j} \end{array}\)
16) \(\vecs r(t)= \cos t \,\hat{\mathbf{i}}+\sin 3t \,\hat{\mathbf{j}}\), at \(t = 0\), at \(t = \frac{\pi}{4}\), and at \(t = \frac{\pi}{2}\)
17) \(\vecs r(t)= \ln t \,\hat{\mathbf{i}}+(t-2)\,\hat{\mathbf{j}}\), at \(t = 1\) and at \(t = 2\)
- Answer
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\(\begin{array}{lll} \, & t = 1: & t =2: \\
\vecs r(t) = (\ln t) \, \mathbf{\hat i} + (t - 2) \, \mathbf{\hat j}, & \vecs r(1) = - \mathbf{\hat j}, & \vecs r(2) = (\ln 2)\,\mathbf{\hat i} \\
\vecs v(t) = \dfrac{1}{t}\,\mathbf{\hat i} + \mathbf{\hat j}, & \vecs v(1) = \mathbf{\hat i} + \mathbf{\hat j}, & \vecs v(2) = \frac{1}{2}\mathbf{\hat i} + \mathbf{\hat j} \\
\vecs a(t) = -\dfrac{1}{t^2} \, \mathbf{\hat i}, & \vecs a(1) = -\mathbf{\hat i}, & \vecs a(2) = -\frac{1}{4}\,\mathbf{\hat i} \end{array}\)
In questions 18 - 24, find the velocity, speed, and acceleration of a particle with the given position function. Remember that the speed is the magnitude of the velocity represented by \(‖\vecs v(t)‖\) or \(‖\vecs r′(t)‖\) .
18) \(\vecs r(t)=e^{2t} \,\hat{\mathbf{i}}+\sin t \,\hat{\mathbf{j}}\)
19) \(\vecs r(t)=\cos t^3 \,\hat{\mathbf{i}}+\sin t^3 \,\hat{\mathbf{j}}\)
- Answer
- \(\vecs v(t)=-3t^2\sin t^3 \,\hat{\mathbf{i}}+3t^2\cos t^3 \,\hat{\mathbf{j}}, \quad \text{speed}(t)=‖\vecs v(t)‖=3t^2, \quad \vecs a(t) = \left( -6t\sin t^3 - 9t^4\cos t^3 \right) \,\hat{\mathbf{i}}+\left( 6t\cos t^3 - 9t^4\sin t^3 \right) \,\hat{\mathbf{j}}\)
20) \(\vecs r(t)=⟨e^t,e^{−t},0⟩\)
21) \(\vecs r(t)=⟨t+ \cos t,t− \sin t⟩\)
- Answer
- \(\vecs v(t)=⟨1− \sin t,1−\cos t⟩, \quad \text{speed}(t)=‖\vecs v(t)‖=\sqrt{3−2( \sin t+\cos t)}, \quad \vecs a(t) = ⟨- \cos t, \sin t⟩\)
22) \(\vecs r(t)=\dfrac{2t−1}{2t+1} \,\hat{\mathbf{i}}+\ln(1−4t^2) \,\hat{\mathbf{j}}\)
23) \(\vecs r(t)=\cos 3t \,\hat{\mathbf{i}}+\sin 3t \,\hat{\mathbf{j}} + 0.5t \,\hat{\mathbf{k}} \)
- Answer
- \(\vecs v(t)=-3\sin 3t \,\hat{\mathbf{i}}+3\cos 3t \,\hat{\mathbf{j}} +0.5 \,\mathbf{\hat k}, \quad \text{speed}(t)=‖\vecs v(t)‖=\sqrt{9.25}\text{ units/sec}, \quad \vecs a(t) = -9\cos 3t \,\hat{\mathbf{i}}-9\sin 3t \,\hat{\mathbf{j}}\)
24) \(\vecs r(t)=e^{-t} \,\hat{\mathbf{i}}+(\ln t) \,\hat{\mathbf{j}}+(\sin 7t)\,\hat{\mathbf{k}}\)
25) Consider the position vector for a particle to be \(\vecs r(t)=t \,\hat{\mathbf{i}}+t^2\,\hat{\mathbf{j}}+t^3 \,\hat{\mathbf{k}}\). The graph is shown here:
a. Find the velocity vector at any time.
b. Find the speed of the particle at time \(t=2\) sec.
c. Find the acceleration at time \(t=2\) sec.
- Answer
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a. \(\vecs v(t)=\hat{\mathbf{i}}+2t\,\hat{\mathbf{j}}+3t^2 \,\hat{\mathbf{k}}\)
b. \(\sqrt{161}\) units/sec
Note that \(\text{speed}(t) = \|\vecs v(t)\| = \sqrt{1^2 + (2t)^2 + (3t^2)^2} = \sqrt{1 + 4t^2 + 9t^4}.\)
Hence, \(\text{speed}(2) = \sqrt{1 + 16 + 9(16)} = \sqrt{161}\) units/sec.
c. Since \(\vecs a(t)=2\,\hat{\mathbf{j}}+6t \,\hat{\mathbf{k}},\) \(\vecs a(2) = 2\,\hat{\mathbf{j}}+12 \,\hat{\mathbf{k}}\)
26) A particle travels along the path of an ellipse with the equation \(\vecs r(t)=\cos t \,\hat{\mathbf{i}}+2 \sin t \,\hat{\mathbf{j}}+0 \,\hat{\mathbf{k}}\). Find the following:
a. Velocity of the particle
- Answer
- \(\vecs v(t)=⟨−\sin t,2 \cos t,0⟩\)
b. Speed of the particle at \(t=\frac{π}{4}\)
c. Acceleration of the particle at \(t=\frac{π}{4}\)
- Answer
- \(\vecs a(t)=⟨−\frac{\sqrt{2}}{2},−\sqrt{2},0⟩\)
27) Show that if the speed of a particle traveling along a curve represented by a vector-valued function is constant, then the velocity function is always perpendicular to the acceleration function.
- Answer
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\(\begin{align*} ‖\vecs v(t)‖ \; & = k \\ \vecs v(t)·\vecs v(t) \; & = k^2 \\ \frac{d}{dt}\Big(\vecs v(t)·\vecs v(t)\Big) \; & =\frac{d}{dt}\Big(k^2\Big)=0 \\ \vecs v(t)·\vecs v′(t)+\vecs v′(t)·\vecs v(t) \; & = 0 \\ 2\vecs v(t)·\vecs v′(t) \; & =0 \\ \vecs v(t)·\vecs v′(t) \; & = 0\end{align*}\)
The last statement implies that the velocity and acceleration are perpendicular or orthogonal.
28) Given the vector-valued function \(\vecs r(t)=⟨\tan t,\sec t,0⟩\) (graph is shown here), find the following:
a. Velocity
- Answer
- \(\vecs v(t)=\langle \sec ^2 t,\,\sec t \tan t\rangle\)
b. Speed
- Answer
- \(‖\vecs v(t)‖=\sqrt{\sec ^4 t+\sec ^2 t \tan ^2 t}=\sqrt{(\sec ^2 t)(\sec ^2 t+\tan ^2 t)}\)
c. Acceleration
- Answer
- \(\vecs a(t)=\langle 2\sec ^2 t\tan t,\,\sec t \tan^2 t+\sec^3 t\rangle\)
29) Find the minimum speed of a particle traveling along the curve \(\vecs r(t)=⟨t+\cos t,t−\sin t⟩\), where \(t∈[0,2π)\). Then also find its maximum speed on this interval.
- Answer
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Min. speed is \(\sqrt{3-2\sqrt{2}}\approx 0.41421\) when \(t=\tfrac{\pi}{4}\).
Max. speed is \(\sqrt{3+2\sqrt{2}}\approx 2.41421\) when \(t=\tfrac{5\pi}{4}\).
For questions 30 - 31, consider a particle that moves on a circular path of radius \(b\) according to the function \(\vecs r(t)=b \cos(\omega t) \,\hat{\mathbf{i}}+b\sin(\omega t) \,\hat{\mathbf{j}}\) , where \(\omega\) is the angular velocity, \(\dfrac{d \theta}{dt}\) .
30) Show that the speed of the particle is proportional to the angular velocity.
31) Find the velocity function and show that \(\vecs v(t)\) is always orthogonal to \(\vecs r(t)\).
- Answer
- \(\vecs r'(t)=−b \omega \sin( \omega t)\,\hat{\mathbf{i}}+b \omega \cos(\omega t)\,\hat{\mathbf{j}}\). To show orthogonality, note that \(\vecs r'(t)⋅\vecs r(t)=0\).
Smoothness of Vector-Valued Functions
For questions 32 - 40,
a. Determine any values of \(t\) at which \(\vecs r\) is not smooth.
b. Determine the open intervals on which \(\vecs r\) is smooth.
c. Graph the vector-valued function and describe its behavior at the points where it is not smooth.
32) \(\vecs r(t) = \langle 3t, 5t^2 - 1\rangle\)
33) \(\vecs r(t)= t^3\,\hat{\mathbf{i}}+5t^2 \,\hat{\mathbf{j}}\)
- Answer
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a. \(\vecs r\) is not smooth at \(t = 0\), since \(\vecs r'(0) = \vecs 0\).
b. \(\vecs r\) is smooth on the open intervals \((-\infty, 0)\) and \((0, \infty)\).
c. There is a cusp when \(t = 0\).
34) \(\vecs r(t)=\langle 5,\, 2 \sin (t),\, \cos(t)\rangle\)
35) \(\vecs r(t) = \langle t^3 - 3t^2, 7\rangle\)
- Answer
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a. \(\vecs r\) is not smooth at \(t = 0\) and \(t = 2\), since \(\vecs r'(0) = \vecs 0\) and \(\vecs r'(2) = \vecs 0\).
b. \(\vecs r\) is smooth on the open intervals \((-\infty, 0)\), \((0, 2)\), and \((2, \infty)\).
c. The motion on the curve reverses along the same path at both \(t = 0\) and \(t = 2\).
36) \(\vecs r(t)=t^2 \,\hat{\mathbf{i}}+t^3 \,\hat{\mathbf{j}}−5e^{−4t} \,\hat{\mathbf{k}}\)
37) \(\vecs r(t)=\left\langle \ln(t^2+4t+5), \,\dfrac{t^3}{3} - 4t,\, 5\right\rangle\)
- Answer
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a. \(\vecs r\) is not smooth at \(t = -2\), since \(\vecs r'(-2) = \vecs 0\).
Since the domain of \(\vecs r\) is \((-\infty, \infty)\), this is all we have to remove.
b. \(\vecs r\) is smooth on the open intervals \((-\infty, -2)\) and \((-2, \infty)\).
c. There is a cusp when \(t = -2\).
38) \(\vecs r(t) = \left( 5\cos t - \cos 5t\right) \,\hat{\mathbf{i}} + \left( 5\sin t - \sin 5t \right) \,\hat{\mathbf{j}}\), for \(0 \le t \le 2\pi\)
39) \(\vecs r(t)=\sqrt{t^3 + 9t^2} \,\hat{\mathbf{i}}+\left(t^2 +12t\right) \,\hat{\mathbf{j}}+7\,\hat{\mathbf{k}}\)
- Answer
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a. The domain of \(\vecs r\) is \([-9, \infty)\).
And \(\vecs r\) is not smooth at \(t = -6\), since \(\vecs r'(-6) = \vecs 0\).
The domain of \(\vecs r'\) is \((-9, \infty)\), since \(\vecs r'\) is undefined at \(t = -9\).
b. \(\vecs r\) is smooth on the open intervals \((-9, -6)\) and \((-6, \infty)\).
c. There is a cusp when \(t = -6\).
40) \(\vecs r(t)= \cos^3 t\,\hat{\mathbf{i}}+\sin t \,\hat{\mathbf{j}}\), for \(0 \le t \le 2\pi\)
- Answer
-
a. The domain of \(\vecs r\) is \((-\infty, \infty)\).
\(\vecs r'(t) = -3(\cos^2 t)(\sin t)\,\hat{\mathbf{i}}+\cos t \,\hat{\mathbf{j}}\). It's domain is also \((-\infty, \infty)\).
But note that both components have a factor of \(\cos t\), so both components will be \(0\) when \(\cos t = 0\).
Therefore, \(\vecs r\) is not smooth at \(t = \frac{\pi}{2}\) and at \(t = \frac{3\pi}{2}\), since \(\vecs r'\left( \frac{\pi}{2}\right) = \vecs 0\) and \(\vecs r'\left( \frac{3\pi}{2}\right) = \vecs 0\). Note then that \(\vecs r\) is not smooth for any odd multiple of \(\frac{\pi}{2}\), that is for \(t = \frac{(2n + 1)\pi}{2}\), for any integer value \(n\).
b. \(\vecs r\) is smooth on the open intervals \(\left(\dfrac{(2n - 1)\pi}{2}, \dfrac{(2n + 1)\pi}{2}\right)\), for any integer value \(n\).
c. There is a cusp when \(t = \dfrac{(2n + 1)\pi}{2}\), for any integer value \(n\).
Properties of the Derivative
For questions, 41 - 43, evaluate each expression given that \(\vecs r(t)=t \,\hat{\mathbf{i}}+t^2 \,\hat{\mathbf{j}}−t^4 \,\hat{\mathbf{k}} \) and \(\vecs s(t)=\sin(t) \,\hat{\mathbf{i}}+e^t \,\hat{\mathbf{j}}+ \cos(t) \,\hat{\mathbf{k}} \)
41) \(\dfrac{d}{dt}\big[\vecs r(t^2)\big]\)
- Answer
- \(\dfrac{d}{dt}\big[\vecs r(t^2)\big] = ⟨2t,4t^3,−8t^7⟩\)
42) \(\dfrac{d}{dt}\big[t^2⋅\vecs s(t)\big]\)
43) \(\dfrac{d}{dt}\big[\vecs r(t)⋅\vecs s(t)\big]\)
- Answer
- \(\dfrac{d}{dt}\big[\vecs r(t)⋅\vecs s(t)\big]=\sin t+2te^t−4t^3 \cos t+t\cos t+t^2e^t+t^4\sin t\)
44) Find \(\vecs r'(t)⋅\vecs r''(t) \; for \; \vecs r(t)=−3t^5 \,\hat{\mathbf{i}}+5t \,\hat{\mathbf{j}}+2t^2 \,\hat{\mathbf{k}}\).
- Answer
- \(\vecs r'(t)⋅\vecs r''(t) = 900t^7+16t\)
45) Given \(\vecs r(t)=t \,\hat{\mathbf{i}}+3t \,\hat{\mathbf{j}}+t^2 \,\hat{\mathbf{k}}\) and \(\vecs u(t)=4t \,\hat{\mathbf{i}}+t^2 \,\hat{\mathbf{j}}+t^3 \,\hat{\mathbf{k}}\), find \(\frac{d}{dt}(\vecs r(t) \times \vecs u(t))\).
46) Evaluate \(\dfrac{d}{dt}\big[\vecs u(t) \times\vecs u′(t)\big]\) given \(\vecs u(t)=t^2 \,\hat{\mathbf{i}}−2t \,\hat{\mathbf{j}}+\,\hat{\mathbf{k}}\).
- Answer
- \(\dfrac{d}{dt}\big[\vecs u(t) \times\vecs u′(t)\big] = 0 \,\hat{\mathbf{i}} +2 \,\hat{\mathbf{j}}+4t \,\hat{\mathbf{k}}\)
47) Given \(\vecs r(t)=t \,\hat{\mathbf{i}}+2\sin t \,\hat{\mathbf{j}}+2 \cos t \,\hat{\mathbf{k}}\) and \(\vecs u(t)=\dfrac{1}{t} \,\hat{\mathbf{i}}+2 \sin t \,\hat{\mathbf{j}}+2 \cos t \,\hat{\mathbf{k}}\), find the following:
a. \(\vecs r(t) \times \vecs u(t)\)
- Answer
- \(\vecs r(t) \times \vecs u(t) = \left\langle 0, \;2(\cos t)\left(\frac{1}{t}-t\right),\; 2 (\sin t)\left(t- \frac{1}{t}\right)\right\rangle\)
b. \(\frac{d}{dt}\big(\vecs r(t) \times \vecs u(t)\big)\)
- Answer
- \(\frac{d}{dt}\big(\vecs r(t) \times \vecs u(t)\big) = \left\langle 0, \;2(\sin t)\left(t− \frac{1}{t}\right)−2 (\cos t)\left(1+ \frac{1}{t^2}\right),\;2 \left(\sin t\right)\left(1+ \frac{1}{t^2}\right)+2 \left(\cos t\right)\left(t−\frac{1}{t}\right)\right\rangle\)
c. Now, use the product rule for the derivative of the cross product of two vectors and show this result is the same as the answer for the preceding problem.
Unit Tangent Vectors
For questions 48 - 51, find a unit tangent vector at the indicated value of \(t\) .
48) \(\vecs r(t)=3t^3 \,\hat{\mathbf{i}}+2t^2 \,\hat{\mathbf{j}}+\dfrac{1}{t} \,\hat{\mathbf{k}}; \quad t=1\)
49) \(\vecs r(t)=t \,\hat{\mathbf{i}}+\sin(2t) \,\hat{\mathbf{j}}+\cos(3t) \,\hat{\mathbf{k}}; \quad t=\frac{π}{3}\)
- Answer
-
\(\vecs r'\left(\frac{π}{3}\right) = \langle 1, \,-1,0\rangle\) is a tangent vector, so a unit tangent vector would be:
\(\frac{1}{\sqrt{2}}⟨1,−1,0⟩ \quad = \quad \langle \frac{\sqrt{2}}{2}, \,-\frac{\sqrt{2}}{2},\, 0\rangle\)
50) \(\vecs r(t)=\cos(2t) \,\hat{\mathbf{i}}+2 \sin t \,\hat{\mathbf{j}}+t^2 \,\hat{\mathbf{k}};\quad t=\frac{π}{2}\)
51) \(\vecs r(t)=3e^t \,\hat{\mathbf{i}}+2e^{−3t} \,\hat{\mathbf{j}}+4e^{2t} \,\hat{\mathbf{k}}; \quad t= \ln(2)\)
- Answer
-
\(\vecs r'(\ln(2))= ⟨6,−\frac{3}{4},32⟩\) is a tangent vector, so a unit tangent vector would be:
\(\dfrac{1}{\sqrt{1060.5625}}\left\langle 6,−\frac{3}{4},32\right\rangle \quad = \quad \left\langle \dfrac{24\sqrt{16969}}{16969}, -\dfrac{12\sqrt{16969}}{67876}, \frac{128\sqrt{16969}}{16969}\right\rangle\)
For questions 52 - 58, find the unit tangent vector \(\vecs T(t)\) for the following parameterized curves.
52) \(\vecs r(t)=t \,\hat{\mathbf{i}}+3t \,\hat{\mathbf{j}}+t^2 \,\hat{\mathbf{k}}\)
53) \(\vecs r(t)=6 \,\hat{\mathbf{i}}+\cos(3t) \,\hat{\mathbf{j}}+3\sin(4t) \,\hat{\mathbf{k}}, \quad 0≤t<2π\)
- Answer
- \(\vecs T(t)=\dfrac{1}{\sqrt{9\sin ^2 (3t)+144\cos ^2 (4t)}}⟨0,−3\sin(3t),12\cos(4t)⟩\)
54) \(\vecs r(t)=⟨t \cos t,t \sin t⟩\)
55) \(\vecs r(t)=⟨t+1,2t+1,2t+2⟩\)
- Answer
- \(\vecs T(t)=\frac{1}{3} ⟨1,2,2⟩\)
56) \(\vecs r(t)=\cos t \,\hat{\mathbf{i}}+\sin t \,\hat{\mathbf{j}}+\sin t \,\hat{\mathbf{k}}, \quad 0≤t<2π\). Two views of this curve are presented here:
57) \(\vecs r(t)=\left\langle t,\dfrac{1}{t}\right\rangle\). The graph is shown here:
- Answer
- \(\vecs T(t)=\left\langle\dfrac{t^2}{\sqrt{t^4+1}},\frac{-1}{\sqrt{t^4+1}}\right\rangle\)
58) \(\vecs r(t)=3 \cos(4t) \,\hat{\mathbf{i}}+3 \sin(4t) \,\hat{\mathbf{j}}+5t\,\hat{\mathbf{k}},\quad 1 \le t \le 2\)
- Answer
- \(\vecs T(t)=−\frac{12}{13} \sin(4t) \,\hat{\mathbf{i}}+ \frac{12}{13}\cos (4t) \,\hat{\mathbf{j}}+\frac{5}{13} \,\hat{\mathbf{k}}\)
59) A particle travels along the path of a helix with the equation \(\vecs r(t)= \cos(t) \,\hat{\mathbf{i}}+\sin(t) \,\hat{\mathbf{j}}+t \,\hat{\mathbf{k}}\). See the graph presented here:
Find the following:
a. Velocity of the particle at any time
- Answer
- \(\vecs v(t)=⟨−\sin t,\cos t,1⟩\)
b. Speed of the particle at any time
c. Acceleration of the particle at any time
- Answer
- \(\vecs a(t)=−\cos t \,\hat{\mathbf{i}}− \sin t \,\hat{\mathbf{j}}+0 \,\hat{\mathbf{k}}\)
d. Find the unit tangent vector for the helix.
Integration of Vector-Valued Functions
Evaluate the following integrals:
60) \(\displaystyle \int \left(e^t \,\hat{\mathbf{i}}+\sin t \,\hat{\mathbf{j}}+ \frac{1}{2t−1} \,\hat{\mathbf{k}}\right)\, dt\)
61) \(\displaystyle \int_0^1 \vecs r(t)\,dt\), where \(\vecs r(t)=\left\langle\sqrt[3]{t},\dfrac{1}{t+1},e^{−t}\right\rangle\)
- Answer
- \(\frac{3}{4}\,\hat{\mathbf{i}}+\ln(2) \,\hat{\mathbf{j}}+(1−\frac{1}{e}) \,\hat{\mathbf{k}}\)
62) Evaluate \(\displaystyle \int_0^3‖t\,\hat{\mathbf{i}}+t^2\,\hat{\mathbf{j}}‖dt\).
- Answer
- \(\frac{1}{3}(10^{\frac{3}{2}}−1)\)
63) The acceleration function, initial velocity, and initial position of a particle are
\[\begin{align*} \vecs a(t)&=−5 \cos t \,\hat{\mathbf{i}}−5\sin t \,\hat{\mathbf{j}}, \\ \vecs v(0)&=9 \,\hat{\mathbf{i}}+2 \,\hat{\mathbf{j}}, \quad \text{and} \\ \vecs r(0)&=5 \,\hat{\mathbf{i}} \end{align*}\]
Find \(\vecs v(t)\) and \(\vecs r(t)\).
- Answer
-
\(\vecs v(t) = \left(9 - 5\sin t\right)\,\hat{\mathbf{i}} + \left(-3 + 5\cos t\right)\,\hat{\mathbf{j}}\)
\(\vecs r(t) = \left(9t + 5\cos t\right)\,\hat{\mathbf{i}} + \left(-3t + 5\sin t\right)\,\hat{\mathbf{j}}\)
64) Find the antiderivative of \(\vecs r'(t)=\cos(2t) \,\hat{\mathbf{i}}−2\sin t \,\hat{\mathbf{j}}+\dfrac{1}{1+t^2} \,\hat{\mathbf{k}}\) that satisfies the initial condition \(\vecs r(0)=3 \,\hat{\mathbf{i}}−2 \,\hat{\mathbf{j}}+\,\hat{\mathbf{k}}\).
65) An object starts from rest at point \(P(1,2,0)\) and moves with an acceleration of \(\vecs a(t)=\,\hat{\mathbf{j}}+2 \,\hat{\mathbf{k}}\), where \(‖\vecs a(t)‖\) is measured in feet per second per second. Find the location of the object after \(t=2\) sec.
Contributors:
-
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .
- Paul Seeburger (Monroe Community College) created questions 11 - 19.


20: Answer Key Last updated Save as PDF Page ID 52378 Ernstmeyer & Christman (Eds.) Chippewa Valley Technical College via OpenRN 20.1: Chapter 1 (Scope of Practice) 20.2: Chapter 2 (Communication) 20.3: Chapter 3 (Diverse Patients) 20.4: Chapter 4 (Nursing Process) 20.5: Chapter 5 (Safety) 20.6: Chapter 6 (Cognitive Impairments) 20.7: Chapter 7 (Sensory Impairments) 20.8: Chapter 8 (Oxygenation) 20.9: Chapter 9 (Infection) 20.10: Chapter 10 (Integumentary) 20.11: Chapter 11 (Comfort) 20.12: Chapter 12 (Sleep and Rest) 20.13: Chapter 13 (Mobility) 20.14: Chapter 14 (Nutrition) 20.15: Chapter 15 (Fluids and Electrolytes) 20.16: Chapter 16 (Elimination) 20.17: Chapter 17 (Grief and Loss) 20.18: Chapter 18 (Spirituality) 20.19: Chapter 19 (Care of the Older Adult)


10.2: Two Population Means with Unknown Standard Deviations
-
- Last updated
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- The two independent samples are simple random samples from two distinct populations.
-
For the two distinct populations:
- if the sample sizes are small, the distributions are important (should be normal)
- if the sample sizes are large, the distributions are not important (need not be normal)
The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch \(t\)-test. The degrees of freedom formula was developed by Aspin-Welch.
The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, \(\bar{X}_{1} - \bar{X}_{2}\), and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.
Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error , of the difference in sample means , \(\bar{X}_{1} - \bar{X}_{2}\).
The standard error is:
\[\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}\]
The test statistic ( t -score) is calculated as follows:
\[\dfrac{(\bar{x}-\bar{x}) - (\mu_{1} - \mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}\]
where:
- \(s_{1}\) and \(s_{2}\), the sample standard deviations, are estimates of \(\sigma_{1}\) and \(\sigma_{1}\), respectively.
- \(\sigma_{1}\) and \(\sigma_{2}\) are the unknown population standard deviations.
- \(\bar{x}_{1}\) and \(\bar{x}_{2}\) are the sample means. \(\mu_{1}\) and \(\mu_{2}\) are the population means.
The number of degrees of freedom (\(df\)) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The \(df\) are not always a whole number. The test statistic calculated previously is approximated by the Student's t -distribution with \(df\) as follows:
Degrees of freedom
\[df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1}-1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2} + \left(\dfrac{1}{n_{2}-1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}\]
When both sample sizes \(n_{1}\) and \(n_{2}\) are five or larger, the Student's t approximation is very good. Notice that the sample variances \((s_{1})^{2}\) and \((s_{2})^{2}\) are not pooled. (If the question comes up, do not pool the variances.)
It is not necessary to compute the degrees of freedom by hand. A calculator or computer easily computes it.
Example \(\PageIndex{1}\): Independent groups
The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table \(\PageIndex{1}\). Each populations has a normal distribution.
| Sample Size | Average Number of Hours Playing Sports Per Day | Sample Standard Deviation | |
|---|---|---|---|
| Girls | 9 | 2 | 0.8660.866 |
| Boys | 16 | 3.2 | 1.00 |
Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.
Answer
The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, \(\mu_{g}\) is the population mean for girls and \(\mu_{b}\) is the population mean for boys. This is a test of two independent groups, two population means.
Random variable: \(\bar{X}_{g} - \bar{X}_{b} =\) difference in the sample mean amount of time girls and boys play sports each day.
- \(H_{0}: \mu_{g} = \mu_{b}\)
- \(H_{0}: \mu_{g} - \mu_{b} = 0\)
- \(H_{a}: \mu_{g} \neq \mu_{b}\)
- \(H_{a}: \mu_{g} - \mu_{b} \neq 0\)
The words "the same" tell you \(H_{0}\) has an "=". Since there are no other words to indicate \(H_{a}\), assume it says "is different." This is a two-tailed test.
Distribution for the test: Use \(t_{df}\) where \(df\) is calculated using the \(df\) formula for independent groups, two population means. Using a calculator, \(df\) is approximately 18.8462. Do not pool the variances.
Calculate the p -value using a Student's t -distribution: \(p\text{-value} = 0.0054\)
Graph:
\[s_{g} = 0.866\]
\[s_{b} = 1\]
So,
\[\bar{x}_{g} - \bar{x}_{b} = 2 - 3.2 = -1.2\]
Half the \(p\text{-value}\) is below –1.2 and half is above 1.2.
Make a decision: Since \(\alpha > p\text{-value}\), reject \(H_{0}\). This means you reject \(\mu_{g} = \mu_{b}\). The means are different.
Press
STAT
. Arrow over to
TESTS
and press
4:2-SampTTest
. Arrow over to Stats and press
ENTER
. Arrow down and enter
2
for the first sample mean, \(\sqrt{0.866}\) for Sx1,
9
for n1,
3.2
for the second sample mean,
1
for Sx2, and
16
for n2. Arrow down to μ1: and arrow to
does not equal
μ2. Press
ENTER
. Arrow down to Pooled: and
No
. Press
ENTER
. Arrow down to
Calculate
and press
ENTER
. The \(p\text{-value}\) is \(p = 0.0054\), the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw.
Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).
Exercise \(\PageIndex{1}\)
Two samples are shown in Table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.
| Sample Size | Sample Mean | Sample Standard Deviation | |
|---|---|---|---|
| Population A | 25 | 5 | 1 |
| Population B | 16 | 4.7 | 1.2 |
Answer
The \(p\text{-value}\) is \(0.4125\), which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.
When the sum of the sample sizes is larger than \(30 (n_{1} + n_{2} > 30)\) you can use the normal distribution to approximate the Student's \(t\).
Example \(\PageIndex{2}\)
A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.
- Is this a test of two means or two proportions?
- Are the populations standard deviations known or unknown?
- Which distribution do you use to perform the test?
- What is the random variable?
- What are the null and alternate hypotheses? Write the null and alternate hypotheses in words and in symbols.
- Is this test right-, left-, or two-tailed?
- What is the \(p\text{-value}\)?
- Do you reject or not reject the null hypothesis?
Solutions
- two means
- unknown
- Student's t
- \(\bar{X}_{A} - \bar{X}_{B}\)
- \(H_{0}: \mu_{A} \leq \mu_{B}\) and \(H_{a}: \mu_{A} > \mu_{B}\)
-
right
- g. 0.1928
- h. Do not reject.
- i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B.
Exercise \(\PageIndex{2}\)
A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.
- Are the population standard deviations known?
- Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?
Answer
- They are unknown.
- The \(p\text{-value} = 0.0878\). At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.
Example \(\PageIndex{3}\)
A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table \(\PageIndex{3}\) and Table \(\PageIndex{4}\).
| 67.6 | 41.2 | 85.3 | 55.9 | 82.4 | 91.2 | 73.5 | 94.1 | 64.7 | 64.7 |
| 70.6 | 38.2 | 61.8 | 88.2 | 70.6 | 58.8 | 91.2 | 73.5 | 82.4 | 35.5 |
| 94.1 | 88.2 | 64.7 | 55.9 | 88.2 | 97.1 | 85.3 | 61.8 | 79.4 | 79.4 |
| 77.9 | 95.3 | 81.2 | 74.1 | 98.8 | 88.2 | 85.9 | 92.9 | 87.1 | 88.2 |
| 69.4 | 57.6 | 69.4 | 67.1 | 97.6 | 85.9 | 88.2 | 91.8 | 78.8 | 71.8 |
| 98.8 | 61.2 | 92.9 | 90.6 | 97.6 | 100 | 95.3 | 83.5 | 92.9 | 89.4 |
Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:
- Is this a test of two means or two proportions?
- Are the population standard deviations known or unknown?
- Which distribution do you use to perform the test?
- What is the random variable?
- What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
- Is this test right, left, or two tailed?
- What is the \(p\text{-value}\)?
- Do you reject or not reject the null hypothesis?
- At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
(See the conclusion in Example, and write yours in a similar fashion)
Be careful not to mix up the information for Group 1 and Group 2!
Answer
- two means
- unknown
- Student's \(t\)
- \(\bar{X}_{1} - \bar{X}_{2}\)
-
- \(H_{0}: \mu_{1} = \mu_{2}\) Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.
- \(H_{a}: \mu_{1} < \mu_{2}\) Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.
- left-tailed
-
\(p\text{-value} = 0.0011\)
Figure \(\PageIndex{3}\).
- Reject the null hypothesis
-
The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class.
At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.
First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to \(\mu_{1}\): and arrow to \(\neq \mu_{1}\) (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.
Cohen's Standards for Small, Medium, and Large Effect Sizes
Cohen's \(d\) is a measure of effect size based on the differences between two means. Cohen’s \(d\), named for United States statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes.
| Size of effect | \(d\) |
|---|---|
| Small | 0.2 |
| medium | 0.5 |
| Large | 0.8 |
Cohen's \(d\) is the measure of the difference between two means divided by the pooled standard deviation: \(d = \dfrac{\bar{x}_{2}-\bar{x}_{2}}{s_{\text{pooled}}}\) where \(s_{pooled} = \sqrt{\dfrac{(n_{1}-1)s^{2}_{1} + (n_{2}-1)s^{2}_{2}}{n_{1}+n_{2}-2}}\)
Example \(\PageIndex{4}\)
Calculate Cohen’s d for Example. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem.
Answer
\(\mu_{1} = 4 s_{1} = 1.5 n_{1} = 11\)
\(\mu_{2} = 3.5 s_{2} = 1 n_{2} = 9\)
\(d = 0.384\)
The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small indicating that there is not a significant difference between them.
Example \(\PageIndex{5}\)
Calculate Cohen’s \(d\) for Example. Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem.
Answer
\(d = 0.834\); Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference.
Example 10.2.6
Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the northeast and in the west as identified by Nasdaq on May 24, 2013 are listed in Table and Table, respectively.
| 94.2 | 75.2 | 69.6 | 52.0 | 48.0 | 41.9 | 36.4 | 33.4 | 31.5 | 27.6 |
| 77.3 | 71.9 | 67.5 | 50.6 | 46.2 | 38.4 | 35.2 | 33.0 | 28.7 | 26.5 |
| 76.3 | 71.7 | 56.3 | 48.7 | 43.2 | 37.6 | 33.7 | 31.8 | 28.5 | 26.0 |
| 126.0 | 70.6 | 65.2 | 51.4 | 45.5 | 37.0 | 33.0 | 29.6 | 23.7 | 22.6 |
| 116.1 | 70.6 | 58.2 | 51.2 | 43.2 | 36.0 | 31.4 | 28.7 | 23.5 | 21.6 |
| 78.2 | 68.2 | 55.6 | 50.3 | 39.0 | 34.1 | 31.0 | 25.3 | 23.4 | 21.5 |
Is there a difference in the weighted alpha of the top 30 stocks of banks in the northeast and in the west? Test at a 5% significance level. Answer the following questions:
- Is this a test of two means or two proportions?
- Are the population standard deviations known or unknown?
- Which distribution do you use to perform the test?
- What is the random variable?
- What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.
- Is this test right, left, or two tailed?
- What is the \(p\text{-value}\)?
- Do you reject or not reject the null hypothesis?
- At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______.
- Calculate Cohen’s d and interpret it.
Answer
- two means
- unknown
- Student’s-t
- \(\bar{X}_{1} - \bar{X}_{2}\)
-
- \(H_{0}: \mu_{1} = \mu_{2}\) Null hypothesis: the means of the weighted alphas are equal.
- \(H_{a}: \mu_{1} \neq \mu_{2}\) Alternative hypothesis : the means of the weighted alphas are not equal.
- two-tailed
- \(p\text{-value} = 0.8787\)
- Do not reject the null hypothesis
-
This indicates that the trends in stocks are about the same in the top 30 banks in each region.
Figure \(\PageIndex{4}\).
5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean weighted alphas for the banks in the northeast and the west are different - \(d = 0.040\), Very small, because 0.040 is less than Cohen’s value of 0.2 for small effect size. The size of the difference of the means of the weighted alphas for the two regions of banks is small indicating that there is not a significant difference between their trends in stocks.
References
- Data from Graduating Engineer + Computer Careers. Available online at www.graduatingengineer.com
- Data from Microsoft Bookshelf .
- Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013).
- “List of current United States Senators by Age.” Wikipedia. Available online at en.Wikipedia.org/wiki/List_of...enators_by_age (accessed June 17, 2013).
- “Sectoring by Industry Groups.” Nasdaq. Available online at www.nasdaq.com/markets/barcha...&base=industry (accessed June 17, 2013).
- “Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013).
- “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013).
Review
Two population means from independent samples where the population standard deviations are not known
- Random Variable: \(\bar{X}_{1} - \bar{X}_{2} =\) the difference of the sampling means
- Distribution: Student's t -distribution with degrees of freedom (variances not pooled)
Formula Review
Standard error: \[SE = \sqrt{\dfrac{(s_{1}^{2})}{n_{1}} + \dfrac{(s_{2}^{2})}{n_{2}}}\]
Test statistic ( t -score): \[t = \dfrac{(\bar{x}_{1}-\bar{x}_{2}) - (\mu_{1}-\mu_{2})}{\sqrt{\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}}}\]
Degrees of freedom:
\[df = \dfrac{\left(\dfrac{(s_{1})^{2}}{n_{1}} + \dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}}{\left(\dfrac{1}{n_{1} - 1}\right)\left(\dfrac{(s_{1})^{2}}{n_{1}}\right)^{2}} + \left(\dfrac{1}{n_{2} - 1}\right)\left(\dfrac{(s_{2})^{2}}{n_{2}}\right)^{2}\]
where:
- \(s_{1}\) and \(s_{2}\) are the sample standard deviations, and n 1 and n 2 are the sample sizes.
- \(x_{1}\) and \(x_{2}\) are the sample means.
Cohen’s \(d\) is the measure of effect size:
\[d = \dfrac{\bar{x}_{1} - \bar{x}_{2}}{s_{\text{pooled}}}\]
where
\[s_{\text{pooled}} = \sqrt{\dfrac{(n_{1} - 1)s^{2}_{1} + (n_{2} - 1)s^{2}_{2}}{n_{1} + n_{2} - 2}}\]
Glossary
- Degrees of Freedom (\(df\))
- the number of objects in a sample that are free to vary.
- Standard Deviation
- A number that is equal to the square root of the variance and measures how far data values are from their mean; notation: \(s\) for sample standard deviation and \(\sigma\) for population standard deviation.
- Variable (Random Variable)
-
a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters \(X, Y, Z,\)... Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters \(x, y, z,\).... For example, if \(X\) is the number of children in a family, then \(x\) represents a specific integer 0, 1, 2, 3, .... Variables in statistics differ from variables in intermediate algebra in the two following ways.
- The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if \(X =\) hair color, then the domain is {black, blond, gray, green, orange}.
- We can tell what specific value x of the random variable \(X\) takes only after performing the experiment.


10.3: Use Multiplication Properties of Exponents (Part 2)
Simplify Expressions Using the Product to a Power Property
We will now look at an expression containing a product that is raised to a power. Look for a pattern.
| (2x) 3 | |
| What does this mean? | 2x • 2x • 2x |
| We group the like factors together. | 2 • 2 • 2 • x • x • x |
| How many factors of 2 and of x? | 2 3 • x 3 |
| Notice that each factor was raised to the power. | (2x) 3 is 2 3 • x 3 |
| We write: | $$\begin{split} &(2x)^{3} \\ &2^{3} \cdot x^{3} \end{split}$$ |
The exponent applies to each of the factors. This leads to the Product to a Power Property for Exponents.
If a and b are real numbers and m is a whole number, then
\[(ab)^{m} = a^{m} b^{m} \tag{10.2.27}\]
To raise a product to a power, raise each factor to that power.
An example with numbers helps to verify this property:
\[\begin{split} (2 \cdot 3)^{2} &\stackrel{?}{=} 2^{2} \cdot 3^{2} \\ 6^{2} &\stackrel{?}{=} 4 \cdot 9 \\ 36 &\stackrel{?}{=} 36\; \checkmark \end{split}\]
Simplify: (−11x) 2 .
Solution
| Use the Power of a Product Property, (ab) m = a m b m . | $$(-11)^{\textcolor{red}{2}} x^{\textcolor{red}{2}} \tag{10.2.28}$$ |
| Simplify. | $$121x^{2} \tag{10.2.29}$$ |
Simplify: (−14x) 2 .
- Answer
-
196x 2
Simplify: (−12a) 2 .
- Answer
-
144a 2
Simplify: (3xy) 3 .
Solution
| Raise each factor to the third power. | $$3^{\textcolor{red}{3}} x^{\textcolor{red}{3}} y^{\textcolor{red}{3}} \tag{10.2.30}$$ |
| Simplify. | $$27x^{3} y^{3} \tag{10.2.31}$$ |
Simplify: (−4xy) 4 .
- Answer
-
256x 4 y 4
Simplify: (6xy) 3 .
- Answer
-
216x 3 y 3
Simplify Expressions by Applying Several Properties
We now have three properties for multiplying expressions with exponents. Let’s summarize them and then we’ll do some examples that use more than one of the properties.
If a, b are real numbers and m, n are whole numbers, then
| Product Property | a m • a n = a m + n |
| Power Property | (a m ) n = a m • n |
| Product to a Power Property | (ab) m = a m b m |
Simplify: (x 2 ) 6 (x 5 ) 4 .
Solution
| Use the Power Property. | x 12 • x 20 |
| Add the exponents. | x 32 |
Simplify: (x 4 ) 3 (x 7 ) 4 .
- Answer
-
x 40
Simplify: (y 9 ) 2 (y 8 ) 3 .
- Answer
-
y 42
Simplify: (−7x 3 y 4 ) 2 .
Solution
| Take each factor to the second power. | (−7) 2 (x 3 ) 2 (y 4 ) 2 |
| Use the Power Property. | 49x 6 y 8 |
Simplify: (−8x 4 y 7 ) 3 .
- Answer
-
-512x 12 y 21
Simplify: (−3a 5 b 6 ) 4 .
- Answer
-
81a 20 b 24
Simplify: (6n) 2 (4n 3 ).
Solution
| Raise 6n to the second power. | 6 2 n 2 • 4n 3 |
| Simplify. | 36n 2 • 4n 3 |
| Use the Commutative Property. | 36 • 4 • n 2 • n 3 |
| Multiply the constants and add the exponents. | 144n 5 |
Notice that in the first monomial, the exponent was outside the parentheses and it applied to both factors inside. In the second monomial, the exponent was inside the parentheses and so it only applied to the n.
Simplify: (7n) 2 (2n 12 ).
- Answer
-
98n 14
Simplify: (4m) 2 (3m 3 ).
- Answer
-
48m 5
Simplify: (3p 2 q) 4 (2pq 2 ) 3 .
Solution
| Use the Power of a Product Property. | 3 4 (p 2 ) 4 q 4 • 2 3 p 3 (q 2 ) 3 |
| Use the Power Property. | 81p 8 q 4 • 8p 3 q 6 |
| Use the Commutative Property. | 81 • 8 • p 8 • p 3 • q 4 • q 6 |
| Multiply the constants and add the exponents for each variable. | 648p 11 q 10 |
Simplify: (u 3 v 2 ) 5 (4uv 4 ) 3 .
- Answer
-
64u 18 v 22
Simplify: (5x 2 y 3 ) 2 (3xy 4 ) 3 .
- Answer
-
675x 7 y 18
Multiply Monomials
Since a monomial is an algebraic expression, we can use the properties for simplifying expressions with exponents to multiply the monomials.
Multiply: (4x 2 )(−5x 3 ).
Solution
| Use the Commutative Property to rearrange the factors. | 4 • (−5) • x 2 • x 3 |
| Multiply. | −20x 5 |
Multiply: (7x 7 )(−8x 4 ).
- Answer
-
-56x 11
Multiply: (−9y 4 )(−6y 5 ).
- Answer
-
54y 9
Multiply: \(\left(\dfrac{3}{4} c^{3} d\right)\)(12cd 2 ).
Solution
| Use the Commutative Property to rearrange the factors. | \(\dfrac{3}{4}\) • 12 • c 3 • c • d • d 2 |
| Multiply. | 9c 4 d 3 |
Multiply: \(\left(\dfrac{4}{5} m^{4} n^{3} d\right)\)(15mn 3 ).
- Answer
-
12m 5 n 6
Multiply: \(\left(\dfrac{2}{3} p^{5} q d\right)\)(18p 6 q 7 ).
- Answer
-
12p 11 q 8
Exponent Properties
Exponent Properties 2
Practice Makes Perfect
Simplify Expressions with Exponents
In the following exercises, simplify each expression with exponents.
- 4 5
- 10 3
- \(\left(\dfrac{1}{2}\right)^{2}\)
- \(\left(\dfrac{3}{5}\right)^{2}\)
- (0.2) 3
- (0.4) 3
- (−5) 4
- (−3) 5
- −5 4
- −3 5
- −10 4
- −2 6
- \(- \left(\dfrac{2}{3}\right)^{3}\)
- \(- \left(\dfrac{1}{4}\right)^{4}\)
- −0.5 2
- −0.1 4
Simplify Expressions Using the Product Property of Exponents
In the following exercises, simplify each expression using the Product Property of Exponents.
- x 3 • x 6
- m 4 • m 2
- a • a 4
- y 12 • y
- 3 5 • 3 9
- 5 10 • 5 6
- z • z 2 • z 3
- a • a 3 • a 5
- x a • x 2
- y p • y 3
- y a • y b
- x p • x q
Simplify Expressions Using the Power Property of Exponents
In the following exercises, simplify each expression using the Power Property of Exponents.
- (u 4 ) 2
- (x 2 ) 7
- (y 5 ) 4
- (a 3 ) 2
- (10 2 ) 6
- (2 8 ) 3
- (x 15 ) 6
- (y 12 ) 8
- (x 2 ) y
- (y 3 ) x
- (5 x ) y
- (7 a ) b
Simplify Expressions Using the Product to a Power Property
In the following exercises, simplify each expression using the Product to a Power Property.
- (5a) 2
- (7x) 2
- (−6m) 3
- (−9n) 3
- (4rs) 2
- (5ab) 3
- (4xyz) 4
- (−5abc) 3
Simplify Expressions by Applying Several Properties
In the following exercises, simplify each expression.
- (x 2 ) 4 • (x 3 ) 2
- (y 4 ) 3 • (y 5 ) 2
- (a 2 ) 6 • (a 3 ) 8
- (b 7 ) 5 • (b 2 ) 6
- (3x) 2 (5x)
- (2y) 3 (6y)
- (5a) 2 (2a) 3
- (4b) 2 (3b) 3
- (2m 6 ) 3
- (3y 2 ) 4
- (10x 2 y) 3
- (2mn 4 ) 5
- (−2a 3 b 2 ) 4
- (−10u 2 v 4 ) 3
- \(\left(\dfrac{2}{3} x^{2} y \right)^{3}\)
- \(\left(\dfrac{7}{9} p q^{4} \right)^{2}\)
- (8a 3 ) 2 (2a) 4
- (5r 2 ) 3 (3r) 2
- (10p 4 ) 3 (5p 6 ) 2
- (4x 3 ) 3 (2x 5 ) 4
- \(\left(\dfrac{1}{2} x^{2} y^{3} \right)^{4}\) (4x 5 y 3 ) 2
- \(\left(\dfrac{1}{3} m^{3} n^{2} \right)^{4}\) (9m 8 n 3 ) 2
- (3m 2 n) 2 (2mn 5 ) 4
- (2pq 4 ) 3 (5p 6 q) 2
Multiply Monomials
In the following exercises, multiply the following monomials.
- (12x 2 )(−5x 4 )
- (−10y 3 )(7y 2 )
- (−8u 6 )(−9u)
- (−6c 4 )(−12c)
- \(\left(\dfrac{1}{5} r^{8} \right)\) (20r 3 )
- \(\left(\dfrac{1}{4} a^{5} \right)\) (36a 2 )
- (4a 3 b)(9a 2 b 6 )
- (6m 4 n 3 )(7mn 5 )
- \(\left(\dfrac{4}{7} x y^{2} \right)\) (14xy 3 )
- \(\left(\dfrac{5}{8} u^{3} v \right)^{3}\) (24u 5 v)
- \(\left(\dfrac{2}{3} x^{2} y \right) \left(\dfrac{3}{4} x y^{2} \right)\)
- \(\left(\dfrac{3}{5} m^{3} n^{2} \right) \left(\dfrac{5}{9} m^{2} n^{3} \right)\)
Everyday Math
- Email Janet emails a joke to six of her friends and tells them to forward it to six of their friends, who forward it to six of their friends, and so on. The number of people who receive the email on the second round is 6 2 , on the third round is 6 3 , as shown in the table. How many people will receive the email on the eighth round? Simplify the expression to show the number of people who receive the email.
| Round | Number of people |
|---|---|
| 1 | 6 |
| 2 | 6 2 |
| 3 | 6 3 |
| … | … |
| 8 | ? |
- Salary Raul’s boss gives him a 5% raise every year on his birthday. This means that each year, Raul’s salary is 1.05 times his last year’s salary. If his original salary was $40,000, his salary after 1 year was $40,000(1.05), after 2 years was $40,000(1.05) 2 , after 3 years was $40,000(1.05) 3 , as shown in the table below. What will Raul’s salary be after 10 years? Simplify the expression, to show Raul’s salary in dollars.
| Year | Salary |
|---|---|
| 1 | $40,000(1.05) |
| 2 | $40,000(1.05) 2 |
| 3 | $40,000(1.05) 3 |
| … | … |
| 10 | ? |
Writing Exercises
- Use the Product Property for Exponents to explain why x • x = x 2 .
- Explain why −5 3 = (−5) 3 but −5 4 ≠ (−5) 4 .
- Jorge thinks \(\left(\dfrac{1}{2}\right)^{2}\) is 1. What is wrong with his reasoning?
- Explain why x 3 • x 5 is x 8 , and not x 15 .
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>."


1.1: Why It Matters- Role of Buisness
Why discuss the role of business in society?

As you embark on your study of business, you may be thinking that so much of what you will learn in school isn’t applicable to your career or future—you aren’t really going into business, for instance. Here’s a challenge that may change your mind.
Stop what you are doing and take a minute to look around you. What do you see?  You might be at a local coffee shop, hanging out with some friends who are going to help you study. Or maybe you’re sitting on the beach, reading this on your tablet or phone while you listen to the sounds of the ocean and children playing in the sand.
Now, look around again but this time consider everything within your view and ask yourself what all of these things have in common? If you said that they are all the product of business, then you’re right! How can that be, you ask? Business is everywhere, in everything we touch, we eat, we see, we smell, and we feel. Oh, not always directly, but in one way or another business is there. It’s like the air that we breathe—mostly invisible, but always present.
The next part of the challenge is this: As you work through this first section, keep trying to think of something, anything , that you can say with certainty has no relationship to business. We will check back later and see what you came up with!