9.3: Deposit Insurance FAQs
Q: What is the FDIC?
A: The FDIC (Federal Deposit Insurance Corporation) is an independent agency of the United States government that protects you against the loss of your insured deposits if an FDIC-insured bank or savings association fails. FDIC insurance is backed by the full faith and credit of the United States government.
Q: What is deposit insurance?
A: FDIC deposit insurance covers the depositors of a failed FDIC-insured depository institution dollar-for-dollar, principal plus any interest accrued or due to the depositor, through the date of default, up to at least $250,000. For example, if a person had a CD account in her name alone with a principal balance of $195,000 and $3,000 in accrued interest, the full $198,000 would be insured, since principal plus interest did not exceed the $250,000 insurance limit for single ownership accounts.
Q: What happens when a bank fails?
A: Though unlikely, bank failures do occur and the FDIC responds in two capacities. First, as the insurer of the bank’s deposits, the FDIC pays insurance to depositors up to the insurance limit. Historically, the FDIC pays insurance within a few days after a bank closing, usually the next business day, by either (1) providing each depositor with a new account at another insured bank in an amount equal to the insured balance of their account at the failed bank, or (2) by issuing a check to each depositor for the insured balance of their account at the failed bank.
Some deposits that exceed $250,000 and are linked to trust documents or deposits established by a third party broker may have a short wait so that their accounts can be reviewed to determine the amount of deposit insurance coverage available to them. The amount of time involved depends on how long it takes for the depositor to provide supplemental information to the FDIC so that we can complete the insurance determination.
Second, as the receiver of the failed bank, the FDIC assumes the task of selling/collecting the assets of the failed bank and settling its debts, including claims for deposits in excess of the insured limit. If a depositor has uninsured funds they receive the insured portion of their funds quickly, as described above. They may also, however, recover some portion of their uninsured funds (their remaining claim on the failed bank) from the proceeds from the sale of failed bank assets. It can take several years to sell off the assets of a failed bank. As assets are sold, however, depositors who had uninsured funds usually receive periodic payments (on a pro-rata “cents on the dollar” basis) on their remaining claim.
Q: How can I get deposit insurance?
A: Depositors do not need to apply for FDIC insurance. Coverage is automatic whenever a deposit account is opened at an FDIC-insured bank. If you want your funds insured by the FDIC, simply make sure you are placing your funds in a deposit account at an FDIC-insured bank and that your deposit does not exceed the insurance limit for that ownership category.
Q: How do I find out if a bank is FDIC-insured?
A: To determine if a bank is FDIC-insured, you can ask a bank representative, look for the FDIC sign at your bank, call the FDIC at<PHONE_NUMBER>, or you can use the FDIC’s BankFind tool. BankFind allows you to access detailed information about all FDIC-insured institutions, including branch locations, the bank’s official website address, the current operating status of your bank, and the regulator to contact for additional information and assistance.
Coverage
Q: How much deposit insurance coverage do I qualify for?
A: The standard deposit insurance amount is $250,000 per depositor, per FDIC-insured bank, per ownership category. For a basic category-by-category overview of FDIC deposit insurance coverage, you can use the Account Categories tool .
The “Your Insured Deposits” brochure also includes more comprehensive information and examples of deposit insurance coverage for various ownership categories. You can also access the FDIC’s Electronic Deposit Insurance Estimator ( EDIE ) to get details about your specific situation.
Q: Is every financial product at a bank covered by the FDIC?
A: No, FDIC deposit insurance coverage depends on whether your chosen financial product is a deposit product. The FDIC covers the traditional types of bank deposit accounts – including checking and savings accounts, money market deposit accounts (MMDAs), and certificates of deposit (CDs). Investment products that are not deposits, such as mutual funds, annuities, life insurance policies and stocks and bonds are not covered by FDIC deposit insurance.
For a more comprehensive list of financial products that are insured by the FDIC and financial products that are not insured by the FDIC, visit Accounts Covered by the FDIC
Q: Can I have more than $250,000 of deposit insurance coverage at one FDIC-insured bank?
A: Yes. The FDIC insures deposits according to the ownership category in which the funds are insured and how the accounts are titled. The standard deposit insurance coverage limit is $250,000 per depositor, per FDIC-insured bank, per ownership category. Deposits held in different ownership categories are separately insured, up to at least $250,000, even if held at the same bank. For example, a revocable trust account (including living trusts and informal revocable trusts commonly referred to as payable on death (POD) accounts) with one owner naming three unique beneficiaries can be insured up to $750,000. See the Your Insured Deposit brochure for details.
Q: Can I check to see if my accounts are fully covered?
A: Yes, you can get detailed information about your specific deposit insurance coverage by accessing the FDIC’s Electronic Deposit Insurance Estimator ( EDIE ) and entering information about your accounts. You can also call the FDIC at 1-877-ASK-FDIC (1-877-275-3342) and ask to speak to an FDIC deposit insurance specialist.
Contributors and Attributions
- Authored by : Federal Deposit Insurance Corporation. Located at : https://www.fdic.gov/deposit/deposits/faq.html . License : Public Domain: No Known Copyright


10.6: Labor Struggles, Accommodations, and Strategies of Survival
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- Huan Piñón
- University of California Press
Creative professionals face an array of challenges in their mostly unstable work settings across the region. These involve a lack of steady jobs and an absence of work benefits such as health insurance, retirement funds, paid vacations, and maternity leave. Most workers are nonunionized, and their employers are themselves subordinate to the power of national networks and multinational conglomerates. Both workers and management sense the precariousness of their situation. Their only leverage resides in the growing demand for content and the fierce competition for talent. Given that, the relationship between these rather small business entities and their creative labor is shaped by complex and intertwined conditions of professionalism, emotional attachment, creative styles, and in many cases “family”-style ties and relationships.
There are several sizes and modalities of production indies bring to the audiovisual realm. Many houses produce advertising and corporate and institutional content as a way to maintain a healthy income. They also keep their payrolls trim, commonly anchored by three to ten permanent office employees and a network of freelance creatives. Indies typically rely on the prestige and connections of particular professionals, mostly the producers themselves. So, while a company might be initially founded by a famous actor, director, writer, producer, or venture capitalist, most commonly its ongoing operations are led by a producer who is the owner, co-owner, or CEO of the company.
While varying in size, indies nevertheless tend to have similar staffing patterns. They routinely hire administrative personnel for the office, including accountants, administrators, secretaries, and office assistants. In larger companies, especially those producing fictional programming, there is also a second group of permanent employees, mostly technicians and manual laborers. In some cases, there is an editor on staff and a few other below-the-line professionals who enjoy permanent contracts and legal benefits such as health benefits, retirement funds, vacations, and compensation in case of a layoff. Yet they employ mostly a nonunionized workforce.
When launching a particular project, the company hires above-the-line creative professionals on temporary contracts at higher wages but with no legal benefits. These professionals are nonunionized, and although many are paid quite well, their work stints are unpredictable, punctuated by regular periods of unemployment without benefits. These professionals include writers, directors, actors, photographers, casting experts, art and costume designers, and many others. They constitute a floating labor force that offers its services either to independent producers or to fill the needs of network in-house production teams on particular projects. However, there are some working conditions that vary by country. In Mexico, freelance laborers take care of their individual health expenses and save for retirement. In Uruguay, individuals are hired through a personnel company that provides benefits, while in Colombia workers have benefits via payroll.
In terms of stability, freelancers are in the most vulnerable position, always pursuing the next gig by offering their services to multiple clients in different audiovisual sectors. At the same time, some workers embrace the flexibility and freedom of being able to change jobs and negotiate working conditions. An executive producer from the Uruguayan indie house Microtime argues, “There are professionals who prefer a freelance status because that allows them to manage their lives. They take vacations when they want. They do not want to ask permission from their boss, producer, and that kind of flexibility works very well for them.” 44 At the same time, this kind of freedom comes with a downside: according to an Argos screenwriter, “I cannot plan my life more than two or three months in advance. If I have earned money from a project, I need to save and take care until I find the next one.” 45 Interviewees explained that forging a good professional reputation is essential to keeping themselves in the labor force. Intelligence, talent, technical and artistic abilities, and work ethic are considered to be essential elements for survival. Work continuity and stability are linked to gaining the producers’ trust.
Interestingly, the small number of highly visible indie productions and their small size tend to create tight relationships within professional circles. Work relations are permeated with emotional ties and a sense of common artistic purpose. Some executive producers from indie houses refer to their permanent staff members and regular freelancers as family, saying they want to take care of professionals who have worked with them for a long time by offering a sense of stability. An artistic director from Del Barrio says it’s “because we know that these workers have families, and they have worked for us too many years. So, yes, there is an emotional element.” 46 Thus, just as emotional and pecuniary relations are intertwined, so too is the artistic sensibility that permeates some of the indie shops. This is often described as a common outlook on how television should be made, a shared understanding in the realm of either aesthetics or ideology. As opposed to the conventions of in-house network productions, these creative teams believe that indie shops are the perfect space to make something different. The professional profiles of the leading indie producers seem to permeate the institutional culture of their production houses, creating personal bonds around common artistic goals, including innovation and brand distinctiveness that set them apart from the dominant network studios.


9.2: Derivation of the Wave Equation
To derive the wave equation in one spacial dimension, we imagine an elastic string that undergoes small amplitude transverse vibrations. We define \(u(x, t)\) to be the vertical displacement of the string from the \(x\)-axis at position \(x\) and time \(t\), and we wish to find the pde satisfied by \(u\). We define \(\rho\) to be the constant mass density of the string, \(T\) the tension of the string, and \(\theta\) the angle between the string and the horizontal line. We consider an infinitesimal string element located between \(x_1\) and \(x_2\), with \(\Delta x = x_2 − x_1\), as shown in Fig. \(\PageIndex{1}\). The governing equations are Newton’s law of motion for the horizontal and vertical acceleration of our infinitesimal string element, and we assume that the string element only accelerates vertically. Therefore, the horizontal forces must balance and we have \[T_2\cos\theta_2=T_1\cos\theta_1.\nonumber\]
The vertical forces result in a vertical acceleration, and with \(u_{tt}\) the vertical acceleration of the string element and \(\rho\sqrt{\Delta x^2+\Delta u^2}=\rho\Delta x\sqrt{1+u_x^2}\) its mass, where we have used \(u_x = \Delta u/\Delta x\), exact as \(\Delta x\to 0\), we have \[\rho\Delta x\sqrt{1+u_x^2}u_{tt}=T_2\sin\theta_2-T_1\sin\theta_1.\nonumber\]
We now make the assumption of small vibrations, that is \(\Delta u ≪ \Delta x\), or equivalently \(u_x ≪ 1\). Note that \([u] = l\) so that \(u_x\) is dimensionless. With this approximation, to leading-order in \(u_x\) we have \[\cos\theta_2=\cos\theta_1=1,\nonumber\] \[\sin\theta_2=u_x(x_2,t),\quad\sin\theta_1=u_x(x_1,t),\nonumber\] and \[\sqrt{1+u_x^2}=1.\nonumber\]
Therefore, to leading order \(T_1 = T_2 = T\), (i.e., the tension in the string is approximately constant), and \[\rho\Delta xu_{tt}=T(u_x(x_2,t)-u_x(x_1,t)).\nonumber\]
Dividing by \(\Delta x\) and taking the limit \(\Delta x\to 0\) results in the wave equation \[u_{tt}=c^2u_{xx},\nonumber\] where \(c^2=T/\rho\). Since \([T]=ml/t^2\) and \([\rho]=m/l\), we have \([c^2]=l^2/t^2\) so that \(c\) has units of velocity.


16.7: Time Reversal in Discrete-Time Chains
The Markov property, stated in the form that the past and future are independent given the present, essentially treats the past and future symmetrically. However, there is a lack of symmetry in the fact that in the usual formulation, we have an initial time 0, but not a terminal time. If we introduce a terminal time, then we can run the process backwards in time. In this section, we are interested in the following questions:
- Is the new process still Markov?
- If so, how does the new transition probability matrix relate to the original one?
- Under what conditions are the forward and backward processes stochastically the same?
Consideration of these questions leads to reversed chains , an important and interesting part of the theory of Markov chains.
Basic Theory
Reversed Chains
Our starting point is a (homogeneous) discrete-time Markov chain \( \bs X = (X_0, X_1, X_2, \ldots) \) with (countable) state space \( S \) and transition probability matrix \( P \). Let \( m \) be a positive integer, which we will think of as the terminal time or finite time horizon . We won't bother to indicate the dependence on \( m \) notationally, since ultimately the terminal time will not matter. Define \( \hat X_n = X_{m-n} \) for \( n \in \{0, 1, \ldots, m\} \). Thus, the process forward in time is \( \bs X = (X_0, X_1, \ldots, X_m) \) while the process backwards in time is \[ \hat{\bs X} = (\hat X_0, \hat X_1, \ldots, \hat X_m) = (X_m, X_{m-1}, \ldots, X_0) \]
For \( n \in \{0, 1, \ldots, m\} \), let \[ \hat{\mathscr F}_n = \sigma\{\hat X_0, \hat X_1, \ldots, \hat X_n\} = \sigma\{X_{m-n}, X_{m - n + 1}, \ldots, X_m\} \] denote the \( \sigma \) algebra of the events of the process \( \hat{\bs X} \) up to time \( n \). So of course, an event for \( \hat{\bs X} \) up to time \( n \) is an event for \( \bs X \) from time \( m - n \) forward. Our first result is that the reversed process is still a Markov chain, but not time homogeneous in general.
The process \( \hat{\bs X} = (\hat X_0, \hat X_1, \ldots, \hat X_m) \) is a Markov chain, but is not time homogenous in general. The one-step transition matrix at time \( n \in \{0, 1, \ldots, m - 1\} \) is given by \[ \P(\hat X_{n+1} = y \mid \hat X_n = x) = \frac{\P(X_{m - n - 1} = y)}{\P(X_{m - n} = x)} P(y, x), \quad (x, y) \in S^2\]
Proof
Let \( A \in \hat{\mathscr F}_n \) and \( x, \, y \in S \). Then \begin{align*} \P(\hat X_{n+1} = y \mid \hat X_n = x, A) & = \frac{\P(\hat X_{n+1} = y, \hat X_n = x, A)}{\P(\hat X_n = x, A)} = \frac{\P(X_{m - n - 1} = y, X_{m - n} = x, A)}{\P(X_{m - n} = x, A)} \\ & = \frac{\P(A \mid X_{m - n - 1} = y, X_{m - n} = x) \P(X_{m - n} = x \mid X_{m - n - 1} = y) \P(X_{m - n - 1} = y)}{\P(A \mid X_{m - n} = x) \P(X_{m - n} = x)} \end{align*} But \( A \in \sigma\{X_{m - n}, \ldots, X_m\} \) and so by the Markov property for \( \bs X \), \[ \P(A \mid X_{m - n - 1} = y, X_{m - n} = x) = \P(A \mid X_{m - n} = x) \] By the time homogeneity of \( \bs X \), \(\P(X_{m - n} = x \mid X_{m - n - 1} = y) = P(y, x)\). Substituting and simplifying gives \[ \P(\hat X_{n+1} = y \mid \hat X_n = x, A) = \frac{\P(X_{m - n - 1} = y)}{\P(X_{m - n} = x)} P(y, x) \]
However, the backwards chain will be time homogeneous if \( X_0 \) has an invariant distribution.
Suppose that \( \bs X \) is irreducible and positive recurrent, with (unique) invariant probability density function \( f \). If \( X_0 \) has the invariant probability distribution, then \( \hat{\bs X} \) is a time-homogeneous Markov chain with transition matrix \( \hat P \) given by \[ \hat P(x, y) = \frac{f(y)}{f(x)} P(y, x), \quad (x, y) \in S^2 \]
Proof
This follows from the result above . Recall that if \( X_0 \) has PDF \( f \), then \( X_k \) has PDF \( f \) for each \( k \in \N \).
Recall that a discrete-time Markov chain is ergodic if it is irreducible, positive recurrent, and aperiodic. For an ergodic chain, the previous result holds in the limit of the terminal time.
Suppose that \( \bs X \) is ergodic, with (unique) invariant probability density function \( f \). Regardless of the distribution of \( X_0 \), \[\P(\hat X_{n+1} = y \mid \hat X_n = x) \to \frac{f(y)}{f(x)} P(y, x) \text{ as } m \to \infty\]
Proof
This follows from the conditional probability above and our study of the limiting behavior of Markov chains. Since \( \bs X \) is ergodic, \( \P(X_k = x) \to f(x) \) as \( k \to \infty \) for every \( x \in S \).
These three results are motivation for the definition that follows. We can generalize by defining the reversal of an irreducible Markov chain, as long as there is a positive, invariant function. Recall that a positive invariant function defines a positive measure on \( S \), but of course not in general a probability distribution.
Suppose that \( \bs X \) is an irreducible Markov chain with transition matrix \( P \), and that \( g: S \to (0, \infty) \) is invariant for \( \bs X \). The reversal of \( \bs X \) with respect to \( g \) is the Markov chain \( \hat{\bs X} = (\hat X_0, \hat X_1, \ldots) \) with transition probability matrix \( \hat P \) defined by \[\hat P(x, y) = \frac{g(y)}{g(x)} P(y, x), \quad (x, y) \in S^2\]
Proof
We need to show that \( \hat P \) is a valid transition probability matrix, so that the definition makes sense. Since \( g \) is invariant for \( \bs X \), \[\sum_{y \in S} \hat P(x, y) = \frac{1}{g(x)} \sum_{y \in S} g(y) P(y, x) = \frac{g(x)}{g(x)} = 1, \quad x \in S\]
Recall that if \( g \) is a positive invariant function for \( \bs X \) then so is \( c g \) for every positive constant \( c \). Note that \( g \) and \( c g \) generate the same reversed chain. So let's consider the cases:
Suppose that \( \bs X \) is an irreducible Markov chain on \( S \).
- If \( \bs X \) is recurrent, then \( \bs X \) always has a positive invariant function that is unique up to multiplication by positive constants. Hence the reversal of a recurrent chain \( \bs X \) always exists and is unique, and so we can refer to the reversal of \( \bs X \) without reference to the invariant function.
- Even better, if \( \bs X \) is positive recurrent, then there exists a unique invariant probability density function, and the reversal of \( \bs X \) can be interpreted as the time reversal (with respect to a terminal time) when \( \bs X \) has the invariant distribution, as in the motivating exercises above.
- If \( \bs X \) is transient, then there may or may not exist a positive invariant function, and if one does exist, it may not be unique (up to multiplication by positive constants). So a transient chain may have no reversals or more than one.
Nonetheless, the general definition is natural, because most of the important properties of the reversed chain follow from the balance equation between the transition matrices \( P \) and \( \hat P \), and the invariant function \( g \): \[ g(x) \hat P(x, y) = g(y) P(y, x), \quad (x, y) \in S^2 \] We will see this balance equation repeated with other objects related to the Markov chains.
Suppose that \( \bs X \) is an irreducible Markov chain with invariant function \( g: S \to (0, \infty) \), and that \( \hat{\bs X} \) is the reversal of \( \bs X \) with respect to \( g \). For \( x, \, y \in S \),
- \( \hat P(x, x) = P(x, x) \)
- \( \hat P(x, y) \gt 0 \) if and only if \( P(y, x) \gt 0 \)
Proof
These results follow immediately from the balance equation \( g(x) \hat P(x, y) = g(y) P(y, x) \) for \( (x, y) \in S^2 \).
From part (b) it follows that the state graphs of \( \bs X \) and \( \hat{\bs X} \) are reverses of each other. That is, to go from the state graph of one chain to the state graph of the other, simply reverse the direction of each edge. Here is a more complicated (but equivalent) version of the balance equation for chains of states:
Suppose again that \( \bs X \) is an irreducible Markov chain with invariant function \( g: S \to (0, \infty) \), and that \( \hat{\bs X} \) is the reversal of \( \bs X \) with respect to \( g \). For every \( n \in \N_+ \) and every sequence of states \( (x_1, x_2, \ldots, x_n, x_{n+1}) \in S^{n+1} \), \[ g(x_1) \hat P(x_1, x_2) \hat P(x_2, x_3) \cdots \hat P(x_n, x_{n+1}) = g(x_{n+1}) P(x_{n+1}, x_n) \cdots P(x_3, x_2) P(x_2, x_1) \]
Proof
This follows from repeated applications of the basic equation. When \( n = 1 \), we have the balance equation itself: \[g(x_1) \hat P(x_1, x_2) = g(x_2) P(x_2, x_1)\] For \( n = 2 \), \[g(x_1) \hat P(x_1, x_2) \hat P(x_2, x_3) = g(x_2)P(x_2, x_1)\hat P(x_2, x_3) = g(x_3)P(x_3, x_2)P(x_2, x_1)\] Continuing in this manner (or using induction) gives the general result.
The balance equation holds for the powers of the transition matrix:
Suppose again that \( \bs X \) is an irreducible Markov chain with invariant function \( g: S \to (0, \infty) \), and that \( \hat{\bs X} \) is the reversal of \( \bs X \) with respect to \( g \). For every \( (x, y) \in S^2 \) and \( n \in \N \), \[ g(x) \hat P^n(x, y) = g(y) P^n (y, x) \]
Proof
When \( n = 0 \), the left and right sides are \( g(x) \) if \( x = y \) and are 0 otherwise. When \( n = 1 \), we have the basic balance equation: \( g(x) \hat P(x, y) = g(y) P(y, x) \). In general, for \( n \in \N_+ \), by the previous result we have \begin{align*} g(x) \hat P^n(x, y) &= \sum_{(x_1, \ldots, x_{n-1}) \in S^{n-1}} g(x) \hat P(x, x_1) \hat P(x_1, x_2) \cdots \hat P(x_{n-1}, y) \\ &= \sum_{(x_1, \ldots, x_{n-1}) \in S^{n-1}} g(y) P(y, x_{n-1}) P(x_{n-1}, x_{n-2}) \cdots P(x_1, x) = g(y) P^n(y, x) \end{align*}
We can now generalize the simple result above .
Suppose again that \( \bs X \) is an irreducible Markov chain with invariant function \( g: S \to (0, \infty) \), and that \( \hat{\bs X} \) is the reversal of \( \bs X \) with respect to \( g \). For \( n \in \N \) and \( (x, y) \in S^2 \),
- \( P^n(x, x) = \hat P^n(x, x) \)
- \( \hat P^n(x, y) \gt 0 \) if and only if \( P^n(y, x) \gt 0 \)
In terms of the state graphs, part (b) has an obvious meaning: If there exists a path of length \( n \) from \( y \) to \( x \) in the original state graph, then there exists a path of length \( n \) from \( x \) to \( y \) in the reversed state graph. The time reversal definition is symmetric with respect to the two Markov chains.
Suppose again that \( \bs X \) is an irreducible Markov chain with invariant function \( g: S \to (0, \infty) \), and that \( \hat{\bs X} \) is the reversal of \( \bs X \) with respect to \( g \). Then
- \( g \) is also invariant for \( \hat{\bs X} \).
- \( \hat{\bs X} \) is also irreducible.
- \( \bs X \) is the reversal of \( \hat{\bs X} \) with respect to \( g \).
Proof
- For \( y \in S \), using the balance equation, \[\sum_{x \in S} g(x) \hat P(x, y) = \sum_{x \in S} g(y) P(y, x) = g(y)\]
- Suppose \( (x, y) \in S^2 \). Since \( \bs X \) is irreducible, there exist \( n \in \N \) with \( P^n(y, x) \gt 0 \). But then from the previous result, \( \hat P^n(x, y) \gt 0 \). Hence \( \hat{\bs X} \) is also irreducible.
- This is clear from the symmetric relationship in the fundamental result.
The balance equation also holds for the potential matrices.
Suppose that \( \bs X \) and \( \hat{\bs X} \) are time reversals with respect to the invariant function \( g: S \to (0, \infty) \). For \( \alpha \in (0, 1] \), the \( \alpha \) potential matrices are related by \[ g(x) \hat R_\alpha(x, y) = g(y) R_\alpha(y, x), \quad (x, y) \in S^2 \]
Proof
This follows easily from the result above and the definition of the potential matrices: \begin{align*} g(x) \hat R_\alpha(x, y) & = g(x) \sum_{n=0}^\infty \alpha^n \hat P^n(x, y) = \sum_{n=0}^\infty \alpha^n g(x) \hat P^n(x, y) \\ & = \sum_{n=0}^\infty \alpha^n g(y) P^n(y, x) = g(y) \sum_{n=0}^\infty \alpha^n P^n(y, x) = g(y) R_\alpha(y, x) \end{align*}
Markov chains that are time reversals share many important properties:
Suppose that \( \bs X \) and \( \hat{\bs X} \) are time reversals. Then
- \( \bs X \) and \( \hat{\bs X} \) are of the same type (transient, null recurrent, or positive recurrent).
- \( \bs X \) and \( \hat{\bs X} \) have the same period.
- \( \bs X \) and \( \hat{\bs X} \) have the same mean return time \( \mu(x) \) for every \( x \in S \).
Proof
Suppose that \( \bs X \) and \( \hat{\bs X} \) are time reversals with respect to the invariant function \( g: S \to (0, \infty) \).
- The expected number of visits to a state \( x \in S \), starting in \( x \), is the same for both chains: \( \hat R(x, x) = R(x, x) \). Hence either both chains are transient (if the common potential is finite) or both chains are recurrent (if the common potential is infinite). If both chains are recurrent then the invariant function \( g \) is unique up to multiplication by positive constants, and both are null recurrent if \( \sum_{x \in S} g(x) = \infty \) and both are positive recurrent if \( \sum_{x \in S} g(x) \lt \infty \).
- This follows since \( P^n(x, x) = \hat P^n(x, x) \) for all \( n \in \N \) and \( x \in S \).
- If both chains are transient or both are null recurrent, then \( \mu(x) = \hat \mu(x) = \infty \) for all \( x \in S \). If both chains are positive recurrent, then for all \( n \in \N \) and \( x \in S \), we have \[\frac{1}{n} \sum_{k = 1}^n P^k(x, x) = \frac{1}{n} \sum_{k = 1}^n \hat P^k(x, x)\] The left side converges to \( 1 / \mu(x) \) as \( n \to \infty \) while the right side converges to \( 1 / \hat \mu(x) \) as \( n \to \infty \).
The main point of the next result is that we don't need to know a-priori that \( g \) is invariant for \( \bs X \), if we can guess \( g \) and \( \hat P \).
Suppose again that \( \bs X \) is irreducible with transition probability matrix \( P \). If there exists a a function \( g: S \to (0, \infty) \) and a transition probability matrix \( \hat P \) such that \( g(x) \hat P(x, y) = g(y) P(y, x) \) for all \( (x, y) \in S^2 \), then
- \( g \) is invariant for \( \bs X \).
- \( \hat P \) is the transition matrix of the reversal of \( \bs X \) with respect to \( g \).
Proof
- Since \( \hat P \) is a transition probability matrix, we have the same computation we have seen before: \[g P(x) = \sum_{y \in S} g(y) P(y, x) = \sum_{y \in S} g(x) \hat P(x, y) = g(x), \quad x \in S\]
- This follows from (a) and the definition.
As a corollary, if there exists a probability density function \( f \) on \( S \) and a transition probability matrix \( \hat P \) such that \( f(x) \hat P(x, y) = f(y) P(y, x) \) for all \( (x, y) \in S^2 \) then in addition to the conclusions above, we know that the chains \( \bs X \) and \( \hat{\bs X} \) are positive recurrent.
Reversible Chains
Clearly, an interesting special case occurs when the transition matrix of the reversed chain turns out to be the same as the original transition matrix. A chain of this type could be used to model a physical process that is stochastically the same, forward or backward in time.
Suppose again that \( \bs X = (X_0, X_1, X_2, \ldots) \) is an irreducible Markov chain with transition matrix \( P \) and invariant function \( g: S \to (0, \infty) \). If the reversal of \( \bs X \) with respect to \( g \) also has transition matrix \( P \), then \( \bs X \) is said to be reversible with respect to \( g \). That is, \( \bs X \) is reversible with respect to \( g \) if and only if \[ g(x) P(x, y) = g(y) P(y, x), \quad (x, y) \in S^2 \]
Clearly if \( \bs X \) is reversible with respect to the invariant function \( g: S \to (0, \infty) \) then \( \bs X \) is reversible with respect to the invariant function \( c g \) for every \( c \in (0, \infty) \). So again, let's review the cases.
Suppose that \( \bs X \) is an irreducible Markov chain on \( S \).
- If \( \bs X \) is recurrent, there exists a positive invariant function that is unique up to multiplication by positive constants. So \( \bs X \) is either reversible or not, and we don't have to reference the invariant function \( g \).
- If \( \bs X \) is positive recurrent then there exists a unique invariant probability density function \( f: S \to (0, 1) \), and again, either \( \bs X \) is reversible or not. If \( \bs X \) is reversible, then \( P \) is the transition matrix of \( \bs X \) forward or backward in time, when the chain has the invariant distribution.
- If \( \bs X \) is transient, there may or may not exist positive invariant functions. If there are two or more positive invariant functions that are not multiplies of one another, \( \bs X \) might be reversible with respect to one function but not the others.
The non-symmetric simple random walk on \( \Z \) falls into the last case. Using the last result in the previous subsection, we can tell whether \( \bs X \) is reversible with respect to \( g \) without knowing a-priori that \( g \) is invariant.
Suppose again that \( \bs X \) is irreducible with transition matrix \( P \). If there exists a function \( g: S \to (0, \infty) \) such that \( g(x) P(x, y) = g(y) P(y, x) \) for all \( (x, y) \in S^2 \), then
- \( g \) is invariant for \( \bs X \).
- \( \bs X \) is reversible with respect to \( g \)
If we have reason to believe that a Markov chain is reversible (based on modeling considerations, for example), then the condition in the previous theorem can be used to find the invariant functions. This procedure is often easier than using the definition of invariance directly. The next two results are minor generalizations:
Suppose again that \( \bs X \) is irreducible and that \( g: S \to (0, \infty) \). Then \( g \) is invariant and \( \bs X \) is reversible with respect to \( g \) if and only if for every \( n \in \N_+ \) and every sequence of states \( (x_1, x_2, \ldots x_n, x_{n+1}) \in S^{n+1} \), \[ g(x_1) P(x_1, x_2) P(x_2, x_3) \cdots P(x_n, x_{n+1}) = g(x_{n+1}) P(x_{n+1}, x_n), \cdots P(x_3, x_2) P(x_2, x_1) \]
Suppose again that \( \bs X \) is irreducible and that \( g: S \to (0, \infty) \). Then \( g \) is invariant and \( \bs X \) is reversible with respect to \( g \) if and only if for every \( (x, y) \in S^2 \) and \( n \in \N_+ \), \[ g(x) P^n(x, y) = g(y) P^n(y, x) \]
Here is the condition for reversibility in terms of the potential matrices.
Suppose again that \( \bs X \) is irreducible and that \( g: S \to (0, \infty) \). Then \( g \) is invariant and \( \bs X \) is reversible with respect to \( g \) if and only if \[ g(x) R_\alpha(x, y) = g(y) R_\alpha(y, x), \quad \alpha \in (0, 1], \, (x, y) \in S^2 \]
In the positive recurrent case (the most important case), the following theorem gives a condition for reversibility that does not directly reference the invariant distribution. The condition is known as the Kolmogorov cycle condition , and is named for Andrei Kolmogorov
Suppose that \( \bs X \) is irreducible and positive recurrent. Then \( \bs X \) is reversible if and only if for every sequence of states \( (x_1, x_2, \ldots, x_n) \), \[ P(x_1, x_2) P(x_2, x_3) \cdots P(x_{n-1}, x_n) P(x_n, x_1) = P(x_1, x_n) P(x_n, x_{n-1}) \cdots P(x_3, x_2) P(x_2, x_1) \]
Proof
Suppose that \( \bs X \) is reversible. Applying the chain result above to the sequence \( (x_1, x_2, \ldots, x_n, x_1) \) gives the Kolmogorov cycle condition. Conversely, suppose that the Kolmogorov cycle condition holds, and let \( f \) denote the invariant probability density function of \( \bs X \). From the cycle condition we have \( P(x, y) P^k(y, x) = P(y, x)P^k(x, y) \) for every \( (x, y) \in S \) and \( k \in \N_+ \). Averaging over \( k \) from 1 to \( n \) gives \[ P(x, y) \frac{1}{n} \sum_{k=1}^n P^k(y, x) = P(y, x) \frac{1}{n} \sum_{k = 1}^n P^k(x, y), \quad (x, y) \in S^2, \; n \in \N_+ \] Letting \( n \to \infty \) gives \(f(x) P(x, y) = f(y) P(y, x) \) for \( (x, y) \in S^2 \), so \( \bs X \) is reversible.
Note that the Kolmogorov cycle condition states that the probability of visiting states \( (x_2, x_3, \ldots, x_n, x_1) \) in sequence, starting in state \( x_1 \) is the same as the probability of visiting states \( (x_n, x_{n-1}, \ldots, x_2, x_1) \) in sequence, starting in state \( x_1 \). The cycle condition is also known as the balance equation for cycles .
Examples and Applications
Finite Chains
Recall the general two-state chain \( \bs X \) on \( S = \{0, 1\} \) with the transition probability matrix \[ P = \left[ \begin{matrix} 1 - p & p \\ q & 1 - q \end{matrix} \right] \] where \( p, \, q \in (0, 1) \) are parameters. The chain \( \bs X \) is reversible and the invariant probability density function is \( f = \left( \frac{q}{p + q}, \frac{p}{p + q} \right) \).
Proof
All we have to do is note that \[\left[\begin{matrix} q & p \end{matrix}\right] \left[ \begin{matrix} 1 - p & p \\ q & 1 - q \end{matrix} \right] = \left[\begin{matrix} q & p \end{matrix}\right]\]
Suppose that \( \bs X \) is a Markov chain on a finite state space \( S \) with symmetric transition probability matrix \( P \). Thus \( P(x, y) = P(y, x) \) for all \( (x, y) \in S^2 \). The chain \( \bs X \) is reversible and that the uniform distribution on \( S \) is invariant.
Proof
All we have to do is note that \( \bs{1}(x) P(x, y) = \bs{1}(y) P(y, x) \) where \( \bs{1} \) is the constant function 1 on \( S \).
Consider the Markov chain \( \bs X \) on \( S = \{a, b, c\} \) with transition probability matrix \( P \) given below:
\[ P = \left[ \begin{matrix} \frac{1}{4} & \frac{1}{4} & \frac{1}{2} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{matrix} \right] \]- Draw the state graph of \( \bs X \) and note that the chain is irreducible.
- Find the invariant probability density function \( f \).
- Find the mean return time to each state.
- Find the transition probability matrix \( \hat P \) of the time-reversed chain \( \hat{\bs X} \).
- Draw the state graph of \( \hat{\bs X} \).
Answer
- \( f = \left( \frac{6}{17}, \frac{6}{17}, \frac{5}{17}\right) \)
- \( \mu = \left( \frac{17}{6}, \frac{17}{6}, \frac{17}{5} \right) \)
- \( \hat P = \left[ \begin{matrix} \frac{1}{4} & \frac{1}{3} & \frac{5}{12} \\ \frac{1}{4} & \frac{1}{3} & \frac{5}{12} \\ \frac{3}{5} & \frac{2}{5} & 0 \end{matrix} \right] \)
Special Models
Read the discussion of reversibility for the Ehrenfest chains.
Read the discussion of reversibility for the Bernoulli-Laplace chain.
Read the discussion of reversibility for the random walks on graphs.
Read the discussion of time reversal for the reliability chains.
Read the discussion of reversibility for the birth-death chains.


3.21: Putting It All Together
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In this section of Grammar Essentials, you have learned about the 20 most common errors students make and how you can avoid them. Hopefully, you have also learned that, if you have had these errors marked on your papers, you aren’t alone. The key is having awareness, as this can help you when you edit.
To help you review, check out the video below, which provides a basic overview of what you have learned about the 20 most common errors students in the United States make.
Note
Because the use of the singular they is new, you may find some people who disagree, but most academics now agree that the singular they is correct. Please always refer to your style guide, as some style guides still recommend not using the singular they .


12.11: Set a Primary Key
Learning Outcomes
- Set the primary key
A primary key can be set for a table in one of three ways. Open a table in Design view and:
- With the field to be used as the primary key selected, click the Primary Key button key in the Tools group.
- Right-click the appropriate field and choose Primary Key from the shortcut menu.
- Allow Access to automatically create the primary key by simply saving the table.
In this example, Field One was highlighted and Primary Key was selected in the Menu bar to designate it as the Primary Key.
Contributors and Attributions
CC licensed content, Original
- Set a Primary Key. Authored by : Robert Danielson. Provided by : Lumen Learning. License : CC BY: Attribution


1.10.6: End-of-Chapter Material
-
- Last updated
- Save as PDF
Key takeaways for this Chapter
- It is important to be physical fit to be able to enjoy life.
- Being active everyday and purposefully exercising are important for your physical fitness.
- It is generally accepted that physical activity and exercise are beneficial for everyone, however it is important to communicate with your doctor.
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You need to do cardiorespiratory endurance exercise to support and build your heart and lungs.
- Cardio exercise intensity should be moderate to vigorous which can be measured by your perceived intensity or by measuring your heart rate.
- Cardio in any length is beneficial with a goal of a cumulative total of 150-300 minutes total from at least three days.
- Cardio exercises include any exercises where the body’s large muscles move in a rhythmic manner for a sustained period of time.
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You need to do strength training exercises to maintain and build your skeletal muscle.
- You must challenge your muscles by overloading them. You can do this by increasing the weight used or increasing the repetitions and sets.
- Make sure to strength train all major muscles groups at least two days/week.
- Stretching is important for flexibility.
Media Attributions
- Exercise zones Fox and Haskell © Morgoth666 is licensed under a CC BY-SA (Attribution ShareAlike) license
References
- U.S. Department of Health and Human Services. Physical Activity Guidelines for Americans, 2nd edition. Washington, DC: U.S. Department of Health and Human Services; 2018. Retrieved from https://health.gov/sites/default/fil...on.pdf#page=32
- WHO guidelines on physical activity and sedentary behaviour. Geneva: World Health Organization; 2020. Licence: CC BY-NC-SA 3.0 IGO. Retrieved from https://www.who.int/publications/i/item/9789240015128
- Global action plan on physical activity 2018–2030: more active people for a healthier world. Geneva: World Health Organization; 2018. Licence: CC BY-NC-SA 3.0 IGO. retrieved from https://www.who.int/news-room/initiatives/gappa
- Caspersen, C. J., Powell, K. E., & Christenson, G. M. (1985). Physical activity, exercise, and physical fitness: definitions and distinctions for health-related research. Public health reports (Washington, D.C. : 1974) , 100 (2), 126–131. Retrieved from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1424733/
- Tips to Help You Exercise More, Get Active, NHLBI, NIH. (2013, February 13). U.S. Department of Health & Human Services. https://www.nhlbi.nih.gov/health/edu...ing-active.htm
- American College of Sports Medicine. (2021, March 18). Stretching and Flexibility Guidelines Update . ACSM. https://www.acsm.org/all-blog-posts/...delines-update
- CDC - How Much Sleep Do I Need? - Sleep and Sleep Disorders. (2017, March 2). Centers for Disease Control and Prevention. https://www.cdc.gov/sleep/about_slee...uch_sleep.html
- Troy, D. (2021, April 2). Healthy Sleep Habits. Sleep Education. https://sleepeducation.org/healthy-s...-sleep-habits/
- Sleep Health Topics. (2021, June 18). National Sleep Foundation. https://www.thensf.org/sleep-health-topics/
- Shepard, J. W., Jr, Buysse, D. J., Chesson, A. L., Jr, Dement, W. C., Goldberg, R., Guilleminault, C., Harris, C. D., Iber, C., Mignot, E., Mitler, M. M., Moore, K. E., Phillips, B. A., Quan, S. F., Rosenberg, R. S., Roth, T., Schmidt, H. S., Silber, M. H., Walsh, J. K., & White, D. P. (2005). History of the development of sleep medicine in the United States. Journal of clinical sleep medicine : JCSM : official publication of the American Academy of Sleep Medicine, 1(1), 61–82. Retrieved from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2413168/
- Sleep Education by AASM. (2021, December 3). Learn about Sleep Disorders. Sleep Education. https://sleepeducation.org/sleep-disorders/


6.2: Networks
A network is a connection of vertices through edges. The internet is an example of a network with computers as the vertices and the connections between these computers as edges.
A spanning subgraph is a graph that joins all of the vertices of a more complex graph, but does not create a circuit
This is a graph showing how six cities are linked by roads. This graph has many spanning subgraphs but two examples are shown below.
This graph spans all of the cities (vertices) of the original graph, but does not contain any circuits.
This graph spans all of the cities (vertices) of the original graph, but does not contain any circuits.
A tree is a graph that is connected and has no circuits. Therefore, a spanning subgraph is a tree and the examples of spanning subgraphs in Example \(\PageIndex{1}\) above are also trees.
- If a graph is a tree, there is one and only one path joining any two vertices. Conversely, if there is one and only one path joining any two vertices of a graph, the graph must be a tree.
- In a tree, every edge is a bridge.
- A tree with N vertices must have N-1 edges.

Consider the spanning subgraph highlighted in green shown in Figure \(\PageIndex{2}\).
- Tree Property 1
Look at the vertices Appleville and Heavytown. Since the graph is a tree, there is only one path joining these two cities. Also, since there is only one path between any two cities on the whole graph, then the graph must be a tree.
- Tree Property 2
Since the graph is a tree, notice that every edge of the graph is a bridge, which is an edge such that if it were removed the graph would become disconnected.
- Tree Property 3
Since the graph is a tree and it has six vertices, it must have N – 1 or six – 1 = five edges.
- Tree Property 4
Since the graph is connected and has six vertices and five edges, it must be a tree.
All of the graphs shown below are trees and they all satisfy the tree properties.
A minimum spanning tree is the tree that spans all of the vertices in a problem with the least cost (or time, or distance).
The above is a weighted graph where the numbers on each edge represent the cost of each edge. We want to find the minimum spanning tree of this graph so that we can find a network that will reach all vertices for the least total cost.
This is the minimum spanning tree for the graph with a total cost of 51.
Since some graphs are much more complicated than the previous example, we can use Kruskal’s Algorithm to always be able to find the minimum spanning tree for any graph.
- Find the cheapest link in the graph. If there is more than one, pick one at random. Mark it in red.
- Find the next cheapest link in the graph. If there is more than one, pick one at random. Mark it in red.
- Continue doing this as long as the next cheapest link does not create a red circuit.
- You are done when the red edges span every vertex of the graph without any circuits. The red edges are the MST (minimum spanning tree).
Suppose that it is desired to install a new fiber optic cable network between the six cities (A, B, C, D, E, and F) shown above for the least total cost. Also, suppose that the fiber optic cable can only be installed along the roadways shown above. The weighted graph above shows the cost (in millions of dollars) of installing the fiber optic cable along each roadway. We want to find the minimum spanning tree for this graph using Kruskal’s Algorithm.
Step 1: Find the cheapest link of the whole graph and mark it in red. The cheapest link is between B and C with a cost of four million dollars.
Step 2: Find the next cheapest link of the whole graph and mark it in red. The next cheapest link is between A and C with a cost of six million dollars.
Step 3: Find the next cheapest link of the whole graph and mark it in red as long as it does not create a red circuit. The next cheapest link is between C and E with a cost of seven million dollars.
Step 4: Find the next cheapest link of the whole graph and mark it in red as long as it does not create a red circuit. The next cheapest link is between B and D with a cost of eight million dollars.
Step 5: Find the next cheapest link of the whole graph and mark it in red as long as it does not create a red circuit. The next cheapest link is between A and B with a cost of nine million dollars, but that would create a red circuit so we cannot use it. Therefore, the next cheapest link after that is between E and F with a cost of 12 million dollars, which we are able to use. We cannot use the link between C and D which also has a cost of 12 million dollars because it would create a red circuit.
This was the last step and we now have the minimum spanning tree for the weighted graph with a total cost of $37,000,000.


39.4: Transport of Gases in Human Bodily Fluids
Skills to Develop
- Describe how oxygen is bound to hemoglobin and transported to body tissues
- Explain how carbon dioxide is transported from body tissues to the lungs
Once the oxygen diffuses across the alveoli, it enters the bloodstream and is transported to the tissues where it is unloaded, and carbon dioxide diffuses out of the blood and into the alveoli to be expelled from the body. Although gas exchange is a continuous process, the oxygen and carbon dioxide are transported by different mechanisms.
Transport of Oxygen in the Blood
Although oxygen dissolves in blood, only a small amount of oxygen is transported this way. Only 1.5 percent of oxygen in the blood is dissolved directly into the blood itself. Most oxygen—98.5 percent—is bound to a protein called hemoglobin and carried to the tissues.
Hemoglobin
Hemoglobin , or Hb, is a protein molecule found in red blood cells (erythrocytes) made of four subunits: two alpha subunits and two beta subunits (Figure \(\PageIndex{1}\)). Each subunit surrounds a central heme group that contains iron and binds one oxygen molecule, allowing each hemoglobin molecule to bind four oxygen molecules. Molecules with more oxygen bound to the heme groups are brighter red. As a result, oxygenated arterial blood where the Hb is carrying four oxygen molecules is bright red, while venous blood that is deoxygenated is darker red.
It is easier to bind a second and third oxygen molecule to Hb than the first molecule. This is because the hemoglobin molecule changes its shape, or conformation, as oxygen binds. The fourth oxygen is then more difficult to bind. The binding of oxygen to hemoglobin can be plotted as a function of the partial pressure of oxygen in the blood (x-axis) versus the relative Hb-oxygen saturation (y-axis). The resulting graph—an oxygen dissociation curve —is sigmoidal, or S-shaped (Figure \(\PageIndex{2}\)). As the partial pressure of oxygen increases, the hemoglobin becomes increasingly saturated with oxygen.
Art Connection
The kidneys are responsible for removing excess H + ions from the blood. If the kidneys fail, what would happen to blood pH and to hemoglobin affinity for oxygen?
Factors That Affect Oxygen Binding
The oxygen-carrying capacity of hemoglobin determines how much oxygen is carried in the blood. In addition to \(\text{P}_{\text{O}_2}\), other environmental factors and diseases can affect oxygen carrying capacity and delivery.
Carbon dioxide levels, blood pH, and body temperature affect oxygen-carrying capacity (Figure \(\PageIndex{2}\)). When carbon dioxide is in the blood, it reacts with water to form bicarbonate (\(\text{HCO}_3^-\)) and hydrogen ions (H + ). As the level of carbon dioxide in the blood increases, more H + is produced and the pH decreases. This increase in carbon dioxide and subsequent decrease in pH reduce the affinity of hemoglobin for oxygen. The oxygen dissociates from the Hb molecule, shifting the oxygen dissociation curve to the right. Therefore, more oxygen is needed to reach the same hemoglobin saturation level as when the pH was higher. A similar shift in the curve also results from an increase in body temperature. Increased temperature, such as from increased activity of skeletal muscle, causes the affinity of hemoglobin for oxygen to be reduced.
Diseases like sickle cell anemia and thalassemia decrease the blood’s ability to deliver oxygen to tissues and its oxygen-carrying capacity. In sickle cell anemia , the shape of the red blood cell is crescent-shaped, elongated, and stiffened, reducing its ability to deliver oxygen (Figure \(\PageIndex{3}\)). In this form, red blood cells cannot pass through the capillaries. This is painful when it occurs. Thalassemia is a rare genetic disease caused by a defect in either the alpha or the beta subunit of Hb. Patients with thalassemia produce a high number of red blood cells, but these cells have lower-than-normal levels of hemoglobin. Therefore, the oxygen-carrying capacity is diminished.
Transport of Carbon Dioxide in the Blood
Carbon dioxide molecules are transported in the blood from body tissues to the lungs by one of three methods: dissolution directly into the blood, binding to hemoglobin, or carried as a bicarbonate ion. Several properties of carbon dioxide in the blood affect its transport. First, carbon dioxide is more soluble in blood than oxygen. About 5 to 7 percent of all carbon dioxide is dissolved in the plasma. Second, carbon dioxide can bind to plasma proteins or can enter red blood cells and bind to hemoglobin. This form transports about 10 percent of the carbon dioxide. When carbon dioxide binds to hemoglobin, a molecule called carbaminohemoglobin is formed. Binding of carbon dioxide to hemoglobin is reversible. Therefore, when it reaches the lungs, the carbon dioxide can freely dissociate from the hemoglobin and be expelled from the body.
Third, the majority of carbon dioxide molecules (85 percent) are carried as part of the bicarbonate buffer system . In this system, carbon dioxide diffuses into the red blood cells. Carbonic anhydrase (CA) within the red blood cells quickly converts the carbon dioxide into carbonic acid (H 2 CO 3 ). Carbonic acid is an unstable intermediate molecule that immediately dissociates into bicarbonate ions (\(\text{HCO}_3^-\)) and hydrogen (H + ) ions. Since carbon dioxide is quickly converted into bicarbonate ions, this reaction allows for the continued uptake of carbon dioxide into the blood down its concentration gradient. It also results in the production of H + ions. If too much H + is produced, it can alter blood pH. However, hemoglobin binds to the free H + ions and thus limits shifts in pH. The newly synthesized bicarbonate ion is transported out of the red blood cell into the liquid component of the blood in exchange for a chloride ion (Cl - ); this is called the chloride shift . When the blood reaches the lungs, the bicarbonate ion is transported back into the red blood cell in exchange for the chloride ion. The H + ion dissociates from the hemoglobin and binds to the bicarbonate ion. This produces the carbonic acid intermediate, which is converted back into carbon dioxide through the enzymatic action of CA. The carbon dioxide produced is expelled through the lungs during exhalation.
The benefit of the bicarbonate buffer system is that carbon dioxide is “soaked up” into the blood with little change to the pH of the system. This is important because it takes only a small change in the overall pH of the body for severe injury or death to result. The presence of this bicarbonate buffer system also allows for people to travel and live at high altitudes: When the partial pressure of oxygen and carbon dioxide change at high altitudes, the bicarbonate buffer system adjusts to regulate carbon dioxide while maintaining the correct pH in the body.
Carbon Monoxide Poisoning
While carbon dioxide can readily associate and dissociate from hemoglobin, other molecules such as carbon monoxide (CO) cannot. Carbon monoxide has a greater affinity for hemoglobin than oxygen. Therefore, when carbon monoxide is present, it binds to hemoglobin preferentially over oxygen. As a result, oxygen cannot bind to hemoglobin, so very little oxygen is transported through the body (Figure \(\PageIndex{4}\)). Carbon monoxide is a colorless, odorless gas and is therefore difficult to detect. It is produced by gas-powered vehicles and tools. Carbon monoxide can cause headaches, confusion, and nausea; long-term exposure can cause brain damage or death. Administering 100 percent (pure) oxygen is the usual treatment for carbon monoxide poisoning. Administration of pure oxygen speeds up the separation of carbon monoxide from hemoglobin.
Summary
Hemoglobin is a protein found in red blood cells that is comprised of two alpha and two beta subunits that surround an iron-containing heme group. Oxygen readily binds this heme group. The ability of oxygen to bind increases as more oxygen molecules are bound to heme. Disease states and altered conditions in the body can affect the binding ability of oxygen, and increase or decrease its ability to dissociate from hemoglobin.
Carbon dioxide can be transported through the blood via three methods. It is dissolved directly in the blood, bound to plasma proteins or hemoglobin, or converted into bicarbonate. The majority of carbon dioxide is transported as part of the bicarbonate system. Carbon dioxide diffuses into red blood cells. Inside, carbonic anhydrase converts carbon dioxide into carbonic acid (H 2 CO 3 ), which is subsequently hydrolyzed into bicarbonate (\(\text{HCO}_3^-\)) and H + . The H + ion binds to hemoglobin in red blood cells, and bicarbonate is transported out of the red blood cells in exchange for a chloride ion. This is called the chloride shift. Bicarbonate leaves the red blood cells and enters the blood plasma. In the lungs, bicarbonate is transported back into the red blood cells in exchange for chloride. The H + dissociates from hemoglobin and combines with bicarbonate to form carbonic acid with the help of carbonic anhydrase, which further catalyzes the reaction to convert carbonic acid back into carbon dioxide and water. The carbon dioxide is then expelled from the lungs.
Art Connections
Figure \(\PageIndex{2}\): The kidneys are responsible for removing excess H + ions from the blood. If the kidneys fail, what would happen to blood pH and to hemoglobin affinity for oxygen?
- Answer
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The blood pH will drop and hemoglobin affinity for oxygen will decrease.
Glossary
- bicarbonate buffer system
- system in the blood that absorbs carbon dioxide and regulates pH levels
- bicarbonate (\(\text{HCO}_3^-\)) ion
- ion created when carbonic acid dissociates into H + and \((\ce{HCO3-})\)
- carbaminohemoglobin
- molecule that forms when carbon dioxide binds to hemoglobin
- carbonic anhydrase (CA)
- enzyme that catalyzes carbon dioxide and water into carbonic acid
- chloride shift
- chloride shift exchange of chloride for bicarbonate into or out of the red blood cell
- heme group
- centralized iron-containing group that is surrounded by the alpha and beta subunits of hemoglobin
- hemoglobin
- molecule in red blood cells that can bind oxygen, carbon dioxide, and carbon monoxide
- oxygen-carrying capacity
- amount of oxygen that can be transported in the blood
- oxygen dissociation curve
- curve depicting the affinity of oxygen for hemoglobin
- sickle cell anemia
- genetic disorder that affects the shape of red blood cells, and their ability to transport oxygen and move through capillaries
- thalassemia
- rare genetic disorder that results in mutation of the alpha or beta subunits of hemoglobin, creating smaller red blood cells with less hemoglobin


1: Introduction
Abortion is often in the news. In the course of writing this essay in early 2019, Kentucky, Mississippi, Ohio, Georgia, Alabama, Missouri and Louisiana passed legislation to outlaw and criminalize abortions starting at six to eight weeks in pregnancy, with more states likely following. Federal law, however, generally permits abortions, so it is unclear what the legal outcome here will be.
Abortion is a political issue—with different political parties tending to have different perspectives on the issue—because abortion is a moral or ethical issue. (These two words, “moral” and “ethical,” mean the same thing.)
Some believe that abortions are typically morally permissible , or not wrong , and so believe that abortions should be legal. If doing something isn’t wrong, it shouldn’t be illegal: criminalizing actions that aren’t wrong is a form of injustice.
Others believe that abortion is morally wrong, that it’s often wrong, maybe nearly always or even always .
Some people argue that even though they believe abortion is wrong, it should remain legal: after all, if every morally wrong action was illegal, we would all be in jail! Seriously though, there are many actions that are morally wrong, even really hurtful, that the government shouldn’t try to prevent or punish. (You can supply the potential examples to make the point. 1 ) People who think abortion is wrong might also think that, for a variety of other reasons, their personal moral views on the issues shouldn’t be made into law for all.
Others argue that abortions are wrong and should be illegal. What types of wrongdoing should be illegal? This question isn’t easy to answer: it’s abstract and general. One answer is that seriously, extremely wrong actions should be illegal . This might seem plausible, since many illegal actions are seriously wrong. But since there are other very wrong actions that shouldn’t be illegal, this answer isn’t perfect.
We argue, however, that abortion should not be illegal because most abortions are not morally wrong (and so they are not seriously or extremely wrong). So the states above are making bad moral and legal moves, to say the least, in trying to criminalize abortions, at least when they are done early in pregnancy, as they usually are. And if federal law changes towards prohibiting abortions, that would be another, more profound step towards injustice.
There is a lot to discuss. Here’s the plan:
- First, we define “abortion.” There are controversies even in stating our topic.
- Second, we give some brief factual, scientific information about how fetuses develop, in terms of the emergence of consciousness, awareness and feeling, briefly explain the moral significance of these psychological characteristics, and review the evidence on when most abortions occur, and why .
- Third, we discuss some common, but bad, arguments. First, we review many common what are called “question-begging” arguments. This type of argument assumes the conclusion it is trying to support, instead of giving genuine reasons to support that conclusion. These arguments are a type of circular reasoning and are no good from the perspective of people who want to think critically and base their beliefs and actions on good arguments.
Next, we discuss arguments that you’d often see as comments on newspaper stories and editorials, and even in those writings themselves. We call these “everyday arguments.” Seeing why these arguments are bad will help us all shift the focus to better arguments.
- Finally, we discuss some of the most important better arguments on the issues, focusing on arguments that professional philosophers tend to focus on. Here we argue that the most influential arguments “against” abortion are weak: they don’t provide good reasons to believe that most abortions are wrong. And we argue that there are good positive reasons to believe that abortion is usually not wrong. These arguments are based on facts about early fetuses completely lacking any consciousness, awareness or feeling, and the insight that the “right to life” is not a right to anyone else’s body. So, we argue that there are good arguments to justify a broadly “pro-choice” perspective.
People often begin discussions of abortion with a lot of “what ifs”: “ What if an abortion is wanted because of rape?” “ What if it’s needed to save a woman’s life?” “ What if there are fetal abnormalities?” “ What if …?”
We want to initially set aside these “what ifs?” to focus on more “ordinary cases” (if there is such a thing) where abortion is considered, not cases like these. We should acknowledge though that even most people who call themselves “pro-life” think that abortion can be permissible if it is genuinely needed to save the woman’s life. This is because if she dies, then the fetus dies also, and so an abortion—which saves one life—would be more “pro-life” than allowing two deaths. We will return to the ethics of abortions due to rape at the end of the essay and briefly discuss the ethics and legality of rare abortions done later in pregnancy, far past the first trimester.
In reading this essay, we encourage trying to think about the issues with an “open mind.” What we mean is to try to consider and evaluate the arguments as if you didn’t already have strong views on the issue that you are committed to. (Maybe you are like this, which can be good: you shouldn’t have firm opinions on issues if you aren’t well-informed on them).
Critical thinking often involves defining words and giving and evaluating reasons: asking questions like “what do you mean?” and “why think that?” It involves stating arguments in their full pattern of reasoning and rigorously evaluating all premises. It involves identifying differing explanations of various moral and scientific facts and trying to determine which explanations are best. It involves thinking about thinking.
Most importantly though, good critical thinking isn’t done with an agenda or to support a point of view: it’s to find a point of view that’s worth supporting. Our perspectives on abortion didn’t develop (we hope!) with an “agenda” in mind beyond believing what’s supported by good arguments, and neither should yours . And views can and should change, in response to understanding better arguments, so our conclusions here are not “set in stone.” New arguments, including responses to the arguments presented here, might change our minds for the better —and the same should be true for all critical thinkers.
Let’s begin!


19.4: Landforms of Alpine Glaciation
A region altered by an alpine glacier is depicted in Figure \(\PageIndex{1}\). Refer to this diagram as you read about the various landforms created by alpine glaciers.
The headwaters of stream tributaries serve as the birth place for alpine glaciers. The headwaters of tributary valleys lie at the highest elevation in the drainage basin of a mountain stream. Here snow accumulates to great thickness and starts to move down valley. As it moves outward from the zone of accumulation, the glacier scours away at the valley sides. The material eroded from the surface is transported the length of the glacier to the zone of ablation, where it is deposited.
At the zone of accumulation where ice is accumulating, the glacier plucks rock from the head of the valley. The erosion creates a huge, amphitheater-like depression at the valley head called a cirque . After the glacier melts away a tarn , or cirque lake , may be found on the floor. Surrounding the floor are the massive curved side and headwalls of the cirque. A headwall crevasse known as a bergschrund forms where snow and compacted firn pull away from ice that remains frozen to the rock of the headwall. As alpine glaciers erode headward, they narrow the upland between ice sheets from opposite directions and a pyramid-shaped peak, called a horn is created. When two adjacent cirques on opposite sides of a divide cut back and remove part of it, a sharp-edge notch or pass called a col is formed.
The massive amount of ice that fills the main valley is from that which accumulates at the head of the valley plus that from all the tributary valleys that enter it. The enormous weight of the ice erodes away at the valley sides. The V-shape of the pre-existing stream valley now takes on the characteristic U-shaped of a glacial trough .
Prior to glaciation, tributary streams entered the main stream at the elevation of the main channel. But during glaciation, tributary glaciers feeding into the main valley are smaller and do not erode their valleys as deep as the glacier that occupies the main valley. Tributary valleys are left hanging at a higher elevation above the main valley floor as a result of more intensive downward cutting by the main valley glacier. These hanging valleys create spectacular waterfalls.
Glaciers in adjacent tributary valleys scour away at the upland between them (an interfluve). As erosion of the interfluve continues, it is narrows into a serrated ridge known as an arête . As mountain glaciers flow down valley they encounter exposed bed rock of varying resistance to erosion. The glacier will erode down into weaker rock but have to flow over the stronger rock. This creates a series of rock steps composed of the more resistant rocks with small depression behind them where the weaker rock is exposed. Often these depressions are occupied with water to form staircase lakes . Connected by a small stream these lakes are also know as paternoster lakes .
Glacial till is deposited along the valley sides as lateral moraine . Till is transported and deposited at the nose of the glacier as an end moraine . The end or terminal moraine marks the furthest advance of the glacier. Behind the terminal moraine are found recessional moraines indicating positions of the glacier front during times of retreat. When lateral moraines merge upon entering a main glacial trough, medial moraines are formed and run the length of the glacier.
Video : "Hanging Glaciers" (Courtesy of UBC Virtual Soil Learning Resources)


5.5: Graphs and Properties of Logarithmic Functions
In this section, you will:
- examine properties of logarithmic functions
- examine graphs of logarithmic functions
- examine the relationship between graphs of exponential and logarithmic functions
Recall that the exponential function \(f(x) = 2^x\) produces this table of values
| \(x\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| \(f(x)\) | 1/8 | 1/4 | 1/2 | 1 | 2 | 4 | 8 |
Since the logarithmic function is an inverse of the exponential, \(g(x)=\log_{2}(x)\) produces the table of values
| \(x\) | 1/8 | 1/4 | 1/2 | 1 | 2 | 4 | 8 |
| \(g(x)\) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
In this second table, notice that
- As the input increases, the output increases.
- As input increases, the output increases more slowly.
- Since the exponential function only outputs positive values, the logarithm can only accept positive values as inputs, so the domain of the log function is \((0, \infty)\).
- Since the exponential function can accept all real numbers as inputs, the logarithm can have any real number as output, so the range is all real numbers or \((-\infty, \infty)\).
Plotting the graph of \(g(x) = \log_{2}(x)\) from the points in the table , notice that as the input values for \(x\) approach zero, the output of the function grows very large in the negative direction, indicating a vertical asymptote at \(x = 0\).
In symbolic notation we write
as \(x \rightarrow 0^{+}\), \(f(x) \rightarrow-\infty\)
and as \(x \rightarrow \infty, f(x) \rightarrow \infty\)
Source: The material in this section of the textbook originates from David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, “ Chapter 4: Exponential and Logarithmic Functions ,” licensed under a Creative Commons CC BY-SA 3.0 license. The material here is based on material contained in that textbook but has been modified by Roberta Bloom, as permitted under this license.
Graphically, in the function \(g(x) = \log_{b}(x)\), \(b > 1\), we observe the following properties:
- The graph has a horizontal intercept at (1, 0)
- The line x = 0 (the y-axis) is a vertical asymptote; as \(x \rightarrow 0^{+}, y \rightarrow-\infty\)
- The graph is increasing if \(b > 1\)
- The domain of the function is \(x > 0\), or (0, \(\infty\))
- The range of the function is all real numbers, or \((-\infty, \infty)\)
However if the base \(b\) is less than 1, 0 < \(b\) < 1, then the graph appears as below.
This follows from the log property of reciprocal bases : \(\log _{1 / b} C=-\log _{b}(C)\)
- The graph has a horizontal intercept at (1, 0)
- The line x = 0 (the y-axis) is a vertical asymptote; as \(x \rightarrow 0^{+}, y \rightarrow \infty\)
- The graph is decreasing if 0 < \(b\) < 1
- The domain of the function is \(x\) > 0, or (0, \(\infty\))
- The range of the function is all real numbers, or \((-\infty, \infty)\)
When graphing a logarithmic function, it can be helpful to remember that the graph will pass through the points (1, 0) and (\(b\), 1).
Finally, we compare the graphs of \(y = b^x\) and \(y = \log_{b}(x)\), shown below on the same axes.
Because the functions are inverse functions of each other, for every specific ordered pair
(\(h\), \(k\)) on the graph of \(y = b^x\), we find the point (\(k\), \(h\)) with the coordinates reversed on the graph of \(y = \log_{b}(x)\).
In other words, if the point with \(x = h\) and \(y = k\) is on the graph of \(y = b^x\), then the point with \(x = k\) and \(y = h\) lies on the graph of \(y = \log_{b} (x)\)
The domain of \(y = b^x\) is the range of \(y = \log_{b} (x)\)
The range of \(y = b^x\) is the domain of \(y = \log_{b} (x)\)
For this reason, the graphs appear as reflections, or mirror images, of each other across the diagonal line \(y=x\). This is a property of graphs of inverse functions that students should recall from their study of inverse functions in their prerequisite algebra class.
| \(\bf{y = b^x}\) , with \(\bf{b>1}\) | \(\bf{y = \log_{b} (x)}\) , with \(\bf{b>1}\) | |
|---|---|---|
| Domain |
1}\)" valign="top" width="188" class="lt-math-38599">
all real numbers |
1}\)" valign="top" width="189" class="lt-math-38599">
all positive real numbers |
| Range |
1}\)" valign="top" width="188" class="lt-math-38599">
all positive real numbers |
1}\)" valign="top" width="189" class="lt-math-38599">
all real numbers |
| Intercepts |
1}\)" valign="top" width="188" class="lt-math-38599">
(0,1) |
1}\)" valign="top" width="189" class="lt-math-38599">
(1,0) |
| Asymptotes |
1}\)" valign="top" width="188" class="lt-math-38599">
Horizontal asymptote is
As \(x \rightarrow-\infty, \: y \rightarrow 0\) |
1}\)" valign="top" width="189" class="lt-math-38599">
Vertical asymptote is
As \(x \rightarrow 0^{+}, \: y \rightarrow-\infty\) |
Source: The material in this section of the textbook originates from David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, “ Chapter 4: Exponential and Logarithmic Functions ,” licensed under a Creative Commons CC BY-SA 3.0 license. The material here is based on material contained in that textbook but has been modified by Roberta Bloom, as permitted under this license.


2.2.4: Therapeutic Relationship
A therapeutic relationship means creating a positive alliance, or relationship, with a patient. Establishing a therapeutic relationship with a patient takes time, practice, and skill. It is important for a HHA/PCA to take the time to establish a good working relationship with their patient. It will make all the difference in their ability to provide good care to their patient and for their patient to feel good about the care they receive.
The relationship between a HHA/PCA and their patient/family is different from the one they have with their own family or friends. The relationship the HHA/PCA has with their own family and friends is reciprocal, or mutual, meaning both they and their family member share goals and work to meet each other’s needs. The relationship they have with their patients and their families is one where they focus on their patient’s needs, problems, and goals. While they may take on many roles with their patients and the patients may come to view the HHA/PCA as “family,” or as a “friend”, it is important to always remember that Home Health Aides/Personal Care Aides are there to care for the patient. The patients cannot work to help the HHA/PCA meet their own needs. This does not mean that Home Health Aides/Personal Care Aides are not friendly with their patients and their families. It just means that they must always focus on their patient and be aware of professional boundaries.
How to Establish a Therapeutic Relationship:
Trust: Especially during the first few interactions with a patient and their family, it is important for Home Health Aides/Personal Care Aides to demonstrate that they are trustworthy. This means showing up when they say they will, performing tasks and skills well, being honest, and demonstrating their concern for their patient’s well-being.
Setting boundaries: When meeting a patient and their family, Home Health Aides/Personal Care Aides should let them know who they are and what their role is. Tell them the tasks they can and cannot perform. For instance, they can say something like, “I am Karen. I will be your HHA/PCA. If it is okay with you, I will be coming here Monday, Wednesday, and Friday to help you bathe, dress, and eat. I am also here to be a support to you and your family.” If a patient/family asks a HHA/PCA to perform a task they are either not trained to do, is outside of their scope of practice, or that they are uncomfortable with, let them know, and inform their supervisor. They can say something to the effect of, “I‘m sorry, I am not able to put a catheter in you. I will let your nurse know that this is something you need help with.”
Confidentiality: It is important to let their patients know that they will protect their privacy. However, they should never tell a patient that they will keep something a secret, especially from their supervisor and the rest of the healthcare team. If a patient asks if they can tell their HHA/PCA a secret and that they not tell their nurse, they should politely let them know they cannot do this as their whole team is there to help care for them. They can say something like, “Mr. Thorpe, I cannot keep a secret from your nurse or doctor. They are here to help care for you. I do have to share what we do and talk about with my supervisor so that she knows you are properly being cared for. However, I will not share what you tell me with anyone outside of your health–care team if you do not wish.”
Empathy: Having empathy for another person means that you understand the thoughts and feelings of the other person. Empathy is different from sympathy. With sympathy, we feel sorry for another person. Home Health Aides/Personal Care Aides should be careful not to feel sorry for their patient as this often comes across as pity, or being condescending (looking down upon a person), which most people do not like. Sympathy is also more of an expression of one’s own feelings, while empathy shows one understands a patient‘s feelings. In expressing sympathy, we are showing agreement with the other person instead of helping them explore their feelings. Demonstrating that they are empathetic, or understanding, of their patient’s thoughts and feelings shows their patient that they care for them and are there to help them.
Scenario: Here is an example of the difference between sympathy and empathy. In this scenario a patient has just told their HHA/PCA that they have been diagnosed with cancer.
- Sympathetic response: “Oh! I am so sorry! That is terrible! I know exactly how you feel. My mother was diagnosed with cancer last year and it has been just awful.” You then go on to describe the situation with your mom and how upsetting it has been to deal with her diagnosis.
- Empathetic response: “How upsetting this must be for you. When my mom was diagnosed with cancer last year I had so many different emotions. What are your thoughts and feelings about your diagnosis?” You then sit with the patient and allow them the time and space to explore their thoughts and feelings. Here, the focus is on the patient and not on you.
Positive regard: Positive regard is respecting another person and demonstrating that you value them as a person, regardless of your differences, and whether or not you agree with their thoughts, feelings, or behaviors. We demonstrate our positive regard for someone by our verbal and nonverbal communication. What we say, how we say it, and our body language all demonstrate our respect for another person. If a patient does something that the HHA/PCA does not approve of or says something that is in disagreement with their own beliefs, it is important for them to convey to the patient that they are not judging them. Home Health Aides/Personal Care Aides should watch their tone of voice, facial expressions, and body language. For example, a patient may tell a HHA/PCA that they have decided to stop receiving life-saving treatments. While they may not agree with their decision, they must convey an understanding of and respect for their decision. “Mr. Wahlman, that must have been a difficult decision for you to make. How are you feeling about it?” This response demonstrates that they respect their right to make a decision and are interested in understanding how they feel about it.


9.2: Use a Problem Solving Strategy (Part 2)
Some number word problems ask you to find two or more numbers. It may be tempting to name them all with different variables, but so far we have only solved equations with one variable. We will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.
One number is five more than another. The sum of the numbers is twenty-one. Find the numbers.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | You are looking for two numbers. |
| Step 3. Name . Choose a variable to represent the first number. What do you know about the second number? Translate. |
Let n = 1 st number. One number is five more than another. x + 5 = 2 nd number |
| Step 4. Translate . Restate the problem as one sentence with all the important information. Translate into an equation. Substitute the variable expressions. |
The sum of the numbers is 21. The sum of the 1st number and the 2nd number is 21.
|
| Step 5. Solve the equation. | $$n + n + 5 = 21 \tag{9.1.15}$$ |
| Combine like terms. | $$2n + 5 = 21 \tag{9.1.16}$$ |
| Subtract five from both sides and simplify. | $$2n = 16 \tag{9.1.17}$$ |
| Divide by two and simplify. | $$n = 8 \qquad 1^{st}\; number \tag{9.1.18}$$ |
| Find the second number too. | $$n + 5 \qquad 2^{nd}\; number \tag{9.1.19}$$ |
| Substitute n = 8. | $$\textcolor{red}{8} + 5 \tag{9.1.20}$$ |
| Step 6. Check : Do these numbers check in the problem? Is one number 5 more than the other? Is thirteen, 5 more than 8? Yes. | $$\begin{split} 13 &\stackrel{?}{=} 8 + 5 \\ 13 &= 13\; \checkmark \end{split}$$ |
| Is the sum of the two numbers 21? | $$\begin{split} 8 + 13 &= 21 \\ 21 &= 21\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The numbers are 8 and 13. |
One number is six more than another. The sum of the numbers is twenty-four. Find the numbers.
- Answer
-
9, 15
The sum of two numbers is fifty-eight. One number is four more than the other. Find the numbers.
- Answer
-
27, 31
The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | two numbers |
| Step 3. Name . Choose a variable. What do you know about the second number? Translate. |
Let n = 1 st number One number is 4 less than the other. n - 4 = 2 nd number |
| Step 4. Translate . Write as one sentence. Translate into an equation. Substitute the variable expressions. |
The sum of two numbers is negative fourteen.
|
| Step 5. Solve the equation. | $$n + n - 4 = -14 \tag{9.1.21}$$ |
| Combine like terms. | $$2n - 4 = -14 \tag{9.1.22}$$ |
| Add 4 to each side and simplify. | $$2n = -10 \tag{9.1.23}$$ |
| Divide by 2. | $$n = -5 \qquad 1^{st}\; number \tag{9.1.24}$$ |
| Substitute n = −5 to find the 2 nd number. | $$\begin{split} n - &4 \qquad 2^{nd}\; number \\ \textcolor{red}{-5} - &4 \\ - &9 \end{split}$$ |
| Step 6. Check : Is −9 four less than −5? | $$\begin{split} -5 - 4 &\stackrel{?}{=} -9\\ -9&= -9\; \checkmark \end{split}$$ |
| Is their sum −14? | $$\begin{split} -5 + (-9) &\stackrel{?}{=} -14 \\ -14 &= -14\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The numbers are −5 and −9. |
The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.
- Answer
-
-8, -15
The sum of two numbers is negative eighteen. One number is 40 more than the other. Find the numbers.
- Answer
-
-29, 11
One number is ten more than twice another. Their sum is one. Find the numbers.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | two numbers |
| Step 3. Name . Choose a variable. One number is ten more than twice another. |
Let x = 1 st number 2x + 10 = 2 nd number |
| Step 4. Translate . Restate as one sentence. | Their sum is one. |
| Translate into an equation. | |
| Step 5. Solve the equation. | $$x + 2x + 10 = 1 \tag{9.1.25}$$ |
| Combine like terms. | $$3x + 10 = 1 \tag{9.1.26}$$ |
| Subtract 10 from each side. | $$3x = -9 \tag{9.1.27}$$ |
| Divide each side by 3 to get the first number. | $$x = -3 \tag{9.1.28}$$ |
| Substitute to get the second number. | $$\begin{split} 2x + &10 \\ 2(\textcolor{red}{-3}) + &10 \\ &4 \end{split}$$ |
| Step 6. Check : Is 4 ten more than twice −3? | $$\begin{split} 2(-3) + 10 &\stackrel{?}{=} 4 \\ -6 + 10 &= 4 \\ 4 &= 4\; \checkmark \end{split}$$ |
| Is their sum 1? | $$\begin{split} -3 + 4 &\stackrel{?}{=} 1 \\ 1 &= 1\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The numbers are −3 and 4. |
One number is eight more than twice another. Their sum is negative four. Find the numbers.
- Answer
-
-4, 0
One number is three more than three times another. Their sum is negative five. Find the numbers.
- Answer
-
-2, -3
Consecutive integers are integers that immediately follow each other. Some examples of consecutive integers are:
\[\ldots 1, 2, 3, 4, \ldots \tag{9.1.29}\]
\[\ldots -10, -9, -8, -7, \ldots \tag{9.1.30}\]
\[\ldots 150, 151, 152, 153, \ldots \tag{9.1.31}\]
Notice that each number is one more than the number preceding it. So if we define the first integer as n, the next consecutive integer is n + 1. The one after that is one more than n + 1, so it is n + 1 + 1, or n + 2.
\[\begin{split} n \qquad &1^{st}\; integer \\ n + 1 \qquad &2^{nd}\; consecutive\; integer \\ n + 2 \qquad &3^{rd}\; consecutive\; integer \end{split}\]
The sum of two consecutive integers is 47. Find the numbers.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | two consecutive integers |
| Step 3. Name . |
Let n = 1 st integer n + 1 = next consecutive integer |
| Step 4. Translate . Restate as one sentence. Translate into an equation. | |
| Step 5. Solve the equation. | $$n + n + 1 = 47 \tag{9.1.32}$$ |
| Combine like terms. | $$2n + 1 = 47 \tag{9.1.33}$$ |
| Subtract 1 from each side. | $$2n = 46 \tag{9.1.34}$$ |
| Divide each side by 2. | $$n = 23 \qquad 1^{st}\; integer \tag{9.1.35}$$ |
| Substitute to get the second number. | $$\begin{split} n + 1& \qquad 2^{nd}\; integer \\ \textcolor{red}{23} + 1& \\ 24& \end{split}$$ |
| Step 6. Check . | $$\begin{split} 23 + 24 &\stackrel{?}{=} 47 \\ 47 &= 47\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The two consecutive integers are 23 and 24. |
The sum of two consecutive integers is 95. Find the numbers.
- Answer
-
47, 48
The sum of two consecutive integers is −31. Find the numbers.
- Answer
-
-15, -16
Find three consecutive integers whose sum is 42.
Solution
| Step 1. Read the problem. | |
| Step 2. Identify what you are looking for. | three consecutive integers |
| Step 3. Name . |
Let n = 1 st integer n + 1 = 2 nd consecutive integer n + 2 = 3 rd consecutive integer |
| Step 4. Translate . Restate as one sentence. Translate into an equation. | |
| Step 5. Solve the equation. | $$n + n + 1 + n + 2 = 42 \tag{9.1.36}$$ |
| Combine like terms. | $$3n + 3 = 42 \tag{9.1.37}$$ |
| Subtract 3 from each side. | $$3n = 39 \tag{9.1.38}$$ |
| Divide each side by 3. | $$n = 13 \qquad 1^{st}\; integer \tag{9.1.39}$$ |
| Substitute to get the second number. | $$\begin{split} n + 1& \qquad 2^{nd}\; integer \\ \textcolor{red}{13} + 1& \\ 24& \end{split}$$ |
| Substitute to get the third number. | $$\begin{split} n + 2& \qquad 3^{rd}\; integer \\ \textcolor{red}{13} + 2& \\ 15& \end{split}$$ |
| Step 6. Check . | $$\begin{split} 13 + 14 + 15 &\stackrel{?}{=} 42 \\ 42 &= 42\; \checkmark \end{split}$$ |
| Step 7. Answer the question. | The three consecutive integers are 13, 14, and 15. |
Find three consecutive integers whose sum is 96.
- Answer
-
31, 32, 33
Find three consecutive integers whose sum is −36.
- Answer
-
-11, -12, -13
Practice Makes Perfect
Use a Problem-solving Strategy for Word Problems
In the following exercises, use the problem-solving strategy for word problems to solve. Answer in complete sentences.
- Two-thirds of the children in the fourth-grade class are girls. If there are 20 girls, what is the total number of children in the class?
- Three-fifths of the members of the school choir are women. If there are 24 women, what is the total number of choir members?
- Zachary has 25 country music CDs, which is one-fifth of his CD collection. How many CDs does Zachary have?
- One-fourth of the candies in a bag of are red. If there are 23 red candies, how many candies are in the bag?
- There are 16 girls in a school club. The number of girls is 4 more than twice the number of boys. Find the number of boys in the club.
- There are 18 Cub Scouts in Troop 645. The number of scouts is 3 more than five times the number of adult leaders. Find the number of adult leaders.
- Lee is emptying dishes and glasses from the dishwasher. The number of dishes is 8 less than the number of glasses. If there are 9 dishes, what is the number of glasses?
- The number of puppies in the pet store window is twelve less than the number of dogs in the store. If there are 6 puppies in the window, what is the number of dogs in the store?
- After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
- Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original weekly salary?
- Tim left a $9 tip for a $50 restaurant bill. What percent tip did he leave?
- Rashid left a $15 tip for a $75 restaurant bill. What percent tip did he leave?
- Yuki bought a dress on sale for $72. The sale price was 60% of the original price. What was the original price of the dress?
- Kim bought a pair of shoes on sale for $40.50. The sale price was 45% of the original price. What was the original price of the shoes?
Solve Number Problems
In the following exercises, solve each number word problem.
- The sum of a number and eight is 12. Find the number.
- The sum of a number and nine is 17. Find the number.
- The difference of a number and twelve is 3. Find the number.
- The difference of a number and eight is 4. Find the number.
- The sum of three times a number and eight is 23. Find the number.
- The sum of twice a number and six is 14. Find the number.
- The difference of twice a number and seven is 17. Find the number.
- The difference of four times a number and seven is 21. Find the number.
- Three times the sum of a number and nine is 12. Find the number.
- Six times the sum of a number and eight is 30. Find the number.
- One number is six more than the other. Their sum is forty-two. Find the numbers.
- One number is five more than the other. Their sum is thirty-three. Find the numbers.
- The sum of two numbers is twenty. One number is four less than the other. Find the numbers.
- The sum of two numbers is twenty-seven. One number is seven less than the other. Find the numbers.
- A number is one more than twice another number. Their sum is negative five. Find the numbers.
- One number is six more than five times another. Their sum is six. Find the numbers.
- The sum of two numbers is fourteen. One number is two less than three times the other. Find the numbers.
- The sum of two numbers is zero. One number is nine less than twice the other. Find the numbers.
- One number is fourteen less than another. If their sum is increased by seven, the result is 85. Find the numbers.
- One number is eleven less than another. If their sum is increased by eight, the result is 71. Find the numbers.
- The sum of two consecutive integers is 77. Find the integers.
- The sum of two consecutive integers is 89. Find the integers.
- The sum of two consecutive integers is −23. Find the integers.
- The sum of two consecutive integers is −37. Find the integers.
- The sum of three consecutive integers is 78. Find the integers.
- The sum of three consecutive integers is 60. Find the integers.
- Find three consecutive integers whose sum is −36.
- Find three consecutive integers whose sum is −3.
Everyday Math
- Shopping Patty paid $35 for a purse on sale for $10 off the original price. What was the original price of the purse?
- Shopping Travis bought a pair of boots on sale for $25 off the original price. He paid $60 for the boots. What was the original price of the boots?
- Shopping Minh spent $6.25 on 5 sticker books to give his nephews. Find the cost of each sticker book.
- Shopping Alicia bought a package of 8 peaches for $3.20. Find the cost of each peach.
- Shopping Tom paid $1,166.40 for a new refrigerator, including $86.40 tax. What was the price of the refrigerator before tax?
- Shopping Kenji paid $2,279 for a new living room set, including $129 tax. What was the price of the living room set before tax?
Writing Exercises
- Write a few sentences about your thoughts and opinions of word problems. Are these thoughts positive, negative, or neutral? If they are negative, how might you change your way of thinking in order to do better?
- When you start to solve a word problem, how do you decide what to let the variable represent?
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
Contributors and Attributions
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Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


5: Plate Tectonics
Identify and compare common rock forming minerals, mineral structures and mineral properties.
- Describe the basic chemistry involved in mineral formation and structures.
- Identify and classify common rock forming minerals.
- Identify the most common elements in the Earth’s crust and their order of abundance.
Plate tectonics is often seen as the missing piece of the puzzle for geologists. Plate tectonics explains, either directly or indirectly, just about every topic discussed in geology. It is the glue that binds everything together.
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- 5.3: Basics of Plate Tectonics
- The theory of plate tectonics revolutionized geology in the 1960s. By 1970, college geology majors were taught a set of ideas that were unheard of by most geology students prior to 1960. The foundation for the new way of comprehending earth processes is the understanding that the outer layer of the earth is the lithosphere rather than simply the crust.
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- 5.7: Theory of Plate Tectonics
- lates of lithosphere move because of convection currents in the mantle. One type of motion is produced by seafloor spreading. Plate boundaries can be located by outlining earthquake epicenters. Plates interact at three types of plate boundaries: divergent, convergent and transform. Most of the Earth’s geologic activity takes place at plate boundaries. At a divergent boundary, volcanic activity produces a mid ocean ridge and small earthquakes.
Thumbnail: Convergent boundary . (CC BY 4.0; ).
Contributors and Attributions
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Original content from Kimberly Schulte (Columbia Basin College) and supplemented by Lumen Learning . The content on this page is copyrighted under a Creative Commons Attribution 4.0 International license.


14.4: The Dividend Discount Model (DDM)- Fixed Dividend or No- Growth Version
W e shall assume that the security’s Intrinsic Value (V) will be equal to the market price (P), i.e., that P = V. Intrinsic value refers to what the security is “ truly ” worth. To the extent that P ≠ V, one would be presented with the opportunity to earn an extraordinary profit (by either buying cheaply or selling “richly”) . Hereforward , we shall use P rather than V, although y ou may see V used elsewhere.
A stock is a kind of perpetuity ; the corporation, as a going concern, is eternal. Dividends will be paid in perpetuity. Unlike bonds, there is no face value to be paid at a specified time in the future. The only cash flows are the dividends.
This formula says that the price of a stock is equal to its aggregated discounted future cash flows, i.e., the dividends discounted to present value. Since, for the moment we assume that dividends are fixed (as in a preferred stock), the foregoing, never-ending equation may be simplified algebraically to the following (as was demonstrated for a perpetuity, only the terms have been modified to accommodate stock):
This simplification works because of the “Law of Limits,” discussed earlier in our TVM section. Interestingly, by transposing, the above formula may be re-formulated as:
Note : D ÷P is the Dividend Yield !
While corporations pay dividends quarterly, the formula looks at annual dividends. Recall that, in the case of a perpetuity, discounting frequency has no effect on Present Value.