3.16: Forms of Doing Business
Learning Outcomes
- Describe the legal implications of a business’ organization on its accounting
As we discussed in the prior section, most small businesses are organized functionally. As a business grows, it tends to gravitate toward a divisional structure, often landing somewhere between the two. In addition to the formal and informal internal structure, however, there is a legal side to the business organization.
Many small businesses in the United States are classified as sole proprietorships, which, just like it sounds, means that one person is the owner and is responsible for all the debts and is the beneficiary of all the profits. A sole proprietorship can have employees, can secure debt, buy assets, sell goods and services, and do anything that any business does. It can even grow large, with revenues in the hundreds of millions of dollars, but that kind of volume is rare is a sole proprietorship, because it takes more business acumen and attention to grow to that size than a single person, even a talented individual, possesses.
The single biggest drawback to a sole proprietorship is that the legal liability for debts and lawsuits (suppose someone slips and falls in your store) lies with the owner. A sole proprietor is not separate from the business, and so the owner’s personal assets are at risk if the business gets into trouble. The proprietor is the business. However, a sole proprietorship is easy to set up, simple to run, and the profits flow through the business to the owner, reported on the proprietor’s Federal and state income tax returns as self-employment income, and is then subject to both the employer and employee portion of Federal payroll taxes, in addition to Federal income tax. The tax issue could be a benefit or a burden, depending on the tax rates on the other forms of doing business.
Adding an additional owner, or two, or even more, transforms a sole proprietorship into a partnership. A general partnership looks for tax purposes much like a sole proprietorship, with the added challenge of splitting profits and losses between partners. There’s no steadfast legal rule on how that is done—it’s up to the partners to figure that out and account for it. There is one additional legal issue worth noting with a general partnership; any general partner may make a business decision (or blunder) that binds all of the general partners. Thus, the idea of the limited partnership was born. Limited partners are sometimes called silent partners, because they invest their money in the business but have no say in how it is run, thus they are sheltered from the general partnership liabilities and decisions. In addition, any time a partner comes on board or leaves, legally the old partnership dissolves and a new one has to be formed, and often a new partner can only be admitted upon approval of the existing partners, making continuity of the partnership sometimes problematic. That’s why every partnership should have a written agreement covering a wide variety of terms, such as admitting a new partner, divvying up profits and losses, distribution to the owner, and dissolution.
Most large companies though are incorporated. The word incorporated literally means “embodied.” In order to form a corporation, the owners file articles of incorporation with a state government and an artificial person is created. A corporation is a separate legal entity, able to buy and sell property, to be sued and to sue others, and to engage in business and commerce. Because it is a separate legal entity run by a board of directors who then delegate the day to day business to the “C suite” of leadership (CEO, CFO, etc.), owners are protected from debts, illegal acts, lawsuits, and all the myriad challenges that sole proprietors and general partners face in running a business. Corporate owners, called shareholders or stockholders, simply invest money in the business in exchange for a return on that investment. Profits are reported and taxed at the corporate level, If the corporation pays out profits to shareholders, the shareholders report that income as taxable (so it is, in essence, taxed twice). Distributions of profits from a corporation are called dividends.
In addition to these three basic forms of doing business, there are a myriad of alternatives and variations, the most notable of which is the Limited Liability Company (LLC). An LLC is much like a partnership. The partners are usually called members, and the LLC can elect to be taxed as a partnership is, with the profits reported on the partner’s individual income tax returns, or as a corporation, where the profits are taxed at the corporate level and distributions to members are dividends that are taxed again at the individual level. Of course, if an LLC or a corporation decides to retain income in the company for operating purposes and future growth and expansion, the double taxation is not an issue. The same holds true for S-Corporations, which are small corporations (100 shareholders or less) that elect to be taxed as a partnership. Other forms of ownership include Professional Service Corporations (PSC), Co-op for farms, factories, and even apartment buildings, and Publicly Traded Partnerships (mostly oil and gas companies.) In addition, some firms, like Kahn Academy, the Red Cross, and many colleges and private school, apply for and are granted to be exempt from tax. These not-for-profit entities have no shareholders, since they are held in the public trust, and the excess of revenues over expenditures have to stay in the organization to further the exempt purpose, like education, health, and social justice. However, most businesses can be categorized into one of the three main categories: corporation, partnership, and sole proprietorship.
| Organization Type | Number of Organizations in U.S. |
|---|---|
| C-Corporations and S-Corporations | 3.9 million |
| Partnerships | 0.7 million |
| Sole Proprietors | 0.9 million |
But why would an investor buy shares of stock in a corporation if the business was not paying out earnings in the form dividends? The answer is growth—specifically, growth in the price of the stock on the open market.
Take FaceBook for example. Mark Zuckerberg started FaceBook in his dorm room in 2004 with no real capital and a single, informal partner—his roommate Eduardo Saverin. As the site grew in popularity, Zuckerberg and Saverin formalized their arrangement by registering in Florida as a Limited Liability Company.
As investors came on board, the company incorporated, still keeping it’s highly centralized management style and functional organizational structure. The initial public offering of stock (IPO) on May 18, 2012 generated $16 billion, with shares selling at just over $38 each. The proceeds from the IPO went directly into the company coffers to fund future expansion, but subsequent trading of those shares of stock issued are between investors who are banking on FaceBook’s continued success (if they are buying) or who are taking their gains and profits (if they are selling). Just under six years later, the stock was trading at a high of over $190 a share. An investor who purchased $10,000 worth of stock during the IPO could have sold that stock in early 2018 for $50,000, making a $40,000 profit for an annualized return on investment of over 32%.
Of course, not every company boasts grand gains. Even the most astute investors lose money on some stocks. That’s why the accounting information that these publicly traded companies provide to the marketplace in the form of quarterly and annual reports is so important. Investors, lenders, and vendors extending credit are relying on that information to make informed judgments and rational financial decisions.
Below is a summary of forms of business ownership and organization:
| Organization by Type | Sole Proprietorship | General Partnership | Limited Partnership | Limited Liability Company (LLC) | C-Corporation | S-Corporation |
|---|---|---|---|---|---|---|
| Ownership | One person | Two or more | At least one general partner plus or one or more limited partners | One or more members | Virtually unlimited number of shareholders | No more than 100 stock holders |
| Capital | Generally comes from personal assets of owner | Generally comes from personal assets of owners and ability to borrow | Limited partners generally provide capital | Members provide capital, limited ability to borrow | Capital can be raised by selling shares on the open market, also debt issuance | Members provide capital, limited ability to borrow |
| Liability | Owner liable for all legal obligations | Partners are jointly and severally liable | General partners are jointly and severally liable but limited partners are not | Members are not liable for LLC legal obligations | Corporation is a separate legal entity. Shareholders are not personally liable for corporate obligations | Corporation is a separate legal entity. Shareholders are not personally liable for corporate obligations |
| Taxation | Net income taxed at the personal level | Net income taxed at the personal level | Net income taxed at the personal level | Can elect to be taxed as a partnership or a corporation | Net income taxed at the corporate level | Net income taxed at the personal level |
| Distributions to owners from profits | Not taxed | Not taxed | Not taxed | Depends on tax election | Taxed as dividends | Not taxed |
| Other | Easy to form and account for, no continuity | Relatively easy to form and account for, some continuity | More difficult to form and account for, some continuity | More difficult to form and account for, some continuity | Most difficult to form and account for, generally continuous | More difficult to form and account for, some continuity |
Practice Questions


2.2: Matrix multiplication and linear combinations
\(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\twovec}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} } \) \(\newcommand{\threevec}[3]{\begin{pmatrix} #1 \\ #2 \\ #3 \end{pmatrix} } \) \(\newcommand{\fourvec}[4]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \end{pmatrix} } \) \(\newcommand{\fivevec}[5]{\begin{pmatrix} #1 \\ #2 \\ #3 \\ #4 \\ #5 \end{pmatrix} } \)The previous section introduced vectors and linear combinations and demonstrated how they provide a means of thinking about linear systems geometrically. In particular, we saw that the vector \(\mathbf b\) is a linear combination of the vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) if the linear system corresponding to the augmented matrix
is consistent.
Our goal in this section is to introduction matrix multiplication, another algebraic operation that connects linear systems and linear combinations.
2.2.1 Matrices
We first thought of a matrix as a rectangular array of numbers. When the number of rows is \(m\) and columns is \(n\text{,}\) we say that the dimensions of the matrix are \(m\times n\text{.}\) For instance, the matrix below is a \(3\times4\) matrix:
We may also think of the columns of a matrix as a collection of vectors. For instance, the matrix above may be represented as
where
In this way, we see that our \(3\times 4\) matrix is the same as a collection of 4 vectors in \(\mathbb R^3\text{.}\)
This means that we may define scalar multiplication and matrix addition operations using the corresponding vector operations.
Preview Activity 2.2.1. Matrix operations.
-
Compute the scalar multiple
\begin{equation*} -3\left[ \begin{array}{rrr} 3 & 1 & 0 \\ -4 & 3 & -1 \\ \end{array} \right]\text{.} \end{equation*}
- Suppose that \(A\) and \(B\) are two matrices. What do we need to know about their dimensions before we can form the sum \(A+B\text{?}\)
-
Find the sum
\begin{equation*} \left[ \begin{array}{rr} 0 & -3 \\ 1 & -2 \\ 3 & 4 \\ \end{array} \right] + \left[ \begin{array}{rrr} 4 & -1 \\ -2 & 2 \\ 1 & 1 \\ \end{array} \right]\text{.} \end{equation*}
-
The matrix \(I_n\text{,}\) which we call the
identity
matrix is the \(n\times n\) matrix whose entries are zero except for the diagonal entries, which are 1. For instance,
\begin{equation*} I_3 = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]\text{.} \end{equation*}
If we can form the sum \(A+I_n\text{,}\) what must be true about the matrix \(A\text{?}\)
-
Find the matrix \(A - 2I_3\) where
\begin{equation*} A = \left[ \begin{array}{rrr} 1 & 2 & -2 \\ 2 & -3 & 3 \\ -2 & 3 & 4 \\ \end{array} \right]\text{.} \end{equation*}
As this preview activity shows, both of these operations are relatively straightforward. Some care, however, is required when adding matrices. Since we need the same number of vectors to add and since the vectors must be of the same dimension, two matrices must have the same dimensions as well if we wish to form their sum.
The identity matrix will play an important role at various points in our explorations. It is important to note that it is a square matrix, meaning it has an equal number of rows and columns, so any matrix added to it must be square as well. Though we wrote it as \(I_n\) in the activity, we will often just write \(I\) when the dimensions are clear.
2.2.2 Matrix-vector multiplication and linear combinations
A more important operation will be matrix multiplication as it allows us to compactly express linear systems. For now, we will work with the product of a matrix and vector, which we illustrate with an example.
Suppose we have the matrix \(A\) and vector \(\mathbf x\) as given below.
Their product will be defined to be the linear combination of the columns of \(A\) using the components of \(\mathbf x\) as weights. This means that
Let's take note of the dimensions of the matrix and vectors. The two components of the vector \(\mathbf x\) are weights used to form a linear combination of the columns of \(A\text{.}\) Since \(\mathbf x\) has two components, \(A\) must have two columns. In other words, the number of columns of \(A\) must equal the dimension of the vector \(\mathbf x\text{.}\)
In the same way, the columns of \(A\) are 3-dimensional so any linear combination of them is 3-dimensional as well. Therefore, \(A\mathbf x\) will be 3-dimensional.
We then see that if \(A\) is a \(3\times2\) matrix, \(\mathbf x\) must be a 2-dimensional vector and \(A\mathbf x\) will be 3-dimensional.
More generally, we have the following definition.
The product of a matrix \(A\) by a vector \(\mathbf x\) will be the linear combination of the columns of \(A\) using the components of \(\mathbf x\) as weights.
If \(A\) is an \(m\times n\) matrix, then \(\mathbf x\) must be an \(n\)-dimensional vector, and the product \(A\mathbf x\) will be an \(m\)-dimensional vector. If
then
The next activity introduces some properties of matrix multiplication.
Activity 2.2.2. Matrix-vector multiplication.
-
Find the matrix product
\begin{equation*} \left[ \begin{array}{rrrr} 1 & 2 & 0 & -1 \\ 2 & 4 & -3 & -2 \\ -1 & -2 & 6 & 1 \\ \end{array} \right] \left[ \begin{array}{r} 3 \\ 1 \\ -1 \\ 1 \\ \end{array} \right]\text{.} \end{equation*}
-
Suppose that \(A\) is the matrix
\begin{equation*} \left[ \begin{array}{rrr} 3 & -1 & 0 \\ 0 & -2 & 4 \\ 2 & 1 & 5 \\ 1 & 0 & 3 \\ \end{array} \right]\text{.} \end{equation*}
If \(A\mathbf x\) is defined, what is the dimension of the vector \(\mathbf x\) and what is the dimension of \(A\mathbf x\text{?}\)
- A vector whose entries are all zero is denoted by \(\zerovec\text{.}\) If \(A\) is a matrix, what is the product \(A\zerovec\text{?}\)
- Suppose that \(I = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\) is the identity matrix and \(\mathbf x=\threevec{x_1}{x_2}{x_3}\text{.}\) Find the product \(I\mathbf x\) and explain why \(I\) is called the identity matrix.
-
Suppose we write the matrix \(A\) in terms of its columns as
\begin{equation*} A = \left[ \begin{array}{rrrr} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n \\ \end{array} \right]\text{.} \end{equation*}
If the vector \(\mathbf e_1 = \left[\begin{array}{r} 1 \\ 0 \\ \vdots \\ 0 \end{array}\right]\text{,}\) what is the product \(A\mathbf e_1\text{?}\)
-
Suppose that
\begin{equation*} A = \left[ \begin{array}{rrrr} 1 & 2 \\ -1 & 1 \\ \end{array} \right], \mathbf b = \left[ \begin{array}{r} 6 \\ 0 \end{array} \right]\text{.} \end{equation*}
Is there a vector \(\mathbf x\) such that \(A\mathbf x = \mathbf b\text{?}\)
Multiplication of a matrix \(A\) and a vector is defined as a linear combination of the columns of \(A\text{.}\) However, there is a shortcut for computing such a product. Let's look at our previous example and focus on the first row of the product.
To find the first component of the product, we consider the first row of the matrix. We then multiply the first entry in that row by the first component of the vector, the second entry by the second component of the vector, and so on, and add the results. In this way, we see that the third component of the product would be obtained from the third row of the matrix by computing \(2(3) + 3(1) = 9\text{.}\)
You are encouraged to evaluate Item a using this shortcut and compare the result to what you found while completing the previous activity.
Activity 2.2.3.
In addition, Sage can find the product of a matrix and vector using the
*
operator. For example,
A = matrix(2,2,[1,2,2,1]) v = vector([3,-1]) A*v
-
Use Sage to evaluate the product Item a yet again.
-
In Sage, define the matrix and vectors
\begin{equation*} A = \left[ \begin{array}{rrr} -2 & 0 \\ 3 & 1 \\ 4 & 2 \\ \end{array} \right], \zerovec = \left[ \begin{array}{r} 0 \\ 0 \end{array} \right], \mathbf v = \left[ \begin{array}{r} -2 \\ 3 \end{array} \right], \mathbf w = \left[ \begin{array}{r} 1 \\ 2 \end{array} \right]\text{.} \end{equation*}
- What do you find when you evaluate \(A\zerovec\text{?}\)
- What do you find when you evaluate \(A(3\mathbf v)\) and \(3(A\mathbf v)\) and compare your results?
- What do you find when you evaluate \(A(\mathbf v+\mathbf w)\) and \(A\mathbf v + A\mathbf w\) and compare your results?
- If \(I=\left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\) is the \(3\times3\) identity matrix, what is the product \(IA\text{?}\)
This activity demonstrates several general properties satisfied by matrix multiplication that we record here.
If \(A\) is a matrix, \(\mathbf v\) and \(\mathbf w\) vectors, and \(c\) a scalar, then
- \(A\zerovec = \zerovec\text{.}\)
- \(A(c\mathbf v) = cA\mathbf v\text{.}\)
- \(A(\mathbf v+\mathbf w) = A\mathbf v + A\mathbf w\text{.}\)
2.2.3 Matrix-vector multiplication and linear systems
So far, we have begun with a matrix \(A\) and a vector \(\mathbf x\) and formed their product \(A\mathbf x = \mathbf b\text{.}\) We would now like to turn this around: beginning with a matrix \(A\) and a vector \(\mathbf b\text{,}\) we will ask if we can find a vector \(\mathbf x\) such that \(A\mathbf x = \mathbf b\text{.}\) This will naturally lead back to linear systems.
To see the connection between the matrix equation \(A\mathbf x = \mathbf b\) and linear systems, let's write the matrix \(A\) in terms of its columns \(\mathbf v_i\) and \(\mathbf x\) in terms of its components.
We know that the matrix product \(A\mathbf x\) forms a linear combination of the columns of \(A\text{.}\) Therefore, the equation \(A\mathbf x = \mathbf b\) is merely a compact way of writing the equation for the weights \(c_i\text{:}\)
We have seen this equation before: Remember that Proposition 2.1.7 says that the solutions of this equation are the same as the solutions to the linear system whose augmented matrix is
This gives us three different ways of looking at the same solution space.
If \(A=\left[\begin{array}{rrrr} \mathbf v_1& \mathbf v_2& \ldots\mathbf v_n \end{array}\right]\) and \(\mathbf x=\left[ \begin{array}{r} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array}\right] \text{,}\) then the following are equivalent.
- The vector \(\mathbf x\) satisfies \(A\mathbf x = \mathbf b \text{.}\)
-
The vector \(\mathbf b\) is a linear combination of the columns of \(A\) with weights \(x_j\text{:}\)
\begin{equation*} x_1\mathbf v_1 + x_2\mathbf v_2 + \ldots + x_n\mathbf v_n = \mathbf b\text{.} \end{equation*}
-
The components of \(\mathbf x\) form a solution to the linear system corresponding to the augmented matrix
\begin{equation*} \left[\begin{array}{rrrr|r} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n & \mathbf b \end{array}\right]\text{.} \end{equation*}
When the matrix \(A = \left[\begin{array}{rrrr} \mathbf v_1& \mathbf v_2& \ldots& \mathbf v_n\end{array}\right]\text{,}\) we will frequently write
and say that we augment the matrix \(A\) by the vector \(\mathbf b\text{.}\)
We may think of \(A\mathbf x = \mathbf b\) as merely giving a notationally compact way of writing a linear system. This form of the equation, however, will allow us to focus on important features of the system that determine its solution space.
Describe the solution space of the equation
By Proposition 2.2.4, the solution space to this equation is the same as the equation
which is the same as the linear system corresponding to
We will study the solutions to this linear system by finding the reduced row echelon form of the augmented matrix:
This gives us the system of equations
The variable \(x_3\) is free so we may write the solution space parametrically as
Since we originally asked to describe the solutions to the equation \(A\mathbf x = \mathbf b\text{,}\) we will express the solution in terms of the vector \(\mathbf x\text{:}\)
This shows that the solutions \(\mathbf x\) may be written in the form \(\mathbf v + x_3\mathbf w\text{,}\) for appropriate vectors \(\mathbf v\) and \(\mathbf w\text{.}\) Geometrically, the solution space is a line in \(\mathbb R^3\) through \(\mathbf v\) moving parallel to \(\mathbf w\text{.}\)
Activity 2.2.4. The equation \(A\mathbf x = \mathbf b\).
-
Consider the linear system
\begin{equation*} \begin{alignedat}{4} 2x & {}+{} & y & {}-{} & 3z & {}={} & 4 \\ -x & {}+{} & 2y & {}+{} & z & {}={} & 3 \\ 3x & {}-{} & y & & & {}={} & -4 \\ \end{alignedat}\text{.} \end{equation*}
Identify the matrix \(A\) and vector \(\mathbf b\) to express this system in the form \(A\mathbf x = \mathbf b\text{.}\)
-
If \(A\) and \(\mathbf b\) are as below, write the linear system corresponding to the equation \(A\mathbf x=\mathbf b\text{.}\)
\begin{equation*} A = \left[\begin{array}{rrr} 3 & -1 & 0 \\ -2 & 0 & 6 \end{array} \right], \mathbf b = \left[\begin{array}{r} -6 \\ 2 \end{array} \right] \end{equation*}
and describe the solution space.
-
Describe the solution space of the equation
\begin{equation*} \left[ \begin{array}{rrrr} 1 & 2 & 0 & -1 \\ 2 & 4 & -3 & -2 \\ -1 & -2 & 6 & 1 \\ \end{array} \right] \mathbf x = \left[\begin{array}{r} -1 \\ 1 \\ 5 \end{array} \right]\text{.} \end{equation*}
- Suppose \(A\) is an \(m\times n\) matrix. What can you guarantee about the solution space of the equation \(A\mathbf x = \zerovec\text{?}\)
2.2.4 Matrix products
In this section, we have developed some algebraic operations on matrices with the aim of simplifying our description of linear systems. We will now introduce a final operation, the product of two matrices, that will become important when we study linear transformations in Section 2.5 .
Given matrices \(A\) and \(B\text{,}\) we will form their product \(AB\) by first writing \(B\) in terms of its columns:
We then define
Given the matrices
we have
It is important to note that we can only multiply matrices if the dimensions of the matrices are compatible. More specifically, when constructing the product \(AB\text{,}\) the matrix \(A\) multiplies the columns of \(B\text{.}\) Therefore, the number of columns of \(A\) must equal the number of rows of \(B\text{.}\) When this condition is met, the number of rows of \(AB\) is the number of rows of \(A\text{,}\) and the number of columns of \(AB\) is the number of columns of \(B\text{.}\)
Activity 2.2.5.
Consider the matrices
- Suppose we want to form the product \(AB\text{.}\) Before computing, first explain how you know this product exists and then explain what the dimensions of the resulting matrix will be.
- Compute the product \(AB\text{.}\)
-
Sage can multiply matrices using the
*
operator. Define the matrices \(A\) and \(B\) in the Sage cell below and check your work by computing \(AB\text{.}\) - Are you able to form the matrix product \(BA\text{?}\) If so, use the Sage cell above to find \(BA\text{.}\) Is it generally true that \(AB = BA\text{?}\)
-
Suppose we form the three matrices.
\begin{equation*} A = \left[\begin{array}{rr} 1 & 2 \\ 3 & -2 \\ \end{array}\right], B = \left[\begin{array}{rr} 0 & 4 \\ 2 & -1 \\ \end{array}\right], C = \left[\begin{array}{rr} -1 & 3 \\ 4 & 3 \\ \end{array}\right]\text{.} \end{equation*}
Compare what happens when you compute \(A(B+C)\) and \(AB + AC\text{.}\) State your finding as a general principle.
- Compare the results of evaluating \(A(BC)\) and \((AB)C\) and state your finding as a general principle.
-
When we are dealing with real numbers, we know if \(a\neq 0\) and \(ab = ac\text{,}\) then \(b=c\text{.}\) Define matrices
\begin{equation*} A = \left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \\ \end{array}\right], B = \left[\begin{array}{rr} 3 & 0 \\ 1 & 3 \\ \end{array}\right], C = \left[\begin{array}{rr} 1 & 2 \\ 2 & 2 \\ \end{array}\right] \end{equation*}
and compute \(AB\) and \(AC\text{.}\)
If \(AB = AC\text{,}\) is it necessarily true that \(B = C\text{?}\) -
Again, with real numbers, we know that if \(ab = 0\text{,}\) then either \(a = 0\) or \(b=0\text{.}\) Define
\begin{equation*} A = \left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \\ \end{array}\right], B = \left[\begin{array}{rr} 2 & -4 \\ -1 & 2 \\ \end{array}\right] \end{equation*}
and compute \(AB\text{.}\)
If \(AB = 0\text{,}\) is it necessarily true that either \(A=0\) or \(B=0\text{?}\)
This activity demonstrated some general properties about products of matrices, which mirror some properties about operations with real numbers.
Properties of Matrix-matrix Multiplication.
If \(A\text{,}\) \(B\text{,}\) and \(C\) are matrices such that the following operations are defined, it follows that
- Associativity:
-
\(A(BC) = (AB)C\text{.}\)
- Distributivity:
-
\(A(B+C) = AB+AC\text{.}\)
\((A+B)C = AC+BC\text{.}\)
At the same time, there are a few properties that hold for real numbers that do not hold for matrices.
Things to be careful of.
The following properties hold for real numbers but not for matrices.
- Commutativity:
-
It is not generally true that \(AB = BA\text{.}\)
- Cancellation:
-
It is not generally true that \(AB = AC\) implies that \(B = C\text{.}\)
- Zero divisors:
-
It is not generally true that \(AB = 0\) implies that either \(A=0\) or \(B=0\text{.}\)
Summary
In this section, we have found an especially simple way to express linear systems using matrix multiplication.
- If \(A\) is an \(m\times n\) matrix and \(\mathbf x\) an \(n\)-dimensional vector, then \(A\mathbf x\) is the linear combination of the columns of \(A\) using the components of \(\mathbf x\) as weights. The vector \(A\mathbf x\) is \(m\)-dimensional.
- The solution space to the equation \(A\mathbf x = \mathbf b\) is the same as the solution space to the linear system corresponding to the augmented matrix \(\left[ \begin{array}{r|r} A & \mathbf b \end{array}\right]\text{.}\)
- If \(A\) is an \(m\times n\) matrix and \(B\) is an \(n\times p\) matrix, we can form the product \(AB\text{,}\) which is an \(m\times p\) matrix whose columns are the products of \(A\) and the columns of \(B\text{.}\)
Exercises 2.2.6Exercises
Consider the system of linear equations
- Find the matrix \(A\) and vector \(\mathbf b\) that expresses this linear system in the form \(A\mathbf x=\mathbf b\text{.}\)
-
Give a description of the solution space to the equation \(A\mathbf x = \mathbf b\text{.}\)
Suppose that \(A\) is a \(135\times2201\) matrix. If \(A\mathbf x\) is defined, what is the dimension of \(\mathbf x\text{?}\) What is the dimension of \(A\mathbf x\text{?}\)
Suppose that \(A \) is a \(3\times2\) matrix whose columns are \(\mathbf v_1\) and \(\mathbf v_2\text{;}\) that is,
- What is the dimension of the vectors \(\mathbf v_1\) and \(\mathbf v_2\text{?}\)
- What is the product \(A\twovec{1}{0}\) in terms of \(\mathbf v_1\) and \(\mathbf v_2\text{?}\) What is the product \(A\twovec{0}{1}\text{?}\) What is the product \(A\twovec{2}{3}\text{?}\)
-
Suppose that
\begin{equation*} A\twovec{1}{0} = \threevec{3}{-2}{1}, A\twovec{0}{1} = \threevec{0}{3}{2}\text{.} \end{equation*}
What is the matrix \(A\text{?}\)
Shown below are vectors \(\mathbf v_1\) and \(\mathbf v_2\text{.}\) Suppose that the matrix \(A\) is
- What are the dimensions of the matrix \(A\text{?}\)
-
On the plot above, indicate the vectors
\begin{equation*} A\twovec{1}{0}, A\twovec{2}{3}, A\twovec{0}{-3}\text{.} \end{equation*}
- Find all vectors \(\mathbf x\) such that \(A\mathbf x=\mathbf b\text{.}\)
- Find all vectors \(\mathbf x\) such that \(A\mathbf x = \zerovec\text{.}\)
Suppose that
- Describe the solution space to the equation \(A\mathbf x = \zerovec\text{.}\)
- Find a \(3\times2\) matrix \(B\) with no zero entries such that \(AB = 0\text{.}\)
Consider the matrix
-
Find the product \(A\mathbf x\) where
\begin{equation*} \mathbf x = \fourvec{1}{-2}{0}{2}\text{.} \end{equation*}
-
Give a description of the vectors \(\mathbf x\) such that
\begin{equation*} A\mathbf x = \threevec{-1}{15}{17}\text{.} \end{equation*}
- Find the reduced row echelon form of \(A\) and identify the pivot positions.
- Can you find a vector \(\mathbf b\) such that \(A\mathbf x=\mathbf b\) is inconsistent?
- For a general 3-dimensional vector \(\mathbf b\text{,}\) what can you say about the solution space of the equation \(A\mathbf x = \mathbf b\text{?}\)
The operations that we perform in Gaussian elimination can be accomplished using matrix multiplication. This observation is the basis of an important technique that we will investigate in a subsequent chapter.
Let's consider the matrix
-
Suppose that
\begin{equation*} S = \left[\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
Verify that \(SA\) is the matrix that results when the second row of \(A\) is scaled by a factor of 7. What matrix \(S\) would scale the third row by -3?
-
Suppose that
\begin{equation*} P = \left[\begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
Verify that \(PA\) is the matrix that results from interchanging the first and second rows. What matrix \(P\) would interchange the first and third rows?
-
Suppose that
\begin{equation*} L_1 = \left[\begin{array}{rrr} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
Verify that \(L_1A\) is the matrix that results from multiplying the first row of \(A\) by \(-2\) and adding it to the second row. What matrix \(L_2\) would multiply the first row by 3 and add it to the third row?
-
When we performed Gaussian elimination, our first goal was to perform row operations that brought the matrix into a triangular form. For our matrix \(A\text{,}\) find the row operations needed to find a row equivalent matrix \(U\) in triangular form. By expressing these row operations in terms of matrix multiplication, find a matrix \(L\) such that \(LA = U\text{.}\)
In this exercise, you will construct the inverse of a matrix, a subject that we will investigate more fully in the next chapter. Suppose that \(A\) is the \(2\times2\) matrix:
-
Find the vectors \(\mathbf b_1\) and \(\mathbf b_2\) such that the matrix \(B=\left[\begin{array}{rr} \mathbf b_1 & \mathbf b_2 \end{array}\right]\) satisfies
\begin{equation*} AB = I = \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array}\right]\text{.} \end{equation*}
-
In general, it is not true that \(AB = BA\text{.}\) Check that it is true, however, for the specific \(A\) and \(B\) that appear in this problem.
- Suppose that \(\mathbf x = \twovec{x_1}{x_2}\text{.}\) What do you find when you evaluate \(I\mathbf x\text{?}\)
-
Suppose that we want to solve the equation \(A\mathbf x = \mathbf b\text{.}\) We know how to do this using Gaussian elimination; let's use our matrix \(B\) to find a different way:
\begin{equation*} \begin{aligned} A\mathbf x & {}={} \mathbf b \\ B(A\mathbf x) & {}={} B\mathbf b \\ (BA)\mathbf x & {}={} B\mathbf b \\ I\mathbf x & {}={} B\mathbf b \\ \mathbf x & {}={} B\mathbf b \\ \end{aligned}\text{.} \end{equation*}
In other words, the solution to the equation \(A\mathbf x=\mathbf b\) is \(\mathbf x = B\mathbf b\text{.}\)
Consider the equation \(A\mathbf x = \twovec{5}{-2}\text{.}\) Find the solution in two different ways, first using Gaussian elimination and then as \(\mathbf x = B\mathbf b\text{,}\) and verify that you have found the same result.
Determine whether the following statements are true or false and provide a justification for your response.
- If \(A\mathbf x\) is defined, then the number of components of \(\mathbf x\) equals the number of rows of \(A\text{.}\)
- The solution space to the equation \(A\mathbf x = \mathbf b\) is equivalent to the solution space to the linear system whose augmented matrix is \(\left[\begin{array}{r|r} A & \mathbf b \end{array}\right]\text{.}\)
- If a linear system of equations has 8 equations and 5 unknowns, then the dimensions of the matrix \(A\) in the corresponding equation \(A\mathbf x = \mathbf b\) is \(5\times8\text{.}\)
- If \(A\) has a pivot in every row, then every equation \(A\mathbf x = \mathbf b\) is consistent.
- If \(A\) is a \(9\times5\) matrix, then \(A\mathbf x=\mathbf b\) is inconsistent for some vector \(\mathbf b\text{.}\)
Suppose that \(A\) is an \(4\times4\) matrix and that the equation \(A\mathbf x = \mathbf b\) has a unique solution for some vector \(\mathbf b\text{.}\)
- What does this say about the pivots of the matrix \(A\text{?}\) Write the reduced row echelon form of \(A\text{.}\)
- Can you find another vector \(\mathbf c\) such that \(A\mathbf x = \mathbf c\) is inconsistent?
- What can you say about the solution space to the equation \(A\mathbf x = \zerovec\text{?}\)
- Suppose \(A=\left[\begin{array}{rrrr} \mathbf v_1 & \mathbf v_2 & \mathbf v_3 & \mathbf v_4 \end{array}\right]\text{.}\) Explain why every four-dimensional vector can be written as a linear combination of the vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) \(\mathbf v_3\text{,}\) and \(\mathbf v_4\) in exactly one way.
Define the matrix
-
Describe the solution space to the homogeneous equation \(A\mathbf x = \zerovec\text{.}\) What does this solution space represent geometrically?
-
Describe the solution space to the equation \(A\mathbf x=\mathbf b\) where \(\mathbf b = \threevec{-3}{-4}{1}\text{.}\) What does this solution space represent geometrically and how does it compare to the previous solution space?
-
We will now explain the relationship between the previous two solution spaces. Suppose that \(\mathbf x_h\) is a solution to the homogeneous equation; that is \(A\mathbf x_h=\zerovec\text{.}\) We will also suppose that \(\mathbf x_p\) is a solution to the equation \(A\mathbf x = \mathbf b\text{;}\) that is, \(A\mathbf x_p=\mathbf b\text{.}\)
Use the Linearity Principle expressed in Proposition 2.2.3 to explain why \(\mathbf x_h+\mathbf x_p\) is a solution to the equation \(A\mathbf x = \mathbf b\text{.}\) You may do this by evaluating \(A(\mathbf x_h+\mathbf x_p)\text{.}\)
That is, if we find one solution \(\mathbf x_p\) to an equation \(A\mathbf x = \mathbf b\text{,}\) we may add any solution to the homogeneous equation to \(\mathbf x_p\) and still have a solution to the equation \(A\mathbf x = \mathbf b\text{.}\) In other words, the solution space to the equation \(A\mathbf x = \mathbf b\) is given by translating the solution space to the homogeneous equation by the vector \(\mathbf x_p\text{.}\)
Suppose that a city is starting a bicycle sharing program with bicycles at locations \(B\) and \(C\text{.}\) Bicycles that are rented at one location may be returned to either location at the end of the day. Over time, the city finds that 80% of bicycles rented at location \(B\) are returned to \(B\) with the other 20% returned to \(C\text{.}\) Similarly, 50% of bicycles rented at location \(C\) are returned to \(B\) and 50% to \(C\text{.}\)
To keep track of the bicycles, we form a vector
where \(B_k\) is the number of bicycles at location \(B\) at the beginning of day \(k\) and \(C_k\) is the number of bicycles at \(C\text{.}\) The information above tells us
where
-
Let's check that this makes sense.
- Suppose that there are 1000 bicycles at location \(B\) and none at \(C\) on day 1. This means we have \(\mathbf x_1 = \twovec{1000}{0}\text{.}\) Find the number of bicycles at both locations on day 2 by evaluating \(\mathbf x_2 = A\mathbf x_1\text{.}\)
- Suppose that there are 1000 bicycles at location \(C\) and none at \(B\) on day 1. Form the vector \(\mathbf x_1\) and determine the number of bicycles at the two locations the next day by finding \(\mathbf x_2 = A\mathbf x_1\text{.}\)
- Suppose that one day there are 1050 bicycles at location \(B\) and 450 at location \(C\text{.}\) How many bicycles were there at each location the previous day?
- Suppose that there are 500 bicycles at location \(B\) and 500 at location \(C\) on Monday. How many bicycles are there at the two locations on Tuesday? on Wednesday? on Thursday?
This problem is a continuation of the previous problem.
-
Let us define vectors
\begin{equation*} \mathbf v_1 = \twovec{5}{2}, \mathbf v_2 = \twovec{-1}{1}\text{.} \end{equation*}
Show that
\begin{equation*} A\mathbf v_1 = \mathbf v_1, A\mathbf v_2 = 0.3\mathbf v_2\text{.} \end{equation*} -
Suppose that \(\mathbf x_1 = c_1 \mathbf v_1 + c_2 \mathbf v_2\) where \(c_2\) and \(c_2\) are scalars. Use the Linearity Principle expressed in Proposition 2.2.3 to explain why
\begin{equation*} \mathbf x_{2} = A\mathbf x_1 = c_1\mathbf v_1 + 0.3c_2\mathbf v_2\text{.} \end{equation*}
-
Continuing in this way, explain why
\begin{equation*} \begin{aligned} \mathbf x_{3} = A\mathbf x_2 & {}={} c_1\mathbf v_1 +0.3^2c_2\mathbf v_2 \\ \mathbf x_{4} = A\mathbf x_3 & {}={} c_1\mathbf v_1 +0.3^3c_2\mathbf v_2 \\ \mathbf x_{5} = A\mathbf x_4 & {}={} c_1\mathbf v_1 +0.3^4c_2\mathbf v_2 \\ \end{aligned}\text{.} \end{equation*}
- Suppose that there are initially 500 bicycles at location \(B\) and 500 at location \(C\text{.}\) Write the vector \(\mathbf x_1\) and find the scalars \(c_1\) and \(c_2\) such that \(\mathbf x_1=c_1\mathbf v_1 + c_2\mathbf v_2\text{.}\)
- Use the previous part of this problem to determine \(\mathbf x_2\text{,}\) \(\mathbf x_3\) and \(\mathbf x_4\text{.}\)
- After a very long time, how are all the bicycles distributed?


23.6: Checklist for Secondary IV Solution Administration
Use the checklist below to review the steps for completion of “Secondary IV Solution Administration.” This checklist is used when fluids are already being administered via the primary IV tubing and a second IV solution is administered.
Steps
Disclaimer: Always review and follow agency policy regarding this specific skill.
- Gather supplies: secondary IV fluid/medication, secondary IV tubing, alcohol wipe/scrub hubs, and tubing labels.
- Verify the provider order with the medication administration record (eMAR/MAR).
- Perform the first check of the six rights of medication administration while withdrawing the IV solution and tubing from the medication dispensing unit. Check expiration dates on the fluid and the tubing and verify allergies.
- Verify compatibility of the secondary IV solution with the other IV fluids the patient is concurrently receiving.
- Remove the IV solution from the packaging and gently apply pressure to the bag while inspecting for tears or leaks. Check the color and clarity of the solution.
- Perform the second check of the six rights of medication administration.
- Enter the patient room and greet the patient.
-
Perform safety steps:
- Perform hand hygiene.
- Check the room for transmission-based precautions.
- Introduce yourself, your role, the purpose of your visit, and an estimate of the time it will take.
- Confirm patient ID using two patient identifiers (e.g., name and date of birth).
- Explain the process to the patient.
- Be organized and systematic.
- Use appropriate listening and questioning skills.
- Listen and attend to patient cues.
- Ensure the patient’s privacy and dignity.
- Assess ABCs.
- Perform the third check of the six rights of medication administration at the patient’s bedside.
- If the patient is receiving the medication for the first time, teach the patient and family (if appropriate) about the potential adverse reactions and other concerns related to the medication.
- Remove the secondary IV tubing from the packaging.
- Place the roller clamp to the “off” position.
- Remove the protective sheath from the IV spike and the cover from the tubing port of the IV solution.
- Insert the spike into the IV bag while maintaining sterility.
- Compress and release the drip chamber, filling halfway.
-
Prime the secondary IV tubing. Back priming is considered best practice and is performed using an infusion pump with primary fluids attached:
- Vigorously cleanse the catheter tip on the patient’s IV port with an alcohol pad/scrub hub (or the agency required cleansing agent) for at least five seconds and allow it to dry.
- Connect the secondary tubing to the port closest to the drip chamber. Lower the secondary bag below the primary bag, and allow the fluid from the primary bag to fill secondary tubing. Fill the secondary tubing until it reaches the drip chamber, and then raise the secondary bag above the primary line.
- Hang the secondary IV solution on the IV pole with the primary bag lower than the secondary bag.
- Label the secondary tubing near the drip chamber.
-
Set the infusion rate:
- For infusion pump: Set the volume to be infused and the rate (mL/hr) to be administered based on the provider order.
- For gravity: Set the roller clamp to achieve the appropriate number of drops per minute based on the provider order.
Take time to watch the IV fluid or medication to drip into the drip chamber to ensure the medication or fluid is flowing to the patient.
- Assess the patient’s IV site for signs and symptoms of vein irritation or infiltration after infusion begins. Do not proceed with administering secondary fluids if there are any concerns about the site.
- Assist the patient to a comfortable position, ask if they have any questions, and thank them for their time.
-
Ensure safety measures when leaving the room:
- CALL LIGHT: Within reach
- BED: Low and locked (in lowest position and brakes on)
- SIDE RAILS: Secured
- TABLE: Within reach
- ROOM: Risk-free for falls (scan room and clear any obstacles)
- Perform hand hygiene.
- Document the procedure and assessment findings. Report any concerns according to agency policy.


9.6: Compiling, Batching, and the Download Center
Texts are available for exporting via two processes on the LibreTexts platform: Compiling and Batching.
Compiling
While most users access the content on the LibreTexts directly via our web interface, other mechanisms for dissemination exist as discussed in Chapter 12 of the Construction Guide.
Since compiling is a both a time consuming and CPU consuming activity, to reduce server load, only specific users have access to force compiling in real time. Contact the development team<EMAIL_ADDRESS>) to request this permission.
Initial step: "Compile"
To compile a book access must be set to public using the restrict access selection in the Options menu drop down and the Cover Page menu in the Page Settings must be set. One can choose to make the book downloadable or not.
Using the pdf button the book can now be compiled
Once the text has been compiled, then a new blue "Download" buttons appear on the top right of the book.
If changes are made in a compiled book it must be recompiled so that the changes are reflected in the printed versions. A simple compile is faster but if the structure of the LibreTexts has been changed (pages/chapters added or deleted) a full compile is needed.
Batching
Resources on the LibreTexts platform are considered dynamic and constantly being edited as they are curated or updated. To ensure readers have access to the up to date version of the resources, automatic weekly batch compilation is implemented on each library.
- biology: Saturday @ 0:30
- business: Saturday @12:30
- chemistry: Sunday @ 0:30
- engineering: Sunday @12:30
- espanol: Monday @ 0:30
- geosciences: Monday @12:30
- humanities: Tuesday @ 0:30
- law: Tuesday @12:30 (not implemented yet)
- mathematics: Wednesday @ 0:30
- medicine: Wednesday @12:30
- physics: Thursday @ 0:30
- social sciences: Thursday @12:30
- statistics: Friday @ 0:30
- workforce: Friday @12:30


10.3: Frameworks for Identifying Types of Climate Messages
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In this section, we will examine two frameworks for understanding and identifying relational messages that contribute to communication climate: confirming versus disconfirming messages and supportive versus defensive messages. Many of the common behaviors that can lead to a cold climate are performed in an automatic state of mind, without forethought or strategy. Exploring these types of messages in pairs helps us be better able to mindfully recognize warm climate alternatives that might be better received.
Confirming messages convey another person’s value and worthwhile disconfirming messages may suggest someone is devalued and unimportant. Obviously, most of us like to receive warm confirming messages because they foster emotional safety as well as personal and relational growth. However, it is likely that many of the messages you receive and give are somewhere between the two extremes. Let’s look at three specific types of confirming and disconfirming messages. Consider how each may meet or not meet our social needs.
Another useful framework for understanding communication climate can be found in the six defensive and supportive behavior pairs proposed by psychologist Jack Gibb in 1965, adapted here with some pairs re-named for clarity. These six behaviors are, on the one hand, likely to generate an emotional climate of defensiveness (cold) and are, on the other, likely generate a supportive climate (warm).
In the box below, we define and give examples of each of the six pairs: evaluation/description, manipulation/straightforwardness, control/collaboration, indifference/empathy, superiority/equality, certainty/flexibility. In addition, we propose some possibilities for how climate might be perceived by the recipients of such behavior and why it might be perceived that way.


2.7: Summary
2.1 The Sky Above
The direct evidence of our senses supports a geocentric perspective, with the celestial sphere pivoting on the celestial poles and rotating about a stationary Earth. We see only half of this sphere at one time, limited by the horizon; the point directly overhead is our zenith. The Sun’s annual path on the celestial sphere is the ecliptic—a line that runs through the center of the zodiac, which is the 18-degree-wide strip of the sky within which we always find the Moon and planets. The celestial sphere is organized into 88 constellations, or sectors.
2.2 Ancient Astronomy
Ancient Greeks such as Aristotle recognized that Earth and the Moon are spheres, and understood the phases of the Moon, but because of their inability to detect stellar parallax, they rejected the idea that Earth moves. Eratosthenes measured the size of Earth with surprising precision. Hipparchus carried out many astronomical observations, making a star catalog, defining the system of stellar magnitudes, and discovering precession from the apparent shift in the position of the north celestial pole. Ptolemy of Alexandria summarized classic astronomy in his Almagest ; he explained planetary motions, including retrograde motion, with remarkably good accuracy using a model centered on Earth. This geocentric model, based on combinations of uniform circular motion using epicycles, was accepted as authority for more than a thousand years.
2.3 Astrology and Astronomy
The ancient religion of astrology, with its main contribution to civilization a heightened interest in the heavens, began in Babylonia. It reached its peak in the Greco-Roman world, especially as recorded in the Tetrabiblos of Ptolemy. Natal astrology is based on the assumption that the positions of the planets at the time of our birth, as described by a horoscope, determine our future. However, modern tests clearly show that there is no evidence for this, even in a broad statistical sense, and there is no verifiable theory to explain what might cause such an astrological influence.
2.4 The Birth of Modern Astronomy
Nicolaus Copernicus introduced the heliocentric cosmology to Renaissance Europe in his book De Revolutionibus . Although he retained the Aristotelian idea of uniform circular motion, Copernicus suggested that Earth is a planet and that the planets all circle about the Sun, dethroning Earth from its position at the center of the universe. Galileo was the father of both modern experimental physics and telescopic astronomy. He studied the acceleration of moving objects and, in 1610, began telescopic observations, discovering the nature of the Milky Way, the large-scale features of the Moon, the phases of Venus, and four moons of Jupiter. Although he was accused of heresy for his support of heliocentric cosmology, Galileo is credited with observations and brilliant writings that convinced most of his scientific contemporaries of the reality of the Copernican theory.


2.E: Descriptive Statistics (Exercises)
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- Last updated
- Save as PDF
2.2: Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
Q 2.2.1
Student grades on a chemistry exam were: 77, 78, 76, 81, 86, 51, 79, 82, 84, 99
- Construct a stem-and-leaf plot of the data.
- Are there any potential outliers? If so, which scores are they? Why do you consider them outliers?
Q 2.2.2
The table below contains the 2010 obesity rates in U.S. states and Washington, DC.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
- Use a random number generator to randomly pick eight states. Construct a bar graph of the obesity rates of those eight states.
- Construct a bar graph for all the states beginning with the letter "A."
- Construct a bar graph for all the states beginning with the letter "M."
S 2.2.2
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Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows.
- Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically)
- Press MATH
- Arrow over to PRB
- Press 5:randInt(
- Enter 51,1,8)
Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}.
Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.
For each of the following data sets, create a stem plot and identify any outliers.
Exercise 2.2.7
The miles per gallon rating for 30 cars are shown below (lowest to highest).
19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43
Answer
| Stem | Leaf |
|---|---|
| 1 | 9 9 9 |
| 2 | 0 1 1 5 5 5 6 6 8 9 |
| 3 | 1 1 2 2 3 4 5 6 7 7 8 8 8 8 |
| 4 | 1 3 3 |
The height in feet of 25 trees is shown below (lowest to highest).
25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54
The data are the prices of different laptops at an electronics store. Round each value to the nearest ten.
249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610
Answer
| Stem | Leaf |
|---|---|
| 2 | 5 5 6 7 7 8 |
| 3 | 0 0 1 2 3 3 5 5 5 7 7 9 |
| 4 | 1 6 9 |
| 5 | 6 7 7 |
| 6 | 1 |
The data are daily high temperatures in a town for one month.
61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95
For the next three exercises, use the data to construct a line graph.
Exercise 2.2.8
In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in the Table below.
| Number of times in store | Frequency |
|---|---|
| 1 | 4 |
| 2 | 10 |
| 3 | 16 |
| 4 | 6 |
| 5 | 4 |
Answer
Exercise 2.2.9
In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table .
| Years since last purchase | Frequency |
|---|---|
| 0 | 2 |
| 1 | 8 |
| 2 | 13 |
| 3 | 22 |
| 4 | 16 |
| 5 | 9 |
Exercise 2.2.10
Several children were asked how many TV shows they watch each day. The results of the survey are shown in the Table below.
| Number of TV Shows | Frequency |
|---|---|
| 0 | 12 |
| 1 | 18 |
| 2 | 36 |
| 3 | 7 |
| 4 | 2 |
Answer
Exercise 2.2.11
The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. Table shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.
| Seasons | Number of students | Proportion of population |
|---|---|---|
| Spring | 8 | 24% |
| Summer | 9 | 26% |
| Autumn | 11 | 32% |
| Winter | 6 | 18% |
Using the data from Mrs. Ramirez’s math class supplied in the table above, construct a bar graph showing the percentages.
Answer
Exercise 2.2.12
David County has six high schools. Each school sent students to participate in a county-wide science competition. Table shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.
| High School | Science competition population | Overall student population |
|---|---|---|
| Alabaster | 28.9% | 8.6% |
| Concordia | 7.6% | 23.2% |
| Genoa | 12.1% | 15.0% |
| Mocksville | 18.5% | 14.3% |
| Tynneson | 24.2% | 10.1% |
| West End | 8.7% | 28.8% |
Use the data from the David County science competition supplied in Exercise . Construct a bar graph that shows the county-wide population percentage of students at each school.
Answer
2.3: Histograms, Frequency, Polygons, and Time Series Graphs
Q 2.3.1
Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows:
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 10 | |
| 1 | 12 | |
| 2 | 16 | |
| 3 | 12 | |
| 4 | 8 | |
| 5 | 6 | |
| 6 | 2 | |
| 8 | 2 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 18 | |
| 1 | 24 | |
| 2 | 24 | |
| 3 | 22 | |
| 4 | 15 | |
| 5 | 10 | |
| 7 | 5 | |
| 9 | 1 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0–1 | 20 | |
| 2–3 | 35 | |
| 4–5 | 12 | |
| 6–7 | 2 | |
| 8–9 | 1 |
- Find the relative frequencies for each survey. Write them in the charts.
- Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two.
- In complete sentences, give two reasons why the graphs for Publishers A and B are not identical.
- Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not?
- Make new histograms for Publisher A and Publisher B. This time, make bar widths of two.
- Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer.
Q 2.3.2
Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group.
| Amount($) | Frequency | Rel. Frequency |
|---|---|---|
| 51–100 | 5 | |
| 101–150 | 10 | |
| 151–200 | 15 | |
| 201–250 | 15 | |
| 251–300 | 10 | |
| 301–350 | 5 |
| Amount($) | Frequency | Rel. Frequency |
|---|---|---|
| 100–150 | 5 | |
| 201–250 | 5 | |
| 251–300 | 5 | |
| 301–350 | 5 | |
| 351–400 | 10 | |
| 401–450 | 10 | |
| 451–500 | 10 | |
| 501–550 | 10 | |
| 551–600 | 5 | |
| 601–650 | 5 |
- Fill in the relative frequency for each group.
- Construct a histogram for the singles group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis.
- Construct a histogram for the couples group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis.
-
Compare the two graphs:
- List two similarities between the graphs.
- List two differences between the graphs.
- Overall, are the graphs more similar or different?
- Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x -axis by $50, scale it by $100. Use relative frequency on the y -axis.
-
Compare the graph for the singles with the new graph for the couples:
- List two similarities between the graphs.
- Overall, are the graphs more similar or different?
- How did scaling the couples graph differently change the way you compared it to the singles graph?
- Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences.
S 2.3.2
| Amount($) | Frequency | Relative Frequency |
|---|---|---|
| 51–100 | 5 | 0.08 |
| 101–150 | 10 | 0.17 |
| 151–200 | 15 | 0.25 |
| 201–250 | 15 | 0.25 |
| 251–300 | 10 | 0.17 |
| 301–350 | 5 | 0.08 |
| Amount($) | Frequency | Relative Frequency |
|---|---|---|
| 100–150 | 5 | 0.07 |
| 201–250 | 5 | 0.07 |
| 251–300 | 5 | 0.07 |
| 301–350 | 5 | 0.07 |
| 351–400 | 10 | 0.14 |
| 401–450 | 10 | 0.14 |
| 451–500 | 10 | 0.14 |
| 501–550 | 10 | 0.14 |
| 551–600 | 5 | 0.07 |
| 601–650 | 5 | 0.07 |
- See the tables above
- In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included).
- In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included).
-
Compare the two graphs:
-
Answers may vary. Possible answers include:
- Both graphs have a single peak.
- Both graphs use class intervals with width equal to $50.
-
Answers may vary. Possible answers include:
- The couples graph has a class interval with no values.
- It takes almost twice as many class intervals to display the data for couples.
- Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same.
-
Answers may vary. Possible answers include:
- Check student's solution.
-
Compare the graph for the Singles with the new graph for the Couples:
-
- Both graphs have a single peak.
- Both graphs display 6 class intervals.
- Both graphs show the same general pattern.
- Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different.
-
- Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison.
- Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals.
Q 2.3.3
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows.
| # of movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 5 | ||
| 1 | 9 | ||
| 2 | 6 | ||
| 3 | 4 | ||
| 4 | 1 |
- Construct a histogram of the data.
- Complete the columns of the chart.
Use the following information to answer the next two exercises: Suppose one hundred eleven people who shopped in a special t-shirt store were asked the number of t-shirts they own costing more than $19 each.
Q 2.3.4
The percentage of people who own at most three t-shirts costing more than $19 each is approximately:
- 21
- 59
- 41
- Cannot be determined
S 2.3.4
c
Q 2.3.5
If the data were collected by asking the first 111 people who entered the store, then the type of sampling is:
- cluster
- simple random
- stratified
- convenience
Q 2.3.6
Following are the 2010 obesity rates by U.S. states and Washington, DC.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the \(x\)-axis with the states.
S 2.3.7
Answers will vary.
Exercise 2.3.6
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.
| Data Value (# cars) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
Exercise 2.3.7
What does the frequency column in the Table above sum to? Why?
Answer
65
Exercise 2.3.8
What does the relative frequency column in in the Table above sum to? Why?
Exercise 2.3.9
What is the difference between relative frequency and frequency for each data value in in the Table above ?
Answer
The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value.
Exercise 2.3.10
What is the difference between cumulative relative frequency and relative frequency for each data value?
Exercise 2.3.11
To construct the histogram for the data in in the Table above , determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.
Answer
Answers will vary. One possible histogram is shown:
Exercise 2.3.12
Construct a frequency polygon for the following:
-
Pulse Rates for Women Frequency 60–69 12 70–79 14 80–89 11 90–99 1 100–109 1 110–119 0 120–129 1 -
Actual Speed in a 30 MPH Zone Frequency 42–45 25 46–49 14 50–53 7 54–57 3 58–61 1 -
Tar (mg) in Nonfiltered Cigarettes Frequency 10–13 1 14–17 0 18–21 15 22–25 7 26–29 2
Exercise 2.3.13
Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger.
| Depth of Hunger | Frequency |
|---|---|
| 230–259 | 21 |
| 260–289 | 13 |
| 290–319 | 5 |
| 320–349 | 7 |
| 350–379 | 1 |
| 380–409 | 1 |
| 410–439 | 1 |
Answer
Find the midpoint for each class. These will be graphed on the x -axis. The frequency values will be graphed on the y -axis values.
Exercise 2.3.14
Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?
| Life Expectancy at Birth – Women | Frequency |
|---|---|
| 49–55 | 3 |
| 56–62 | 3 |
| 63–69 | 1 |
| 70–76 | 3 |
| 77–83 | 8 |
| 84–90 | 2 |
| Life Expectancy at Birth – Men | Frequency |
|---|---|
| 49–55 | 3 |
| 56–62 | 3 |
| 63–69 | 1 |
| 70–76 | 1 |
| 77–83 | 7 |
| 84–90 | 5 |
Exercise 2.3.15
Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births.
| Sex/Year | 1855 | 1856 | 1857 | 1858 | 1859 | 1860 | 1861 |
| Female | 45,545 | 49,582 | 50,257 | 50,324 | 51,915 | 51,220 | 52,403 |
| Male | 47,804 | 52,239 | 53,158 | 53,694 | 54,628 | 54,409 | 54,606 |
| Total | 93,349 | 101,821 | 103,415 | 104,018 | 106,543 | 105,629 | 107,009 |
| Sex/Year | 1862 | 1863 | 1864 | 1865 | 1866 | 1867 | 1868 | 1869 |
| Female | 51,812 | 53,115 | 54,959 | 54,850 | 55,307 | 55,527 | 56,292 | 55,033 |
| Male | 55,257 | 56,226 | 57,374 | 58,220 | 58,360 | 58,517 | 59,222 | 58,321 |
| Total | 107,069 | 109,341 | 112,333 | 113,070 | 113,667 | 114,044 | 115,514 | 113,354 |
| Sex/Year | 1871 | 1870 | 1872 | 1871 | 1872 | 1827 | 1874 | 1875 |
| Female | 56,099 | 56,431 | 57,472 | 56,099 | 57,472 | 58,233 | 60,109 | 60,146 |
| Male | 60,029 | 58,959 | 61,293 | 60,029 | 61,293 | 61,467 | 63,602 | 63,432 |
| Total | 116,128 | 115,390 | 118,765 | 116,128 | 118,765 | 119,700 | 123,711 | 123,578 |
Answer
Exercise 2.3.16
The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973.
| Year | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 |
| Police | 260.35 | 269.8 | 272.04 | 272.96 | 272.51 | 261.34 | 268.89 |
| Homicides | 8.6 | 8.9 | 8.52 | 8.89 | 13.07 | 14.57 | 21.36 |
| Year | 1968 | 1969 | 1970 | 1971 | 1972 | 1973 |
| Police | 295.99 | 319.87 | 341.43 | 356.59 | 376.69 | 390.19 |
| Homicides | 28.03 | 31.49 | 37.39 | 46.26 | 47.24 | 52.33 |
- Construct a double time series graph using a common x -axis for both sets of data.
- Which variable increased the fastest? Explain.
- Did Detroit’s increase in police officers have an impact on the murder rate? Explain.
2.4: Measures of the Location of the Data
Q 2.4.1
The median age for U.S. blacks currently is 30.9 years; for U.S. whites it is 42.3 years.
- Based upon this information, give two reasons why the black median age could be lower than the white median age.
- Does the lower median age for blacks necessarily mean that blacks die younger than whites? Why or why not?
- How might it be possible for blacks and whites to die at approximately the same age, but for the median age for whites to be higher?
Q 2.4.2
Six hundred adult Americans were asked by telephone poll, "What do you think constitutes a middle-class income?" The results are in the Table below. Also, include left endpoint, but not the right endpoint.
| Salary ($) | Relative Frequency |
|---|---|
| < 20,000 | 0.02 |
| 20,000–25,000 | 0.09 |
| 25,000–30,000 | 0.19 |
| 30,000–40,000 | 0.26 |
| 40,000–50,000 | 0.18 |
| 50,000–75,000 | 0.17 |
| 75,000–99,999 | 0.02 |
| 100,000+ | 0.01 |
- What percentage of the survey answered "not sure"?
- What percentage think that middle-class is from $25,000 to $50,000?
-
Construct a histogram of the data.
- Should all bars have the same width, based on the data? Why or why not?
- How should the <20,000 and the 100,000+ intervals be handled? Why?
- Find the 40 th and 80 th percentiles
- Construct a bar graph of the data
S 2.4.2
- \(1 - (0.02 + 0.09 + 0.19 + 0.26 + 0.18 + 0.17 + 0.02 + 0.01) = 0.06\)
- \(0.19 + 0.26 + 0.18 = 0.63\)
- Check student’s solution.
-
40
th
percentile will fall between 30,000 and 40,000
80 th percentile will fall between 50,000 and 75,000
- Check student’s solution.
Q 2.4.3
Given the following box plot:
- which quarter has the smallest spread of data? What is that spread?
- which quarter has the largest spread of data? What is that spread?
- find the interquartile range ( IQR ).
- are there more data in the interval 5–10 or in the interval 10–13? How do you know this?
-
which interval has the fewest data in it? How do you know this?
- 0–2
- 2–4
- 10–12
- 12–13
- need more information
Q 2.4.4
The following box plot shows the U.S. population for 1990, the latest available year.
- Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)? How do you know?
- 12.6% are age 65 and over. Approximately what percentage of the population are working age adults (above age 17 to age 65)?
S 2.4.4
- more children; the left whisker shows that 25% of the population are children 17 and younger. The right whisker shows that 25% of the population are adults 50 and older, so adults 65 and over represent less than 25%.
- 62.4%
2.5: Box Plots
Q 2.5.1
In a survey of 20-year-olds in China, Germany, and the United States, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results.
- In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected.
- Have more Americans or more Germans surveyed been to over eight foreign countries?
- Compare the three box plots. What do they imply about the foreign travel of 20-year-old residents of the three countries when compared to each other?
Q 2.5.2
Given the following box plot, answer the questions.
- Think of an example (in words) where the data might fit into the above box plot. In 2–5 sentences, write down the example.
- What does it mean to have the first and second quartiles so close together, while the second to third quartiles are far apart?
S 2.5.2
- Answers will vary. Possible answer: State University conducted a survey to see how involved its students are in community service. The box plot shows the number of community service hours logged by participants over the past year.
- Because the first and second quartiles are close, the data in this quarter is very similar. There is not much variation in the values. The data in the third quarter is much more variable, or spread out. This is clear because the second quartile is so far away from the third quartile.
Q 2.5.3
Given the following box plots, answer the questions.
-
In complete sentences, explain why each statement is false.
- Data 1 has more data values above two than Data 2 has above two.
- The data sets cannot have the same mode.
- For Data 1 , there are more data values below four than there are above four.
- For which group, Data 1 or Data 2, is the value of “7” more likely to be an outlier? Explain why in complete sentences.
Q 2.5.4
A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results.
- In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected for that car series.
- Which group is most likely to have an outlier? Explain how you determined that.
- Compare the three box plots. What do they imply about the age of purchasing a BMW from the series when compared to each other?
- Look at the BMW 5 series. Which quarter has the smallest spread of data? What is the spread?
- Look at the BMW 5 series. Which quarter has the largest spread of data? What is the spread?
- Look at the BMW 5 series. Estimate the interquartile range (IQR).
- Look at the BMW 5 series. Are there more data in the interval 31 to 38 or in the interval 45 to 55? How do you know this?
-
Look at the BMW 5 series. Which interval has the fewest data in it? How do you know this?
- 31–35
- 38–41
- 41–64
S 2.5.4
- Each box plot is spread out more in the greater values. Each plot is skewed to the right, so the ages of the top 50% of buyers are more variable than the ages of the lower 50%.
- The BMW 3 series is most likely to have an outlier. It has the longest whisker.
- Comparing the median ages, younger people tend to buy the BMW 3 series, while older people tend to buy the BMW 7 series. However, this is not a rule, because there is so much variability in each data set.
- The second quarter has the smallest spread. There seems to be only a three-year difference between the first quartile and the median.
- The third quarter has the largest spread. There seems to be approximately a 14-year difference between the median and the third quartile.
- IQR ~ 17 years
- There is not enough information to tell. Each interval lies within a quarter, so we cannot tell exactly where the data in that quarter is concentrated.
- The interval from 31 to 35 years has the fewest data values. Twenty-five percent of the values fall in the interval 38 to 41, and 25% fall between 41 and 64. Since 25% of values fall between 31 and 38, we know that fewer than 25% fall between 31 and 35.
Q 2.5.5
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
| # of movies | Frequency |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 6 |
| 3 | 4 |
| 4 | 1 |
Construct a box plot of the data.
2.6: Measures of the Center of the Data
Q 2.6.1
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table.
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
- What is the best estimate of the average obesity percentage for these countries?
- The United States has an average obesity rate of 33.9%. Is this rate above average or below?
- How does the United States compare to other countries?
Q 2.6.2
The table below gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children?
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16–21.45 | 23 |
| 21.45–26.9 | 4 |
| 26.9–32.35 | 9 |
| 32.35–37.8 | 7 |
| 37.8–43.25 | 6 |
| 43.25–48.7 | 1 |
S 2.6.2
The mean percentage, \(\bar{x} = \frac{1328.65}{50} = 26.75\)
2.7: Skewness and the Mean, Median, and Mode
Q 2.7.1
The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years.
- What does it mean for the median age to rise?
- Give two reasons why the median age could rise.
- For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not?
2.8: Measures of the Spread of the Data
Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005.
- \(\mu = 1000\) FTES
- median = 1,014 FTES
- \(\sigma = 474\) FTES
- first quartile = 528.5 FTES
- third quartile = 1,447.5 FTES
- \(n = 29\) years
Q 2.8.1
A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer.
S 2.8.1
The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or below the median.
Q 2.8.2
75% of all years have an FTES:
- at or below: _____
- at or above: _____
Q 2.8.3
The population standard deviation = _____
S 2.8.3
474 FTES
Q 2.8.4
What percent of the FTES were from 528.5 to 1447.5? How do you know?
Q 2.8.5
What is the IQR ? What does the IQR represent?
S 2.8.5
919
Q 2.8.6
How many standard deviations away from the mean is the median?
Additional Information: The population FTES for 2005–2006 through 2010–2011 was given in an updated report. The data are reported here.
| Year | 2005–06 | 2006–07 | 2007–08 | 2008–09 | 2009–10 | 2010–11 |
| Total FTES | 1,585 | 1,690 | 1,735 | 1,935 | 2,021 | 1,890 |
Q 2.8.7
Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR . Round to one decimal place.
S 2.8.7
- mean = 1,809.3
- median = 1,812.5
- standard deviation = 151.2
- first quartile = 1,690
- third quartile = 1,935
- IQR = 245
Q 2.8.8
Construct a box plot for the FTES for 2005–2006 through 2010–2011 and a box plot for the FTES for 1976–1977 through 2004–2005.
Q 2.8.9
Compare the IQR for the FTES for 1976–77 through 2004–2005 with the IQR for the FTES for 2005-2006 through 2010–2011. Why do you suppose the IQR s are so different?
S 2.8.10
Hint: Think about the number of years covered by each time period and what happened to higher education during those periods.
Q 2.8.11
Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer.
| Student | GPA | School Average GPA | School Standard Deviation |
|---|---|---|---|
| Thuy | 2.7 | 3.2 | 0.8 |
| Vichet | 87 | 75 | 20 |
| Kamala | 8.6 | 8 | 0.4 |
Q 2.8.12
A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer.
S 2.8.12
For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.
Q 2.8.13
An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.
- Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?
- Who is the fastest runner with respect to his or her class? Explain why.
Q 2.8.14
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the table belo2
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
What is the best estimate of the average obesity percentage for these countries? What is the standard deviation for the listed obesity rates? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ obesity rate compared to the average rate? Explain.
S 2.8.14
- \(\bar{x} = 23.32\)
- Using the TI 83/84, we obtain a standard deviation of: \(s_{x} = 12.95\).
- The obesity rate of the United States is 10.58% higher than the average obesity rate.
- Since the standard deviation is 12.95, we see that \(23.32 + 12.95 = 36.27\) is the obesity percentage that is one standard deviation from the mean. The United States obesity rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, while 34% obese, does not have an unusually high percentage of obese people.
Q 2.8.15
The Table below gives the percent of children under five considered to be underweight.
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16–21.45 | 23 |
| 21.45–26.9 | 4 |
| 26.9–32.35 | 9 |
| 32.35–37.8 | 7 |
| 37.8–43.25 | 6 |
| 43.25–48.7 | 1 |
What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain.


Appendix F: Example Emergency Self-Assessment Form
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QUESTION |
YES |
NO |
|---|---|---|
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Have you done an assessment of the types of emergencies your early childcare education program could experience based on your geographic region? |
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Has your early childcare education program developed specific procedures for preparedness, response, and recovery for each type of emergency indicated as a possibility, including how to be informed and how to communicate? |
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|
Does your plan account for practicing and revising your emergency response? |
||
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Do you have a list of emergency contact information for first responders that is visible in your facility |
||
|
Do you have updated emergency contact information for each child in your early childcare education program? |
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Do you have an emergency kit that is updated monthly and has enough supplies to last for up to 72 hours, including emergency contact information for each child in your early childcare education program and daily attendance sheets? |
||
|
Does your program have plans in place to train children, families, and staff (if applicable) about emergency-preparedness plans and procedures? |
||
|
Does your plan address continuation of services for children and families? |
||
|
Does your plan include how to address the mental health and emotional needs of children, families, and staff (if applicable) before, during, and after an emergency? |
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|
Plans to address any questions that receive a NO rating: |


12.5: Summary
The topics covered in this chapter can be summarized as follows:
| Section | Summary |
|---|---|
| 12.1 Stress and Strain | Stress within rocks—which includes compression, extension and shearing—typically originates from plate-boundary processes. Rock that is stressed responds with either elastic or plastic strain, and may eventually break. The way a rock responds to stress depends on its composition and structure, the rate at which strain is applied, and also to the temperature of the rock body and the presence of water. |
| 12.2 Folding | Folding is generally a plastic response to compressive stress, although some brittle behaviour can happen during folding. An upward fold is an antiform. A downward fold is a synform. The axis of a fold can be vertical, inclined, or even horizontal. If we know that the folded beds have not been overturned, then we can use the more specific terms: anticline and syncline. |
| 12.3 Fracturing and Faulting | Fractures (joints) typically form during extension, but can also form during compression. Faulting, which involves the displacement of rock, can take place during compression or extension, as well as during shearing at transform boundaries. Thrust faulting is a special form of reverse faulting. |
| 12.4 Measuring Geological Structures | It is important to be able to measure the strike and dip of planar surfaces, such as a bedding planes, fractures or faults. Special symbols are used to show the orientation of structural features on geological maps. |
- What types of plate boundaries are most likely to contribute to the following?: a) compression, b) extension, and c) shearing.
- Explain the difference between elastic strain and plastic strain.
- List some of the factors that influence whether a rock will deform (in either an elastic or plastic manner) or break when placed under stress.
- Label the types of folds in this diagram, and label any of the important features of the folds.
- Explain why fractures are common in volcanic rocks.
- What is the difference between a normal fault and a reverse fault, and under what circumstances would you expect these to form?
- What type of fault would you expect to see near to a transform plate boundary?
-
i) Describe in words the
general
attitude (strike and dip) of these beds.
ii) Which of these beds is the oldest?
iii) What is “a” and what is its attitude?
iv) What is “b” and what is its attitude?
v) Which of these terms applies to “b”: “left lateral” or “right lateral”?
Answers to Review Questions can be found in Appendix 2 .
Media Attributions
- Figures A, B: © Steven Earle. CC BY.


2.24: Solving Quadratic Equations Using the Square Root Property
Solving Quadratic Equations of the Form \(ax^{2}=k\) using the Square Root Property
We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation \(x^{2}=9\).
\(x^{2}=9\)
Put the equation in standard form.
\(x^{2}-9=0\)
Factor the difference of squares.
\((x-3)(x+3)=0\)
Use the Zero Produce Property.
\(x-3=0 \quad x-3=0\)
Solve each equation.
\(x=3 \quad x=-3\)
We can easily use factoring to find the solutions of similar equations, like \(x^{2}=16\) and \(x^{2}=25\), because \(16\) and \(25\) are perfect squares. In each case, we would get two solutions, \(x=4, x=-4\) and \(x=5, x=-5\)
But what happens when we have an equation like \(x^{2}=7\)? Since \(7\) is not a perfect square, we cannot solve the equation by factoring.
Previously we learned that since \(169\) is the square of \(13\), we can also say that \(13\) is a square root of \(169\). Also, \((-13)^{2}=169\), so \(−13\) is also a square root of \(169\). Therefore, both \(13\) and \(−13\) are square roots of \(169\). So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:
If \(n^{2}=m\), then \(n\) is a square root of \(m\).
Since these equations are all of the form \(x^{2}=k\), the square root definition tells us the solutions are the two square roots of \(k\). This leads to the Square Root Property .
If \(x^{2}=k\), then
\(x=\sqrt{k} \quad\) or \(\quad x=-\sqrt{k} \quad\) or \(\quad x=\pm \sqrt{k}\)
Notice that the Square Root Property gives two solutions to an equation of the form \(x^{2}=k\), the principal square root of \(k\) and its opposite. We could also write the solution as \(x=\pm \sqrt{k}\). We read this as \(x\) equals positive or negative the square root of \(k\).
Now we will solve the equation \(x^{2}=9\) again, this time using the Square Root Property.
\(\begin{aligned} &x^{2} =9 \\ \text { Use the Square Root Property. } \quad& x=\pm \sqrt{9} \\& x =\pm 3 \end{aligned}\)
So \(x=3\) or \(x=-3\)
What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation \(x^{2}=7\).
\(x^{2}=7\)
Use the Square Root Property. \(x=\sqrt{7}, \quad x=-\sqrt{7}\)
We cannot simplify \(\sqrt{7}\), so we leave the answer as a radical.
Solve: \(x^{2}-50=0\)
Solution :
| Step 1 : Isolate the quadratic term and make its coefficient one. | Add \(50\) to both sides to get \(x^{2}\) by itself. | \(\begin{aligned} x^{2}-50 &=0 \\ x^{2} &=50 \end{aligned}\) |
| Step 2 : Use the Square Root Property. | Remember to write the \(\pm\) symbol. | \(x=\pm \sqrt{50}\) |
| Step 3 : Simplify the radical. | Rewrite to show two solutions. | \(\begin{array}{l}{x=\pm \sqrt{25} \cdot \sqrt{2}} \\ {x=\pm 5 \sqrt{2}} \\ {}x=5\sqrt{2}, \:x=-5\sqrt{2}\end{array}\) |
| Step 4 : Check the solutions. | Substitute in \(x=5 \sqrt{2}\) and \(x=-5 \sqrt{2}\) |
\(\begin{array}{r}{x^{2}-50=0} \\ {(\color{red}{5 \sqrt{2}}\color{black}{)}^{2}-50 \stackrel{?}{=} 0} \\ {25 \cdot 2-50 \stackrel{?}{=} 0} \\ {0=0}\end{array}\) \(\begin{array}{r}{x^{2}-50=0} \\ {(\color{red}{-5 \sqrt{2}}\color{black}{)}^{2}-50 \stackrel{?}{=} 0} \\ {25 \cdot 2-50 \stackrel{?}{=} 0} \\ {0=0}\end{array}\) |
Solve: \(y^{2}-27=0\)
- Answer
-
\(y=3 \sqrt{3}, y=-3 \sqrt{3}\)
The steps to take to use the Square Root Property to solve a quadratic equation are listed here.
Solving a Quadratic Equation Using the Square Root Property
- Isolate the quadratic term and make its coefficient one.
- Use Square Root Property.
- Simplify the radical.
- Check the solutions.
In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient \(3\) before using the Square Root Property.
Solve: \(3 z^{2}=108\)
Solution :
| \(3 z^{2}=108\) | |
| The quadratic term is isolated. Divide by \(3\) to make its coefficient \(1\). | \(\frac{3 z^{2}}{3}=\frac{108}{3}\) |
| Simplify. | \(z^{2}=36\) |
| Use the Square Root Property. | \(z=\pm \sqrt{36}\) |
| Simplify the radical. | \(z=\pm 6\) |
| Rewrite to show two solutions. | \(z=6, \quad z=-6\) |
|
Check the solutions: |
Solve: \(2x^{2}=98\)
- Answer
-
\(x=7, x=-7\)
The Square Root Property states ‘If \(x^{2}=k\),’ What will happen if \(k<0\)? This will be the case in the next example.
Solve: \(x^{2}+72=0\)
Solution :
| \(x^{2}+72=0\) | |
| Isolate the quadratic term. | \(x^{2}=-72\) |
| Use the Square Root Property. | \(x=\pm \sqrt{-72}\) |
| Simplify using complex numbers. | \(x=\pm \sqrt{72} i\) |
| Simplify the radical. | \(x=\pm 6 \sqrt{2} i\) |
| Rewrite to show two solutions | \(x=6 \sqrt{2} i, x=-6 \sqrt{2} i\) |
|
Check the solutions: |
Solve: \(q^{2}+24=0\)
- Answer
-
\(c=2 \sqrt{6} i, \quad c=-2 \sqrt{6} i\)
Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.
Solve: \(\frac{2}{3} u^{2}+5=17\)
Solution :
| \(\frac{2}{3} u^{2}+5=17\) | |
| Isolate the quadratic term. | |
| Multiply by \(\frac{3}{2}\) to make the coefficient \(1\). | |
| Simplify. | |
| Use the Square Root Property. | |
| Simplify the radical. | |
| Simplify. | |
| Rewrite to show two solutions. | |
|
Check: |
Solve: \(\frac{3}{4} y^{2}-3=18\)
- Answer
-
\(y=2 \sqrt{7}, y=-2 \sqrt{7}\)
The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .
Solve: \(2 x^{2}-8=41\)
Solution :
| Isolate the quadratic term. | |
| Divide by \(2\) to make the coefficient \(1\). | |
| Simplify. | |
| Use the Square Root Property. | |
| Rewrite the radical as a fraction of square roots. | |
| Rationalize the denominator. | |
| Simplify. | |
| Rewrite to show two solutions. | |
|
Check: We leave te check for you. |
Solve: \(3 t^{2}+6=70\)
- Answer
-
\(t=\frac{8 \sqrt{3}}{3}, \quad t=-\frac{8 \sqrt{3}}{3}\)
Solving Quadratic Equations of the Form \(a(x-h)^{2}=k\) Using the Square Root Property
We can use the Square Root Property to solve an equation of the form \(a(x-h)^{2}=k\) as well. Notice that the quadratic term, \(x\), in the original form \(ax^{2}=k\) is replaced with \((x-h)\).
The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of \(a\), then the Square Root Property can be used on \((x-h)^{2}\).
Solve: \(4(y-7)^{2}=48\)
Solution :
| \(4(y-7)^{2}=48\) | |
| Divide both sides by the coefficient \(4\). | \((y-7)^{2}=12\) |
| Use the Square Root Property on the binomial. | \(y-7=\pm \sqrt{12}\) |
| Simplify the radical. | \(y-7=\pm 2 \sqrt{3}\) |
| Solve for \(y\). | \(y=7 \pm 2 \sqrt{3}\) |
| Rewrite to show two solutions. |
\(y=7+2 \sqrt{3}\)
\(y=7-2 \sqrt{3}\) |
|
Check: |
Solve: \(2(b+2)^{2}=80\)
- Answer
-
\(b=-2+2 \sqrt{10}, \quad b=-2-2 \sqrt{10}\)
Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.
Solve: \(\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}\)
Solution :
\(\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}\)
Use the Square Root Property.
\(x-\frac{1}{3}=\pm \sqrt{\frac{5}{9}}\)
Rewrite the radical as a fraction of square roots.
\(x-\frac{1}{3}=\pm \frac{\sqrt{5}}{\sqrt{9}}\)
Simplify the radical.
\(x-\frac{1}{3}=\pm \frac{\sqrt{5}}{3}\)
Solve for \(x\).
\(x=\frac{1}{3} \pm \frac{\sqrt{5}}{3}\)
Rewrite to show two solutions.
\(x=\frac{1}{3}+\frac{\sqrt{5}}{3}, x=\frac{1}{3}-\frac{\sqrt{5}}{3}\)
Check:
We leave the check for you.
Solve: \(\left(y+\frac{3}{4}\right)^{2}=\frac{7}{16}\)
- Answer
-
\(y=-\frac{3}{4}+\frac{\sqrt{7}}{4}, y=-\frac{3}{4}-\frac{\sqrt{7}}{4}\)
We will start the solution to the next example by isolating the binomial term.
Solve: \(2(x-2)^{2}+3=57\)
Solution :
\(2(x-2)^{2}+3=57\)
Subtract \(3\) from both sides to isolate the binomial term.
\(2(x-2)^{2}=54\)
Divide both sides by \(2\).
\((x-2)^{2}=27\)
Use the Square Root Property.
\(x-2=\pm \sqrt{27}\)
Simplify the radical.
\(x-2=\pm 3 \sqrt{3}\)
Solve for \(x\).
\(x=2 \pm 3 \sqrt{3}\)
Rewrite to show two solutions.
\(x=2+3 \sqrt{3}, x=2-3 \sqrt{3}\)
Check:
We leave the check for you.
Solve: \(3(b+3)^{2}-8=88\)
- Answer
-
\(b=-3+4 \sqrt{2}, \quad b=-3-4 \sqrt{2}\)
Sometimes the solutions are complex numbers.
Solve: \((2 x-3)^{2}=-12\)
Solution :
\((2 x-3)^{2}=-12\)
Use the Square Root Property.
\(2 x-3=\pm \sqrt{-12}\)
Simplify the radical.
\(2 x-3=\pm 2 \sqrt{3} i\)
Add \(3\) to both sides.
\(2 x=3 \pm 2 \sqrt{3} i\)
Divide both sides by \(2\).
\(x=\frac{3 \pm 2 \sqrt{3 i}}{2}\)
Rewrite in standard form.
\(x=\frac{3}{2} \pm \frac{2 \sqrt{3} i}{2}\)
Simplify.
\(x=\frac{3}{2} \pm \sqrt{3} i\)
Rewrite to show two solutions.
\(x=\frac{3}{2}+\sqrt{3} i, x=\frac{3}{2}-\sqrt{3} i\)
Check:
We leave the check for you.
Solve: \((3 r+4)^{2}=-8\)
- Answer
-
\(r=-\frac{4}{3}+\frac{2 \sqrt{2} i}{3}, r=-\frac{4}{3}-\frac{2 \sqrt{2} i}{3}\)
The left sides of the equations in the next two examples do not seem to be of the form \(a(x-h)^{2}\). But they are perfect square trinomials, so we will factor to put them in the form we need.
Solve: \(4 n^{2}+4 n+1=16\)
Solution :
We notice the left side of the equation is a perfect square trinomial. We will factor it first.
| \(4 n^{2}+4 n+1=16\) | |
| Factor the perfect square trinomial. | \((2 n+1)^{2}=16\) |
| Use the Square Root Property. | \(2 n+1=\pm \sqrt{16}\) |
| Simplify the radical. | \(2 n+1=\pm 4\) |
| Solve for \(n\). | \(2 n=-1 \pm 4\) |
| Divide each side by \(2\). | \(\begin{aligned} \frac{2 n}{2} &=\frac{-1 \pm 4}{2} \\ n &=\frac{-1 \pm 4}{2} \end{aligned}\) |
| Rewrite to show two solutions. | \(n=\frac{-1+4}{2}, n=\frac{-1-4}{2}\) |
| Simplify each equation. | \(n=\frac{3}{2}, \quad n=-\frac{5}{2}\) |
|
Check: |
Solve: \(9 m^{2}-12 m+4=25\)
- Answer
-
\(m=\frac{7}{3}, \quad m=-1\)