7.5: Energy- "Time Part" of Momenergy
energy has two partis: rest energy (= mass) plus kinetic energy
What about the "time part" of the momentum-energy of a particle-the part we have called its energy? This is certainly a strange-looking beast! As measured in a particular free-float frame, say the laboratory, this time component as given in equation (7-5) is
Relativistic expression for energy
\[E=m \frac{d t}{d \tau}=\frac{m}{\left(1-v^{2}\right)^{1 / 2}}=m \gamma\]
Compare this with the Newtonian expression for kinetic energy, using \(K\) as the symbol for kinetic energy:
\[K_{\text {Newton }}=\frac{1}{2} m v^{2} \quad \text { [valid for low speed] } \quad \]
How does the relativistic expression for energy, equation \(\PageIndex{1}\), compare with the Newtonian expression for kinetic energy \(\PageIndex{2}\)? To answer this question, first look at the behavior of these two expressions when particle speed equals zero. The Newtonian kinetic energy goes to zero. In contrast, at zero speed \(1 /\left(1-v^{2}\right)^{1 / 2}=1\) and the relativistic value for energy becomes equal to mass of the particle,
Rest energy of a particle equals its mass
\[E_{\text {rest }}=m\]
where \(E_{\text {rest }}\) is called rest energy of the particle. Rest energy of a particle is simply its mass. So the relativistic expression for energy does not go to zero at zero speed, while the Newtonian expression for kinetic energy does go to zero.
Is this an irreconcilable difference? The Newtonian formula does not contain an expression for rest energy, equal to the mass of the particle. But here is the distinction: The relativistic expression gives the value for total energy of the particle, while the Newtonian expression describes kinetic energy only (valid for low speed). However, in Newtonian mechanics any constant potential energy whatever can be added to the energy of a particle without changing the laws that describe its motion. One may think of the zero-speed limit of the relativistic expression for energy as providing this previously undetermined constant.
When we refer to energy of a particle we ordinarily mean total energy of the particle. As measured in a frame in which the particle is at rest, this total energy equals rest energy, the mass of the particle. As measured from frames in which the particle moves, total energy includes not only rest energy but also kinetic energy.
This leads us to define kinetic energy of a particle as energy above and beyond its rest energy:
Kinetic energy defined
\[(\text { energy })=(\text { rest energy })+(\text { kinetic energy }) \nonumber \]
or
\[E=m+K\]
MOTION IN THE \(X-D I R E C T I O N\)
An object of mass 3 kilograms moves 8 meters along the \(x\) -direction in 10 meters of time as measured in the laboratory. What is its energy and momentum? Its rest energy? Its kinetic energy? What value of kinetic energy would Newton predict for this object? Using relativistic expressions, verify that the velocity of this object equals its momentum divided by its energy.
Solution
From the statement of the problem:
\[\begin{aligned} m &=3 \text { kilograms } \\ t &=10 \text { meters } \\ x &=8 \text { meters } \\ y &=0 \text { meters } \\ z &=0 \text { meters } \end{aligned}\]
From this we obtain a value for the speed:
\[v=\frac{x}{t}=\frac{8 \text { meters of distance }}{10 \text { meters of time }}=0.8\]
Use \(v\) to calculate the factor \(1 /\left(1-v^{2}\right)^{1 / 2}\) in equation (7-8):
\[\frac{1}{\left(1-v^{2}\right)^{1 / 2}}=\frac{1}{\left(1-(0.8)^{2}\right)^{1 / 2}}=\frac{1}{(1-0.64)^{1 / 2}}=\frac{1}{(0.36)^{1 / 2}}=\frac{1}{0.6}=\frac{5}{3}\]
From equation (7-11) the energy is
\[E=m /\left(1-v^{2}\right)^{1 / 2}=(3 \text { kilograms })(5 / 3)=5 \text { kilograms }\]
From equation (7-8) momentum has the magnitude
\[p=m v /\left(1-v^{2}\right)^{1 / 2}=(5 / 3) \times(3 \text { kilograms }) \times 0.8=4 \text { kilograms }\]
Rest energy of the particle just equals its mass:
\[E_{\text {rest }}=m=3 \text { kilograms }\]
From equation (7-15) kinetic energy \(K\) equals total energy minus rest energy:
\[K=E-m=5 \text { kilograms }-3 \text { kilograms }=2 \text { kilograms }\]
The Newtonian prediction for kinetic energy is
\[K_{\text {Newton }}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 3 \times(0.8)^{2}=0.96 \text { kilogram }\]
which is a lot smaller than the correct relativistic result. Even at the speed of light, the Newtonian prediction would be \(K_{\text {Newton }}=1.5\) kilogram, whereas relativistic value would increase without limit.
Equation (7-16) says that velocity equals the ratio (magnitude of momentum) \(/(\) energy):
\[v=\frac{p}{E}=\frac{4 \text { kilograms }}{5 \text { kilograms }}=0.8\]
This is the same value as reckoned directly from the given quantities.
From this comes the relativistic expression for kinetic energy \(K\) :
\[K=E-E_{\text {rest }}=E-m=\frac{m}{\left(1-v^{2}\right)^{1 / 2}}-m=m\left[\frac{1}{\left(1-v^{2}\right)^{1 / 2}}-1\right] \quad(7-15)\]
Box 7-2 elaborates the relation between this expression and the Newtonian expression (7.5.2). Notice that if we divide the respective sides of the momentum equation (7-8) by corresponding sides of the energy equation (7.5.1), the result gives particle speed:
\[v=\frac{p}{E}\]
We could have predicted this directly from the first figure in this chapter, Figure 7.1.1. Speed \(v\) is the tilt (slope) of the worldline from the vertical: (space displacement) /(time for this displacement). Momenergy points along the worldline, with space component \(p\) and time component \(E\) . Therefore momenergy slope \(p / E\) equals worldline slope \(v\) .
Conversion to conventional energy units
Still More Units : In order to convert energy in units of mass to energy in conventional units, such as joules, multiply the expressions above by the square of light speed, \(c^{2}\) , and use subscript "conv":
\[\begin{gathered} E_{\text {conv }}=E c^{2}=\frac{m c^{2}}{\left[1-\left(v_{\text {conv }} / c\right)^{2}\right]^{1 / 2}} \quad \text { [good at any speed] } \quad \text { (7-17) } \\ K_{\text {conv }}=\left(E-E_{\text {rest }}\right) c^{2}=m c^{2}\left[\frac{E_{\text {conv rest }}=m c^{2}}{\left[1-\left(v_{\text {conv }} / c\right)^{2}\right]^{1 / 2}}-1\right] \quad \text { [good at any speed] (7-19) } \quad \text { Conversion to conventional } \\ K_{\text {conv Newton }}=\frac{1}{2} m v^{2} c^{2}=\frac{1}{2} m\left(\frac{v_{\text {conv }}}{c}\right)^{2} c^{2}=\frac{1}{2} m v_{\text {conv }}^{2} \quad \text { [low speed only] } \text { (7-20) } \end{gathered}\]
Thus conversion from energy in units of mass to energy in conventional units is always accomplished by multiplying by conversion factor \(c^{2}\) . This is true whether the expression for energy being converted is Newtonian or relativistic. Table \(7-1\) at the end of the chapter summarizes these comparisons.
Equation (7-18) is the most famous equation in all physics. Historically, the factor \(c^{2}\) captured the public imagination because it witnessed to the vast store of energy available in the conversion of even tiny amounts of mass to heat and radiation. The units of \(m c^{2}\) are joules; the units of \(m\) are kilograms. However, we now recognize that joules and kilograms are units different only because of historical accident. The conversion factor \(c^{2}\) , like the factor of conversion from seconds to meters or miles to feet, can today be counted as a detail of convention rather than as a deep new principle.
For each of the following cases, write down the vector in the given frame in the form \(\left[E, t_{x} t_{j} t_{z}\right] .\) four components of the momentum-energy 4- Each particle has mass \(m\) .
a. A particle moves in the positive \(x\) -direction in the laboratory with kinetic energy equal to three times its rest energy.
b. The same particle is observed in a rocket in which its kinetic energy equals its mass.
c. Another particle moves in the \(y\) -direction in the laboratory frame with momentum equal to twice its mass.
d. Yet another particle moves in the negative \(x\) -direction in the laboratory with total energy equal to four times its mass.
e. Still another particle moves with equal \(x, y\) , and \(z\) momentum components in the laboratory and kinetic energy equal to four times its rest energy.
Solution
a. Total energy of the particle equals rest energy \(m\) plus kinetic energy \(3 m\) . Therefore its total energy \(E\) equals \(E=m+3 m=4 m\) . The particle moves along the \(x\) -direction, so \(p_{y}=p_{z}=0\) and \(p_{x}=p\) , the total momentum. Substitute the value of \(E\) into the equation \(m^{2}=E^{2}-p^{2}\) to obtain
\[p^{2}=E^{2}-m^{2}=(4 m)^{2}-m^{2}=16 m^{2}-m^{2}=15 m^{2}\]
Hence \(p_{x}=(15)^{1 / 2} \mathrm{~m}\) .
In summary, the components of the momenergy 4-vector are
\[\left[E, p_{x}, p_{y}, p_{z}\right]=\left[4 m,(15)^{1 / 2} m, 0,0\right]\]
Of course the magnitude of this momenergy 4-vector equals the mass of the particle \(m\) - true whatever its speed, its energy, or its momentum.
b. In this rocket frame, total energy - rest energy plus kinetic energy-has the value \(E=2 m .\) As before, \(p^{2}=E^{2}-m^{2}=(2 m)^{2}-m^{2}=4 m^{2}-m^{2}=3 m^{2}\) . Hence \(p_{x}=3^{1 / 2} m\) and components of the 4-vector are \(\left[E, p_{x}, p_{y}, p_{z}\right]=\left[2 m, 3^{1 / 2} m,\right.\) , \(0,0]\) .
c. In this case \(p_{x}=p_{z}=0\) and \(p_{y}=p=2 m\) . Moreover, \(E^{2}=m^{2}+p^{2}=m^{2}+(2 m)^{2}\) \(=5 m^{2} .\) So, finally, \(\left[E, p_{x}, p_{y}, p_{z}\right]=\left[5^{1 / 2} m, 0,2 m, 0\right]\) .
d. We are given directly that \(E=4 m\) , the same as in part a, except here the particle travels in the negative \(x\) -direction so has negative \(x\) -momentum. Hence:
\[\left[E, p_{x}, p_{y}, p_{z}\right]=\left[4 m,-(15)^{1 / 2} m, 0,0\right]\]
e. Total energy equals \(E=5 m\) . All momentum components have equal value, say
\[p_{x}=p_{y}=p_{z}=P\]
In this case we use the full equation that relates energy, momentum, and mass:
\[\begin{gathered} \left(p_{x}\right)^{2}+\left(p_{y}\right)^{2}+\left(p_{z}\right)^{2}=3 P^{2}=E^{2}-m^{2}=(5 m)^{2}-m^{2}=24 m^{2} \\ \text { or } P^{2}=8 m^{2} \text { and hence }\left[E, p_{x}, p_{y}, p_{z}\right]=\left[5 m, 8^{1 / 2} m, 8^{1 / 2} m, 8^{1 / 2} m\right] . \end{gathered}\]
Energy at relativistic speeds and energy at everyday speeds: How are expressions for these two cases related?
Energy in Terms of Momentum: In the limit of velocities low compared with the speed of light, the relativistically accurate expression for energy \(E=\left(m^{2}\right.\) \(\left.+p^{2}\right)^{1 / 2}\) reduces to \(E=m+p^{2} /(2 m)+\) corrections. To see why and how, and to estimate the corrections, it is convenient to work in dimensionless ratios. Thus we focus on the accurate expression in the form \(E / m=\left[1+(p / m)^{2}\right]^{1 / 2}\) , or even simpler, \(y=[1+x]^{1 / 2}\) , and on the approximation to this result, in the form
\[E / m=1+(1 / 2)(p / m)^{2}+\text { corrections, or } y=1+(1 / 2) x+\text { corrections }\]
Example: \(x=0.21 .\) Then our approximation formula gives \(y=(1.21)^{1 / 2}=1\) \(+0.105+\) a correction. The accurate result is \(y=1.100\) , which is the square root of \(1.21\) . In other words, the correction is negative and extremely small: correction \(=-0.005 .\)
Energy in Terms of Velocity: In the limit of velocities low compared with the speed of light, the relativistically accurate expression for energy \(E=m /(1-\) \(\left.v^{2}\right)^{1 / 2}\) , reduces to \(E=m+(1 / 2) m v^{2}+\) corrections. It is convenient again to work in dimensionless ratios. Thus we focus on the accurate expression in the form \(E / m=\left[1-v^{2}\right]^{-1 / 2}\) , or even simpler, \(y=[1-x]^{-1 / 2}\) , and on the approximation to this result, in the form
\[E / m=1+(1 / 2) v^{2}+\text { corrections, or } y=1+(1 / 2) x+\text { corrections }\]
Example: \(x=0.19 .\) Then our approximation formula gives \(y=1+(1 / 2)\)
\(0.19+\) a correction \(=1.095+\) a correction. The accurate result is \(y=[1-\) \(0.19]^{-1 / 2}=(0.81)^{-1 / 2}=(0.9)^{-1}=1.1111 . . .\) In other words, the correction is positive and small: correction \(=+0.01611\) .
Another example: A jet plane. Take its speed to be exactly \(v=10^{-6}\) . That speed, according to our approximation, brings with it a fractional augmentation of energy, a kinetic energy per unit mass, equal to \((1 / 2) v^{2}=5 X\) \(10^{-13}\) or \(0.0000000000005 .\) In contrast, the accurate expression \(E / \mathrm{m}=\) \(\left[1-v^{2}\right]^{-1 / 2}\) gives the result \(E / m=1.000000000000500000000009375\) \(000000000 \ldots .\) The 5 a little less than halfway down the length of this string of digits is no trifle, as anyone will testify who has seen the consequences of the crash of a jet plane into a skyscraper. However, the 9375 further down the line is approximately a million million times smaller and totally negligible in its practical consequences.
In brief, low speed gives rise to a kinetic energy which, relative to the mass, is given to good approximation by \((1 / 2) v^{2}\) or by \((1 / 2)(p / m)^{2}\) . Moreover, the same one or other unit-free number la "fraction" because it is small compared to unity) automatically reveals to us the order of magnitude of the fractional correction we would have had to make in this fraction itself if we were to have insisted on a perfectly accurate figure for the kinetic energy.
Energy: Time part of momenergy 4-vector Mass: Magnitude of that 4-vector
Central to an understanding of the equation \(E_{\text {rest }}=m\) or its equivalent \(E_{\text {conv rest }}=m c^{2}\) is the subscript "rest." Energy is not the same as mass! Energy is only the time part of the momenergy 4 -vector. Mass is the magnitude of that 4 -vector. The energy of an object, expressed in conventional units, has the value \(m c^{2}\) only when that object is observed from a frame in which it is at rest. Observed from all other free-float frames, the energy of the object is greater than its rest energy, as shown by equation \((7-17)\) .
Figure \(\PageIndex{1}\) compares relativistic and Newtonian predictions for kinetic energy per unit mass as a function of speed. At low speeds the values are indistinguishable (left side of the graph). When a particle moves with high speed, however, so that the factor \(1 /\left(1-v^{2}\right)^{1 / 2}\) has a value much greater than one, relativistic and Newtonian expressions do not yield at all the same value for kinetic energy (right side of the graph). Then one must choose which expression to use in analyzing collisions and other high-speed phenomena. We choose the relativistic expression because it leads to the same value of the total energy of an isolated system before and after any interaction between particles in the system - it leads to conservation of total energy of the system.
Relativity: All forms of energy automatically conserved
All this talk of reconciliation at low speeds obscures an immensely powerful feature of the relativistic expression for total energy of an isolated system of particles. Total energy is conserved in all interactions among particles in the system: elastic and inelastic collisions as well as creations, transformations, decays, and annihilations of particles. In contrast, total kinetic energy of a system calculated using the Newtonian formula for low-speed interactions is conserved only for elastic collisions. Elastic collisions are defined as collisions in which kinetic energy is conserved. In collisions that are not elastic, kinetic energy transforms into heat energy, chemical energy, potential energy, or other forms of energy. For Newtonian mechanics of low-speed particles, each of these forms of energy must be treated separately: Conservation of energy must be invoked as a separate principle, as something beyond Newtonian analysis of mechanical energy.
In relativity, all these energies are included automatically in the single time component of total momenergy of a system - total energy - which is always conserved for an isolated system. Chapter 8 discusses more fully the momenergy of a system of particles and the effects of interactions between particles on the energy and mass of the system.


There are several free or low-cost third-party assessment tools available over the Internet:
-
The Discovery School website “offers teachers of all subjects and array of powerful tools” for assessment (school.discovery.com/teachingtools/teachingto ols.html). Use Puzzlemaker to generate crossword puzzles, word searches, and math squares. Visit the Quiz Center to create and give quizzes. Try the Worksheet Generator to create custom worksheets for your course materials.
-
Higher education and K–12 instructors use Quia (‘key- ah) “to create customized educational software online, built around their own course materials and made available to students over the Web” (
http://www.quia.com/company/quia_web.html
). Quia is a subscription-based service.
-
Half Baked Software, Inc., created Hot Potatoes, a set of applications that allow instructors “to create interactive multiple-choice, short-answer, jumbled-sentence, crossword, matching/ ordering and gap-fill exercises for the World Wide Web” (
http://hotpot.uvic.ca/
). You can also include MP3 audio files and math symbols as part of these assessment activities. Hot Potatoes requires a licensing fee, unless you work for a publicly funded, non-profit educational institution.
-
QuizStar is a free web-based tool for K–12 instructors to create and assign quizzes, manage student results, and allow students “to review the results for further learning” (
http://quizstar.4teachers.org/
).
Some publishers offer assessment tools that accompany textbook activities. For example, Glencoe Online Mathematics provides an Online Study Tools site (www.glencoe.com/sec/math/studytools/index.ph p4).


4.4: Double-Slit Diffraction
-
- Last updated
- Save as PDF
Learning Objectives
By the end of this section, you will be able to:
- Describe the combined effect of interference and diffraction with two slits, each with finite width
- Determine the relative intensities of interference fringes within a diffraction pattern
- Identify missing orders, if any
When we studied interference in Young’s double-slit experiment, we ignored the diffraction effect in each slit. We assumed that the slits were so narrow that on the screen you saw only the interference of light from just two point sources. If the slit is smaller than the wavelength, then Figure 4.3.4a shows that there is just a spreading of light and no peaks or troughs on the screen. Therefore, it was reasonable to leave out the diffraction effect in that chapter. However, if you make the slit wider, Figure 4.3.4b and (c) show that you cannot ignore diffraction. In this section, we study the complications to the double-slit experiment that arise when you also need to take into account the diffraction effect of each slit.
To calculate the diffraction pattern for two (or any number of) slits, we need to generalize the method we just used for a single slit. That is, across each slit, we place a uniform distribution of point sources that radiate Huygens wavelets , and then we sum the wavelets from all the slits. This gives the intensity at any point on the screen. Although the details of that calculation can be complicated, the final result is quite simple:
Two-Slit Diffraction Pattern
The diffraction pattern of two slits of width \(a\) that are separated by a distance d is the interference pattern of two point sources separated by d multiplied by the diffraction pattern of a slit of width \(a\).
In other words, the locations of the interference fringes are given by the equation
\[d \, \sin \, \theta = m \lambda \nonumber \]
the same as when we considered the slits to be point sources, but the intensities of the fringes are now reduced by diffraction effects, according to Equation 4.3.11 . [Note that in the chapter on interference, we wrote \(d \, sin \, \theta = m \lambda\) and used the integer \(m\) to refer to interference fringes. Equation 4.2.1 also uses \(m\), but this time to refer to diffraction minima. If both equations are used simultaneously, it is good practice to use a different variable (such as \(n\)) for one of these integers in order to keep them distinct.]
Interference and diffraction effects operate simultaneously and generally produce minima at different angles. This gives rise to a complicated pattern on the screen, in which some of the maxima of interference from the two slits are missing if the maximum of the interference is in the same direction as the minimum of the diffraction. We refer to such a missing peak as a missing order . One example of a diffraction pattern on the screen is shown in Figure \(\PageIndex{1}\). The solid line with multiple peaks of various heights is the intensity observed on the screen. It is a product of the interference pattern of waves from separate slits and the diffraction of waves from within one slit.
Example \(\PageIndex{1}\): Intensity of the Fringes
Figure \(\PageIndex{1}\) shows that the intensity of the fringe for m=3 is zero, but what about the other fringes? Calculate the intensity for the fringe at m=1 relative to \(I_0\), the intensity of the central peak.
Strategy
Determine the angle for the double-slit interference fringe, using the equation from Interference , then determine the relative intensity in that direction due to diffraction by using Equation 4.3.11 .
Solution
From the chapter on interference, we know that the bright interference fringes occur at \(d \, sin \, \theta = m \lambda\), or
\[\sin \, \theta = \dfrac{m\lambda}{d}. \nonumber \]
From Equation 4.3.11 ,
\[I = I_0 \left(\dfrac{\sin \, \beta}{\beta}\right)^2 \nonumber \]
where
\[\beta = \dfrac{\phi}{2} = \dfrac{\pi a \sin \, \theta}{\lambda}. \nonumber \]
Substituting from above,
\[\beta = \dfrac{\pi a \sin \, \theta}{\lambda} = \dfrac{\pi a}{\lambda} \cdot \dfrac{m\lambda}{d} = \dfrac{m\pi a}{d}. \nonumber \]
For \(a = 2\lambda, \, d = 6 \lambda\), and \(m = 1\),
\[\beta = \dfrac{(1)\pi (2\lambda)}{(6 \lambda)} = \dfrac{\pi}{3}. \nonumber \]
Then, the intensity is
\[I = I_0 \left(\dfrac{\sin \, \beta}{\beta}\right)^2 = I_0 \left(\dfrac{sin \, (\pi/3)}{\pi/3}\right)^2 = 0.684 I_0. \nonumber \]
Significance
Note that this approach is relatively straightforward and gives a result that is almost exactly the same as the more complicated analysis using phasors to work out the intensity values of the double-slit interference (thin line in Figure \(\PageIndex{1}\)). The phasor approach accounts for the downward slope in the diffraction intensity (blue line) so that the peak near m=1 occurs at a value of θ ever so slightly smaller than we have shown here.
Example \(\PageIndex{2}\): Two-Slit Diffraction
Suppose that in Young’s experiment, slits of width 0.020 mm are separated by 0.20 mm. If the slits are illuminated by monochromatic light of wavelength 500 nm, how many bright fringes are observed in the central peak of the diffraction pattern?
Solution
From Equation 4.2.1 , the angular position of the first diffraction minimum is \(\theta \approx sin \, \theta = \dfrac{\lambda}{a} = \dfrac{5.0 \times 10^{-7}m}{2.0 \times 10^{-5}m} = 2.5 \times 10^{-2} rad\).
Using \(d \sin \, \theta = m \lambda\) for \(\theta = 2.5 \times 10^{-2} rad\), we find
\[m = \dfrac{d \, sin \, \theta}{\lambda} = \dfrac{(0.20 \, mm)(2.5 \times 10^{-2} rad)}{(5.0 \times 10^{-7}m)} = 10, \nonumber \]
which is the maximum interference order that fits inside the central peak. We note that \(m=±10\) are missing orders as \(θ\) matches exactly.
Exercise \(\PageIndex{1}\)
For the experiment in Example \(\PageIndex{2}\), show that m=20 is also a missing order.
Solution
From \(d \, \sin \, \theta = m\lambda\), the interference maximum occurs at \(2.87^o\) for \(m = 20\). From Equation 4.2.1 , this is also the angle for the second diffraction minimum. ( Note: Both equations use the index m but they refer to separate phenomena.)
Explore the effects of double-slit diffraction. In this simulation written by Fu-Kwun Hwang, select N=2 using the slider and see what happens when you control the slit width, slit separation and the wavelength. Can you make an order go “missing?”


3.8: The subject field
As we covered in Chapter 1, when beginning your research, you need to think about your research question and pull out the most important keywords to use in your search. These may be technical terms from a certain field of study or they may be general words that describe a topic or concept. Combining these terms to create a custom search is how most of us start our research.
Here are some examples:
- Forever 21 clothing sweatshops scandal
- 20th century art from the United States
- future of electric cars
- job interview etiquette
These sorts of searches work well in a variety of tools (Google, indexes, Quick Search). However, when you need to find a comprehensive set of resources that are all about the same topic, you should make the most of features built into your finding tool, like subject searching.
Subject searching is a type of field search that works a little differently because subject fields work a little differently. Subject fields contain words that describe topics, but these terms are not randomly selected. They come from a preset list put in place to be used with the finding tool. The terms from these lists are called controlled vocabulary . Using standard terms for subjects lets researchers more easily find resources about their desired topic, even though authors of different resources may use different terms for a topic.
Many indexes call their subject headings by different names such as subjects, categories, subject terms or descriptors. Here are some examples of subject headings you might encounter in scholarly indexes, similar to those described above in the keyword searches:
- Clothing trade–corrupt practices
- Art, modern–20th century–United States–Exhibitions
- Hybrid electric cars
- Employment interviewing
When subject searching, you may encounter terms that seem outdated or even offensive. For instance, categorizations related to identities in the LGBTQIA+ community, racial and ethnic groups, and educational ability have historically been problematic. This is one of the negative consequences of having a standardized system for categorizing topics, as it’s often hard to change the labels once they’ve been established. People are working towards change to address these problems, but these older terms may still remain in your search results. This doesn’t mean that the finding tool or library that uses these terms isn’t trying to change them.
Subject searching
As discussed earlier in this chapter, advanced search features typically allow you to search for your keywords in specific areas of the item record (e.g., author, date, type, format). Searching for a specific subject heading is not the best way of using controlled vocabulary, though. After all, most of the time, you won’t know what words a specific finding tool uses for its subjects. Below are two effective ways of finding subject headings for your search: starting from a keyword search or using a thesaurus.
You could begin with a simple keyword search, or try combining search strategies like the ones we’ve described so far. Once you’ve done your search, examine your results list. When you find an item that seems relevant to your topic, open the item record and skim through the abstract to make sure it is actually a good fit for your research. The item that best fits your needs might not be the first item on your list, or even on the first page of results.
Once you’re sure that you’ve found a useful subject heading, evaluate the subject terms listed and how they relate to your topic. You can use subject terms in two different ways:
- Click on a subject term directly from the item record. Doing this will generate a new Advanced Search and retrieve everything in the index on that topic.
- Modify your original search to incorporate the subject terms you found. To do this, add a new line in Advanced Search, type out one of the subject terms you have found, and select the subject field for that line.
Remember, you can always go back and add more keywords, phrases, and subjects to build upon your initial search and explore items that you might not have otherwise found. Experiment with different search techniques and explore how your results change.
Keep in mind that different databases use different subject headings to refer to similar topics. For example, in a psychology index, there may be more psychology-specific subject headings than there would be in an engineering index. For this reason, when you get started with a new finding tool, you’ll probably need to do some initial exploration to discover the subjects that best match your topic.
Using a thesaurus
Another way to find subject headings is by using a thesaurus. In indexes and databases, a thesaurus is a tool that identifies the subject terms a specific database uses. Some databases may use different names for this tool, such as “Subject Terms” or “Subjects.” This tool can be useful when you want to simultaneously explore a variety of subject terms related to your topic. Note: not all databases have a thesaurus tool.
In some databases, the thesaurus button may appear in different places, such as on the advanced search page or on the homepage. For example, in Academic Search Ultimate, the thesaurus is called Subject Terms and is found in the top menu.
After opening this menu, start by searching for a broad term. Let’s say you are interested in finding articles about drug addiction. A search in the thesaurus for “addiction” will lead you to various options such as “drug addiction,” “social media addiction,” and “nicotine addiction.” From this list, you can select the subjects that are the most relevant to your research topic and directly add them to your search.
If there is one subject heading that seems particularly useful, you can click on it to explore broader or narrower subject terms that are related. Keep track of these subject terms until you are done with your project or paper, in case you need to find more resources later.
There are many ways you can use field searching to improve your research. While this section has focused primarily on subject searching, it is often a better choice to use other fields from the start, particularly when you want to find a resource by a specific author or published in a specific date range. Explore the advanced search options that different indexes and databases provide to figure out what works best for your research.


21.6: The Immune Response Against Pathogens
By the end of this section, you will be able to:
- Explain the development of immunological competence
- Describe the mucosal immune response
- Discuss immune responses against bacterial, viral, fungal, and animal pathogens
- Describe different ways pathogens evade immune responses
Now that you understand the development of mature, naïve B cells and T cells, and some of their major functions, how do all of these various cells, proteins, and cytokines come together to actually resolve an infection? Ideally, the immune response will rid the body of a pathogen entirely. The adaptive immune response, with its rapid clonal expansion, is well suited to this purpose. Think of a primary infection as a race between the pathogen and the immune system. The pathogen bypasses barrier defenses and starts multiplying in the host’s body. During the first 4 to 5 days, the innate immune response will partially control, but not stop, pathogen growth. As the adaptive immune response gears up, however, it will begin to clear the pathogen from the body, while at the same time becoming stronger and stronger. When following antibody responses in patients with a particular disease such as a virus, this clearance is referred to as seroconversion (sero- = “serum”). Seroconversion is the reciprocal relationship between virus levels in the blood and antibody levels. As the antibody levels rise, the virus levels decline, and this is a sign that the immune response is being at least partially effective (partially, because in many diseases, seroconversion does not necessarily mean a patient is getting well).
An excellent example of this is seroconversion during HIV disease (Figure \(\PageIndex{1}\)). Notice that antibodies are made early in this disease, and the increase in anti-HIV antibodies correlates with a decrease in detectable virus in the blood. Although these antibodies are an important marker for diagnosing the disease, they are not sufficient to completely clear the virus. Several years later, the vast majority of these individuals, if untreated, will lose their entire adaptive immune response, including the ability to make antibodies, during the final stages of AIDS.
Everyday Connection
Disinfectants: Fighting the Good Fight?
“Wash your hands!” Parents have been telling their children this for generations. Dirty hands can spread disease. But is it possible to get rid of enough pathogens that children will never get sick? Are children who avoid exposure to pathogens better off? The answers to both these questions appears to be no.
Antibacterial wipes, soaps, gels, and even toys with antibacterial substances embedded in their plastic are ubiquitous in our society. Still, these products do not rid the skin and gastrointestinal tract of bacteria, and it would be harmful to our health if they did. We need these nonpathogenic bacteria on and within our bodies to keep the pathogenic ones from growing. The urge to keep children perfectly clean is thus probably misguided. Children will get sick anyway, and the later benefits of immunological memory far outweigh the minor discomforts of most childhood diseases. In fact, getting diseases such as chickenpox or measles later in life is much harder on the adult and are associated with symptoms significantly worse than those seen in the childhood illnesses. Of course, vaccinations help children avoid some illnesses, but there are so many pathogens, we will never be immune to them all.
Could over-cleanliness be the reason that allergies are increasing in more developed countries? Some scientists think so. Allergies are based on an IgE antibody response. Many scientists think the system evolved to help the body rid itself of worm parasites. The hygiene theory is the idea that the immune system is geared to respond to antigens, and if pathogens are not present, it will respond instead to inappropriate antigens such as allergens and self-antigens. This is one explanation for the rising incidence of allergies in developed countries, where the response to nonpathogens like pollen, shrimp, and cat dander cause allergic responses while not serving any protective function.
The Mucosal Immune Response
Mucosal tissues are major barriers to the entry of pathogens into the body. The IgA (and sometimes IgM) antibodies in mucus and other secretions can bind to the pathogen, and in the cases of many viruses and bacteria, neutralize them. Neutralization is the process of coating a pathogen with antibodies, making it physically impossible for the pathogen to bind to receptors. Neutralization, which occurs in the blood, lymph, and other body fluids and secretions, protects the body constantly. Neutralizing antibodies are the basis for the disease protection offered by vaccines. Vaccinations for diseases that commonly enter the body via mucous membranes, such as influenza, are usually formulated to enhance IgA production.
Immune responses in some mucosal tissues such as the Peyer’s patches (see Figure \(\PageIndex{2}\)) in the small intestine take up particulate antigens by specialized cells known as microfold or M cells (Figure 21.27). These cells allow the body to sample potential pathogens from the intestinal lumen. Dendritic cells then take the antigen to the regional lymph nodes, where an immune response is mounted.
Defenses against Bacteria and Fungi
The body fights bacterial pathogens with a wide variety of immunological mechanisms, essentially trying to find one that is effective. Bacteria such as Mycobacterium leprae , the cause of leprosy, are resistant to lysosomal enzymes and can persist in macrophage organelles or escape into the cytosol. In such situations, infected macrophages receiving cytokine signals from Th1 cells turn on special metabolic pathways.
Fungal infections, such as those from Aspergillus , Candida , and Pneumocystis , are largely opportunistic infections that take advantage of suppressed immune responses. Most of the same immune mechanisms effective against bacteria have similar effects on fungi, both of which have characteristic cell wall structures that protect their cells.
Defenses against Parasites
Worm parasites such as helminths are seen as the primary reason why the mucosal immune response, IgE-mediated allergy and asthma, and eosinophils evolved. These parasites were at one time very common in human society. When infecting a human, often via contaminated food, some worms take up residence in the gastrointestinal tract. Eosinophils are attracted to the site by T cell cytokines, which release their granule contents upon their arrival. Mast cell degranulation also occurs, and the fluid leakage caused by the increase in local vascular permeability is thought to have a flushing action on the parasite, expelling its larvae from the body. Furthermore, if IgE labels the parasite, the eosinophils can bind to it by its Fc receptor.
Defenses against Viruses
The primary mechanisms against viruses are NK cells, interferons, and cytotoxic T cells. Antibodies are effective against viruses mostly during protection, where an immune individual can neutralize them based on a previous exposure. Antibodies have no effect on viruses or other intracellular pathogens once they enter the cell, since antibodies are not able to penetrate the plasma membrane of the cell. Many cells respond to viral infections by downregulating their expression of MHC class I molecules. This is to the advantage of the virus, because without class I expression, cytotoxic T cells have no activity. NK cells, however, can recognize virally infected class I-negative cells and destroy them. Thus, NK and cytotoxic T cells have complementary activities against virally infected cells.
Interferons have activity in slowing viral replication and are used in the treatment of certain viral diseases, such as hepatitis B and C, but their ability to eliminate the virus completely is limited. The cytotoxic T cell response, though, is key, as it eventually overwhelms the virus and kills infected cells before the virus can complete its replicative cycle. Clonal expansion and the ability of cytotoxic T cells to kill more than one target cell make these cells especially effective against viruses. In fact, without cytotoxic T cells, it is likely that humans would all die at some point from a viral infection (if no vaccine were available).
Evasion of the Immune System by Pathogens
It is important to keep in mind that although the immune system has evolved to be able to control many pathogens, pathogens themselves have evolved ways to evade the immune response. An example already mentioned is in Mycobacterium tuberculosis , which has evolved a complex cell wall that is resistant to the digestive enzymes of the macrophages that ingest them, and thus persists in the host, causing the chronic disease tuberculosis. This section briefly summarizes other ways in which pathogens can “outwit” immune responses. But keep in mind, although it seems as if pathogens have a will of their own, they do not. All of these evasive “strategies” arose strictly by evolution, driven by selection.
Bacteria sometimes evade immune responses because they exist in multiple strains, such as different groups of Staphylococcus aureus . S. aureus is commonly found in minor skin infections, such as boils, and some healthy people harbor it in their nose. One small group of strains of this bacterium, however, called methicillin-resistant Staphylococcus aureus , has become resistant to multiple antibiotics and is essentially untreatable. Different bacterial strains differ in the antigens on their surfaces. The immune response against one strain (antigen) does not affect the other; thus, the species survives.
Another method of immune evasion is mutation. Because viruses’ surface molecules mutate continuously, viruses like influenza change enough each year that the flu vaccine for one year may not protect against the flu common to the next. New vaccine formulations must be derived for each flu season.
Genetic recombination—the combining of gene segments from two different pathogens—is an efficient form of immune evasion. For example, the influenza virus contains gene segments that can recombine when two different viruses infect the same cell. Recombination between human and pig influenza viruses led to the 2010 H1N1 swine flu outbreak.
Pathogens can produce immunosuppressive molecules that impair immune function, and there are several different types. Viruses are especially good at evading the immune response in this way, and many types of viruses have been shown to suppress the host immune response in ways much more subtle than the wholesale destruction caused by HIV.


<IP_ADDRESS>: Other Privacy Laws
-
- Last updated
- Save as PDF
Restrictions on Record Collecting
In the US, the government has strict guidelines on how much information can be collected about its citizens. Certain classes of information have been restricted by laws over time, and the advent of digital tools has made these restrictions more important than ever.
Children’s Online Privacy Protection Act
Websites that are collecting information from children under the age of thirteen are required to comply with the Children’s Online Privacy Protection Act (COPPA), which is enforced by the Federal Trade Commission (FTC) . To comply with COPPA, organizations must make a good-faith effort to determine the age of those accessing their websites and, if users are under thirteen years old, must obtain parental consent before collecting any information.
Family Educational Rights and Privacy Act
The Family Educational Rights and Privacy Act (FERPA) is a US law that protects the privacy of student education records. In brief, this law specifies that parents have a right to their child’s educational information until the child reaches either the age of eighteen or begins attending school beyond the high school level. At that point, control of the information is given to the child. While this law is not specifically about the digital collection of information on the Internet, the educational institutions that are collecting student information are at a higher risk for disclosing it improperly because of digital technologies.
Health Insurance Portability and Accountability Act
The Health Insurance Portability and Accountability Act of 1996 ( HIPAA) is the law the specifically singles out records related to health care as a special class of personally identifiable information. This law gives patients specific rights to control their medical records, requires health care providers and others who maintain this information to get specific permission in order to share it, and imposes penalties on the institutions that breach this trust. Since much of this information is now shared via electronic medical records, the protection of those systems becomes paramount.
Sidebar: Do Not Track
When it comes to getting permission to share personal information, the US and the EU have different approaches. In the US, the “opt-out” model is prevalent; in this model, the default agreement is that you have agreed to share your information with the organization and must explicitly tell them that you do not want your information shared. There are no laws prohibiting the sharing of your data (beyond some specific categories of data, such as medical records). In the European Union, the “opt-in” model is required to be the default. In this case, you must give your explicit permission before an organization can share your information.
To combat this sharing of information, the Do Not Track initiative was created. As its creators explain [8] :
Do Not Track is a technology and policy proposal that enables users to opt out of tracking by websites they do not visit, including analytics services, advertising networks, and social platforms. At present few of these third parties offer a reliable tracking opt out, and tools for blocking them are neither user-friendly nor comprehensive. Much like the popular Do Not Call registry, Do Not Track provides users with a single, simple, persistent choice to opt out of third-party web tracking.


33.1.5: Phylum Nemertea
-
- Last updated
- Save as PDF
- Boundless
- Boundless
Learning Objectives
- Identify the key features of the Phylum Nemertea
Phylum Nemertea
The Nemertea are colloquially known as ribbon worms. Most species of phylum Nemertea are marine (predominantly benthic or bottom dwellers) with an estimated 900 species known. However, nemertini have been recorded in freshwater and terrestrial habitats as well. Most nemerteans are carnivores, feeding on worms, clams, and crustaceans. Some species are scavengers, while other nemertini species, such as Malacobdella grossa , have also evolved commensalistic relationships with some mollusks. Interestingly, nemerteans have almost no predators, two species are sold as fish bait, and some species have devastated commercial fishing of clams and crabs.
Morphology
Ribbon worms vary in size from 1 cm to several meters. They show bilateral symmetry and remarkable contractile properties. Because of their contractility, they can change their morphological presentation in response to environmental cues. Animals in phylum Nemertea also show a flattened morphology: they are flat from front to back, like a flattened tube. In addition, nemertea are soft, unsegmented animals.
A unique characteristic of this phylum is the presence of a proboscis enclosed in a rhynchocoel. The proboscis serves to capture food and may be ornamented with barbs in some species. The rhynchocoel is a fluid-filled cavity that extends from the head to nearly two-thirds of the length of the gut in these animals. The proboscis may be extended or retracted by the retractor muscle attached to the wall of the rhynchocoel.
Metabolism
The nemertini show a very well-developed digestive system. A mouth opening that is ventral to the rhynchocoel leads into the foregut, followed by the intestine. The intestine is present in the form of diverticular pouches which ends in a rectum that opens via an anus. Gonads are interspersed with the intestinal diverticular pouches, opening outwards via genital pores. A circulatory system consists of a closed loop of a pair of lateral blood vessels. The circulatory system is derived from the coelomic cavity of the embryo. Some animals may also have cross-connecting vessels in addition to lateral ones. Although these are called blood vessels, since they are of coelomic origin, the circulatory fluid is colorless. Some species bear hemoglobin as well as yellow or green pigments. The blood vessels are connected to the rhynchocoel. The flow of fluid in these vessels is facilitated by the contraction of muscles in the body wall. A pair of protonephridia, or primitive kidneys, is present in these animals to facilitate osmoregulation. Gaseous exchange occurs through the skin in the nemertini.
Nervous System
Nemertini have a ganglion or “brain” situated at the anterior end between the mouth and the foregut, surrounding the digestive system as well as the rhynchocoel.  Paired longitudinal nerve cords emerge from the brain ganglia, extending to the posterior end. Ocelli or eyespots are present in pairs, in multiples of two in the anterior portion of the body. It is speculated that the eyespots originate from neural tissue and not from the epidermis.
Reproduction
Animals in phylum Nemertea show sexual dimorphism, although freshwater species may be hermaphroditic. Eggs and sperm are released into the water; fertilization occurs externally. The zygote develops into a special kind of nemertean larvae called a planuliform larva. In some nemertine species, another larva specific to the nemertinis, a pilidium, may develop inside the young worm from a series of imaginal discs. This larval form, characteristically shaped like a deerstalker cap, devours tissues from the young worm for survival before metamorphosing into the adult-like morphology.
Key Points
- The Nemertini are mostly bottom-dwelling marine organisms, although some are found in freshwater and terrestrial habitats.
- Most nemerteans are carnivores, some are scavengers, and others have evolved relationships with some mollusks that are benefit the Nemertean but do not harm the mollusk.
- Nemerteans vary greatly in size and are bilaterally symmetrical; they are unsegmented and resemble a flat tube which can change morphological presentation in response to environmental cues.
- Nemertini have a simple nervous system comprised of a ring of four nerve masses called “ganglia” at the anterior end between the mouth and the foregut from which paired longitudinal nerve cords emerge and extend to the posterior end.
- Nemertini are mostly sexually dimorphic, fertilizing eggs externally by releasing both eggs and sperm into the water; a larva may develop inside the resulting young worm and devour its tissues before metamorphosing into the adult.
Key Terms
- protonephridia : an invertebrate organ which occurs in pairs and removes metabolic wastes from an animal’s body
- rhynchocoel : a cavity which mostly runs above the midline and ends a little short of the rear of the body of a nemertean and extends or retracts the proboscis
- proboscis : an elongated tube from the head or connected to the mouth, of an animal


2.6: Summary
selected template will load here
This action is not available.
Information systems hardware consists of the components of digital technology that you can touch. In this chapter, we focused on the personal computer and its components. We reviewed the personal computer configuration because it has many of the same attributes as other digital computing devices. A personal computer comprises many components, most importantly the CPU, motherboard, RAM, hard disk, removable media, and input/output devices. We also reviewed some personal computer variations, such as the tablet computer, Bluetooth, and the smartphone . By Moore’s Law, these technologies have improved quickly over the years, making today’s computing devices much more powerful than devices just a few years ago. Finally, we discussed two of the consequences of this evolution: t he commoditization of the personal computer and the problem of electronic waste.


3.4: Working with Events
The Complement Rule
In the sets and logic chapters, we saw the complement of set \(A\) and the negation of statement \(A\). We defined the complement of set \(A\) as another set containing the elements in the universal set that are not in set \(A\), and we wrote the complement of \(A\) as \(A^{\prime}\). Similarly, we defined the negation of statement \(A\) as exactly the opposite of statement \(A\) and wrote the negation of \(A\) as \(\sim A\).
In probability, we can think of the complement in a similar way. The complement of event \(A\) contains all outcomes in the sample space that are not in event \(A\). We write the probability of the complement as \(P(A^{\prime})\) or \(P(\text{not } A)\).
 The sample space is \(S=\{1,2,3,4,5,6\}\). If event \(A\) is " rolling a number less than 3, " then \(A=\{1,2\}\). The complement of event \(A^{\prime}\), " not rolling a number less than 3 ," contains all the outcomes in the sample space that are not in \(A\). Thus, \(A^{\prime}=\{3,4,5,6\}\).
A Venn diagram that illustrates the relationship between \(S\), \(A\), and \(A^{\prime}\) is shown below:
Here, \(P(A)=\frac{2}{6}=\frac{1}{3}\) and \(P(A^{\prime})=\frac{4}{6}=\frac{2}{3}\).
When we combining all outcomes in event \(A\) with all outcomes in its complement \(A^{\prime}\), the result is all the outcomes in the sample space of the experiment. Therefore, the probabilities of an event and its complement must add to 1. This formula and its variations are known as the Complement Rule of Probability.
\(P(A)+P(A^{\prime})=1\) or \(P(A)+P(\text{not }A)=1\)
Alternatively, \(P(A)= 1 - P(A^{\prime})\) or \(P(A)= 1 - P(\text{not }A)\)
\(P(A^{\prime})=1-P(A)\) or \(P(\text{not }A) = 1-P(A)\)
As shown in the box, you may find that it is sometimes easier to calculate the probability of an event's complement than it is to calculate the probability of the event itself. Once this is done, the probability of the event, \(P(A)\), is found using the relationship \(1-P(A^{\prime})\).
Suppose you know that the probability of getting the flu this winter is 0.43. What is the probability that you will not get the flu?
Solution
The probability of getting the flu is \(P(F) = 0.43\).
The probability of not getting the flu is \(P(F^{\prime})=1-P(F)=1 - 0.43 = 0.57\).
The probability that a person does not live in an industrialized country of the world is \(\frac{4}{5}\). Find the probability that a person lives in an industrialized country.
Solution
The probability that a person does not live in an industrialized country is \(P(I^{\prime})=\frac{4}{5}\).
The probability that a person lives in an industrialized country is \(P(I) = 1 - P(I^{\prime}) =1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5} \).
A card is drawn from a standard deck of cards. What is the probability that the card is not a King?
Solution
The probability of getting a King is \(P(K) = \frac{4}{52}=\frac{1}{13}\).
Therefore, the probability of not getting a King is \(P(K^{\prime}) = 1 - \frac{1}{13} = \frac{13}{13} - \frac{1}{13}= \frac{12}{13} \).
A card is drawn from a standard deck of cards. What is the probability that the card is not a club?
- Answer
-
\(\frac{3}{4}\)
Combining Events
Many probabilities in real life involve more than one event. If we draw a single card from a deck we might want to know the probability that it is either red or a jack. If we look at a group of students, we might want to know the probability that a single student has brown hair and blue eyes. When we combine two outcomes to make a single event we connect the outcomes with the word “or” or the word “and.” It is very important in probability to pay attention to the words “and” and “or” if they appear in a problem. The word “and” restricts the field of possible outcomes to only those outcomes that satisfy more than one event at the same time or in succession. The word “or” broadens the field of possible outcomes to those that satisfy one or more events.
Combining Events with "Or"
The event "\(A \text{ or } B\)" refers to an event that includes outcomes of \(A\), or \(B\), or both. As we saw with sets and truth tables, the union \(A \cup B\) and disjunction \(A \vee B\) refer to where at least one of \(A\) or \(B\) occur. Let’s begin the development of a rule for \(P(A \text{ or } B)\) by exploring an example.
A box contains 12 slips of paper, each numbered from 1 to 12. The sample space is \(\{1,2,3,4,5,6,7,8,9,10,11,12\}.\) These twelve outcomes are all equally likely. What is the probability of selecting an even number or a number divisible by 3?
Solution
The event in question must meet one or more of the two criteria: selecting an even number or a number divisible by 3. This means we can select a 2, 4, 6, 8, 10, or 12 (an even number) or select a 3, 6, 9, or 12 (a number divisible by 3.) Here,
\(P(\text{even number}) = \frac{6}{12}\) and \(P(\text{divisible by 3}) = \frac{4}{12}\).
It is tempting to add \(P(\text{even number}) + P(\text{number divisible by 3}) = \frac{6}{12} + \frac{4}{12} = \frac{10}{12}\). However, there is an error in this thinking.
To see this error, look at the sample space of this experiment represented by a Venn diagram. We are interested in finding \(P(A \text{ or } B)\), which is the union of two sets in the diagram.
The outcomes 6 and 12 meet both of these criteria because they are even numbers and divisible by 3. Consequently, we have included these outcomes in both \(P(\text{even number})\) and \(P(\text{number divisible by 3})\). To compensate for over-counting the outcomes 6 and 12, we must subtract the probability of these two over-counted outcomes, \(P(\text{even number and divisible by 3}) = \frac{2}{12}\). So,
\(P(\text{even number or number divisible by 3}) = \frac{6}{12} + \frac{4}{12} -\frac{2}{12} = \frac{8}{12} =\frac{2}{3}.\)
Based on the previous example, it appears that \(P(A\ \text{ or } B) = P(A) + P(B) \; – P(A \text{ and } B)\). This thinking should not be new or a surprise. Joining two sets with "or" is a union of those two sets. And, in Chapter 1, we found that \(n(A \cup B) = n(A)+n(B)-n(A \cap B)\).
But, before we state this general rule for finding the probability of \(A \text{ or } B\), we should consider a special case of a Venn diagram with event \(A\) and event \(B\). The figure below is a Venn diagram showing the events are mutually exclusive . The circles do not overlap because there are no outcomes that are common to both events.
In this Venn diagram, event \(A\) could represent all the outcomes of drawing a red card from a deck of cards, and event \(B\) could represent all the outcomes of drawing a club from a deck of cards. These two events have no outcomes in common because there are no cards in a deck that are both red and a club. Therefore, the events have no outcomes in common.
Event \(A\) and event \(B\) are mutually exclusive if the events have no common outcomes.
That is, event \(A\) and event \(B\) are mutually exclusive if and only if \(A \cap B = \text{ { } }\) and \(P(A \text{ and } B) = 0\).
In the case of mutually exclusive events, case \(P(A \text{ and } B) = 0\). There is no over-counting outcomes because there are no outcomes that are in both \(A\) and \(B\). So, when two mutually exclusive events are joined by "or," the formula we use for \(P(A\text{ or }B)\) becomes much simpler:
\( \begin{align*} P(A\text{ or }B) &= P(\text{A}) + P(\text{B}) - P(A \text{ and B}) \\[4pt] &= P(A) + P(B) - 0 \\[4pt] &= P(A) + P(B) \end{align*}\)
The formula for \(P(A \text{ or }B)\) when \(A\) and \(B\) are mutually exclusive is just a special case of the general formula. Since the two events are mutually exclusive, there is no double counting and no need to subtract their overlap.
The box gives the general formula for \(P(A \text { or } B)\) and a more specialized formula when the events are mutually exclusive.
If \(A\) and \(B\) are any events then
\(P(A\ \text{ or } B) = P(A) + P(B) \; – P(A,\text{ and } B)\) or \(P(A \cup B) = P(A) + P(B) \; – P(A \cap B).\)
If \(A\) and \(B\) are mutually exclusive events then
\(P(A \text{ or } B) = P(A) + P(B)\) or \(P(A \cup B) = P(A) + P(B)\)
To apply the probability formulas for "or" scenarios successfully, we should be careful to distinguish whether the two events are mutually exclusive or not. We now look at several examples.
A single card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is a club or a face card (King, Queen, Jack.)
Solution
The events "club" and "face card" are not mutually exclusive because a card can be a club and also a face card. There are 13 cards that are clubs, 12 face cards (J, Q, K in each suit), and 3 cards that are both face cards and clubs.
\( \begin{align*} P(\text{club or face card}) &= P(\text{club}) + P(\text{face card}) - P(\text{club and face card}) \\[4pt] &= \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} \\[4pt] &= \dfrac{22}{52} = \dfrac{11}{26} \approx {0.4231} \end{align*}\)
The probability that the card is a club or a face card is approximately 0.4231 or 42.31%.
A single card is drawn from a well-shuffled deck of 52 cards. Find the probability that the card is a King or an Ace.
Solution
The event "King" and the event "Ace" have no common outcomes so these are mutually exclusive events.
\( \begin{align*} P(\text{King or Ace}) &= P(\text{King}) + P(\text{Ace}) \\[4pt] &= \dfrac{4}{52} + \dfrac{4}{52} \\[4pt] &= \dfrac{8}{52} \\ &= \dfrac{2}{13} \end{align*}\)
The probability of selecting a King or Ace is \(\dfrac{2}{13} \approx 0.1538\), or about \(15.39\%\).
 What's the probability that the die lands on an odd number or a number less than 5?
- Answer
-
\(\frac{5}{6}\)
On a game spinner, the probability of spinning red is \(\frac{1}{3}\), the probability of spinning white is \(\frac{2}{5}\), and the probability of spinning blue is \(\frac{4}{15}\). What is the probability of spinning once and getting white or blue?
- Answer
-
\(\frac{2}{3}\)
An experiment consists of tossing a coin and then rolling a die. Find the probability that the coin lands heads up or the number is five.
Solution
Let H represent heads up and T represent tails up. The sample space for this experiment is
\(S = \{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6\}\).
- There are six ways the coin can land heads up: \(\{H1, H2, H3, H4, H5, H6\}\).
- There are two ways the die can land on five, \(\{H5, T5\}\).
- There is one way for the coin to land heads up and the die to land on five, \(\{H5\}\).
\( \begin{align*} P(\text{heads or five}) &= P(\text{heads}) + P(\text{five}) - P(\text{both heads and five}) \\[4pt] &= \dfrac{6}{12} + \dfrac{2}{12} - \dfrac{1}{12} \\[4pt] &= \dfrac{7}{12} = \approx {0.5833} \end{align*}\)
The probability that the coin lands heads up or the number is five is approximately 0.5833 or 58.33%.
There is a 90% chance of snow, a 20% chance of strong winds, and a 15% chance of both snow and strong winds. What is the probability of snow or strong winds?
Solution
\( \begin{align*} P(\text{snow or strong winds}) &= P(\text{snow}) + P(\text{strong winds}) - P(\text{snow and strong winds}) \\[4pt] &= 0.90 + 0.20 - 0.15 \\[4pt] &= 0.95 \end{align*}\)
The probability of snow or strong winds is 0.95 or 95%.
In a subdivision, there is a 67% chance that a home has a garage and a 32% chance that a home has a patio. There is a 13% chance that a home has both a garage and a patio. What's the chance that a home has a garage or patio?
- Answer
-
86%
The table shows a group of cyclists and the routes they prefer. One cyclist is randomly chosen from this group.
| Gender | Lake Path, \(L\) | Hilly Path, \(H\) | Wooded Path, \(W\) | Total |
|---|---|---|---|---|
| Male, \(M\) | 45 | 38 | 27 | 110 |
| Female, \(F\) | 26 | 52 | 12 | 90 |
| Total | 71 | 90 | 39 | 200 |
- What is the probability of selecting a cyclist who is a female or prefers the wooded path?
- What is the probability of selecting a cyclist or prefers the lake path or the hilly path?
- What is the probability of selecting a cyclist who does not prefer the lake path?
Solution
- This question asks for \(P(F \text{ or } W)\). These events are not mutually exclusive because there are cyclists who are both female and prefer the hilly path:
\(\begin{aligned} P(F \text{ or } W) &= P(F) + P(W) - P(F \text{ and } W) \\ &= \frac{90}{200} + \frac{39}{200} - \frac{12}{200} \\ &= \frac{117}{200} = 0.585 \end{aligned}\)
- This question asks for \(P(L \text{ or } H)\). These events are mutually exclusive because there are no cyclists who prefer both the Lake path and the Hilly path:
\(\begin{aligned} P(L \text{ or } H) &= P(L) + P(H) \\ &= \frac{71}{200} + \frac{90}{200} \\ &= \frac{161}{200} = 0.805 \end{aligned}\)
- This question asks for \(P(L^{\prime})\):
\(\begin{aligned} P(L^{\prime}) &= 1 - P(L) \\ &= 1-\frac{71}{200} \\ &= \frac{129}{200} = 0.645 \end{aligned}\)
A group of people were surveyed about the type of movies they prefer. Suppose a person is chosen at random from this group.
| Gender | Romantic, \(R\) | Action, \(A\) | Horror, \(H\) | Total |
|---|---|---|---|---|
| Male, \(M\) | 8 | 25 | 6 | 39 |
| Female, \(F\) | 12 | 10 | 3 | 25 |
| Total | 20 | 35 | 9 | 64 |
- What is the probability of selecting a person who is male or prefers horror movies?
- What is the probability of selecting a person who prefers romantic or action movies?
- What is the probability of selecting a person who does not prefer romantic movies?
- Answer
-
- \(\frac{42}{64}= \frac{21}{32}\)
- \(\frac{55}{64}\)
- \(\frac{44}{64} = \frac{11}{16}\)
Combining Events with "And"
Sometimes we need to calculate probabilities for compound events that are connected by the word “and.” This requires that we look for outcomes that are common to both events. As we saw with sets and truth tables, the intersection \(A \cap B\) and conjunction \(A \wedge B\) refer to where both \(A\) and \(B\) occur. When working with probabilities, these events may occur at the same time or they could happen in a sequence, such as \(A\) followed by \(B\).
How we calculate the theoretical probability of the event "\(A \text{ and } B\)" depends on whether the two events are independent or dependent . So, we must discuss these concepts before developing a formula.
Two events \(A\) and \(B\) are independent if the probability that \(B\) occurs is the same whether or not \(A\) occurs. If the probability of \(B\) is affected by the occurrence of \(A\), then we say that the events are dependent .
Coin tosses and die rolls are common examples of independent events – tossing heads does not change the probability of tossing heads the next time. The probability of tossing a heads on the second toss is still 1/2 no matter if there was a heads or tails on the first toss. Likewise, rolling a six on a first die does not change the probability that the next roll will be a six.
Another example of independent events is randomly selecting items from a container when the items are selected with replacement . By replacing the item after the first selection, we reset the probability back to what it was before we made the selection. Since the probabilities in the second selection are the same as the first selection, the events are independent.
If we select without replacement , however, we change the total number of possible outcomes, thereby changing the probabilities in subsequent selections. For example, if we select two people from a group for a committee, the first member selected cannot be selected again so the sample space (and therefore the probabilities) for the second selection has changed. So, if we draw without replacement, the events will be dependent.
What about drawing two cards from a deck? This can be a bit tricky. When you draw the first card and put it back before drawing a second card, the sample space for drawing the second card remains 52 cards just as it was for the first card. These events are independent. However, if you draw the first card and put it aside before drawing a second card, the sample space for drawing the second card changes to 51 because one card has been removed. This is an example of events that are dependent.
Two events are independent events if the occurrence of one event has no effect on the probability of the occurrence of the other event. Examples of independent events include tossing coins, rolling dice, spinning spinners, and selecting items from a group when the items are replaced after each selection.
Two events are dependent events if the occurrence of one event has an effect on the probability of the occurrence of the other event. Examples of dependent events include selecting items from a group when the items are not replaced after each selection.
Determine whether each pair of events are independent or dependent.
- Flipping a coin twice and getting tails both times.
- Selecting a president and then a vice president at random from a pool of five equally qualified individuals.
- The event that it will be cold in Washington, DC tomorrow and the event that it will be cold in Baltimore tomorrow.
- Wearing your lucky socks and getting an A on your exam.
Solution
- The probability of getting tails on the first flip is \(\frac{1}{2}\). After flipping tails, the probability of getting tails on the second flip is still \(\frac{1}{2}\). Since the probability of flipping tails on the second flip did not change because we flipped tails on the first flip, the events are independent.
- Since two different people will be put in the role of president and vice president, we are drawing without replacement and the events are therefore dependent.
- If it is cold in Baltimore it is more likely that it will be cold in Washington, so the events are dependent.
- The socks you wear do not have a direct effect on how well you do on your exam, so these events are independent.
Determine whether each pair of events are independent or dependent.
- Rolling a die twice and getting '2' both times.
- Getting two hearts when selecting one card from a well-shuffled deck, setting it aside, and then selecting another card.
- Drawing two red marbles one-by-one from a bucket of marbles where the first marble is replaced before drawing the second marble.
- Answer
-
- independent
- dependent
- independent
To calculate "and" probabilities we always multiply, but we need to determine whether the events are independent or dependent. If they are independent, then we can multiply the individual probabilities of the events because one does not affect the other. If the events are dependent, we must remember that the first event has had an effect on the probability of the second event. A general formula for finding \(P(A \text{ and } B)\) is given below:
For events \(A\) and \(B\),
\( P(\text{A and B}) = P(A) \cdot P(B)^{*}\).
\(^{*}\) If events \(A\) and \(B\) are dependent, \(P(B)\) depends on the fact that \(A\) happened.
So why does this multiplication rule work?  One ball is selected, its color is observed, and then the ball is placed back in the box.  The tree diagram shows the sample space for the experiment.
From the tree diagram, there are 9 equally likely outcomes in the sample space is \(S =\{RR, RB, RW, BR, BB, BW, WR, WB, WW\}\). By counting, we can see that the probability of selecting a red ball followed by a white ball \(P(RW)\) is \(\frac{1}{9}\).
We could have reached this same answer by thinking another way and without listing a sample space.
- We know there are 9 outcomes in the sample space because there are 3 options for the first selection of ball and 3 options for the second selection of ball. This means there are \(3 \times 3=9\) ways to make the selection of two balls together.
- There is only one ball that is red during the first selection and one ball that is white during the second selection. This means there is \(1 \times 1=1\) way to to make the selection of these two colors.
- Therefore, the probability of selecting red ball followed by a white ball is \(\frac{1\times1}{3\times3}\) or \(\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\). We have multiplied \(P(\text{red ball on 1st selection}) \times P(\text{white ball on 2nd selection})\) to get \(P(\text{red ball on 1st selection and white ball on 2nd selection})\).
Suppose a fair coin is tossed.
- If the coin is tossed twice, what is the probability that both tosses land heads up?
- If the coin is tossed four times, what is the probability that all four tosses and heads up?
Solution
The tosses of the coins are independent events. We will use the formula for \(P(A \text{ and } B)\) when \(A\) and \(B\) are independent.
- \(P(\text{two heads in a row}) = P(\text{1st is heads and 2nd is heads})\)
\( = P(\text{1st is heads}) \cdot P(\text{2nd is heads}) \)
\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \)
\( = \dfrac{1}{4}\)
- \(P(\text{four heads in a row}) = P(\text{1st is heads and 2nd is heads and 3rd is heads and 4th is heads})\)
\( = P(\text{1st is heads}) \cdot P(\text{2nd is heads}) \cdot P(\text{3rd is heads}) \cdot P(\text{4th is heads})\)
\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}\)
\( = \left(\dfrac{1}{2}\right)^4\)
\( = \dfrac{1}{16}\)
The probability that all four tosses land heads up is \(\dfrac{1}{16}\).
The game spinner has 6 equal-sized sections. Jeremy will spin the spinner 3 times in a row.
- Find the probability that Jeremy gets blue on all three spins.
- Find the probability that Jeremy gets red on the first spin, blue on the second spin, and red on the third spin.
- Answer
-
- \(\frac{1}{27}\)
- \(\frac{1}{12}\)
A bag contains 5 red and 4 white marbles. Two marbles are selected one by one and colors are recorded. What is the probability that the first marble is red and the second marble is white if
- the first marble is returned to the bag before the second marble is selected.
- the first marble is not returned to the bag before the second marble is selected.
Solution
- Since the first marble is returned to the bag before the second marble is selected, these are independent events.
\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{9} = \dfrac{20}{81}\end{align*}\]
The probability that the first marble is red and the second marble is white is \(\dfrac{20}{81}\).
- Since the first marble not returned to the bag before the second marble is selected, these are dependent events.
\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{8} = \dfrac{20}{72} = \dfrac{5}{18}\end{align*}\]
The probability that the first marble is red and the second marble is white is \(\dfrac{5}{18}\).
A box contains 6 red blocks, 4 green blocks, and 3 black blocks. What is the probability of selecting
- two red blocks in a row if the first red block is put back in the box?
- two red blocks in a row if the first red block is not put back in the box?
- a red block and then a green block if the red block is not put back in the box?
- Answer
-
- \(\frac{36}{169}\)
- \(\frac{5}{26}\)
- \(\frac{2}{13}\)
Abby has an important meeting in the morning. She sets 3 battery-powered alarm clocks just to be safe. If each alarm clock has a 0.03 probability of malfunctioning, what is the probability that all three alarm clocks fail at the same time?
Solution
Since the clocks are battery-powered we can assume that one failing will have no effect on the operation of the other two clocks. The functioning of the clocks is independent.
\[\begin{align*} P(\text{all three fail}) &= P(\text{first fails}) \cdot P(\text{second fails})\cdot P(\text{third fails}) \\[4pt] &= (0.03)(0.03)(0.03) \\[4pt] &= (0.03)^3 \\[4pt] &= 2.7 \times 10^{-5} \end{align*}\]
The probability that all three clocks will fail is approximately 0.000027 or 0.0027%. It is very unlikely that all three alarm clocks will fail.
Approximately 80% of U.S. children are covered by some type of health insurance. Assume that whether one child has health insurance or not has no effect on whether the next child has health insurance. If 4 children are selected at random, find the probability that
- all four of the children have insurance.
- none of the four children have insurance
Solution
- We will assume independence in this scenario.
\(\begin{aligned} P(\text{all four have insurance}) &= P(\text{1st has insurance}) \cdot P(\text{2nd has insurance}) \cdot P(\text{3rd has insurance}) \cdot P(\text{4th has insurance}) \\ &= (0.80)(0.80)(0.80)(0.80) \\ &= (0.80)^4 \\&= 0.4096 \end{aligned}\)
The probability that all four children are covered by insurance is 0.4096 or 40.96%.
- To find the probability that none of the four children have insurance, we will need to use the Complement Rule. If 80% of children are covered by insurance, then 100% - 80% = 20% are not covered by insurance.
\(\begin{aligned} P(\text{none of the four have insurance}) &= P(\text{1st no insurance}) \cdot P(\text{2nd no insurance})\cdot P(\text{3rd no insurance}) \cdot P(\text{4th no insurance}) \\&= (0.20)(0.20)(0.20)(0.20) \\&= (0.20)^5)\\ &= 0.0016 \end{aligned}\)
The probability that none of the four children are covered by insurance is 0.0016 or 0.16%.
According to some studies, it is believed that 15% of people are left-handed. Suppose 3 people are selected at random.
- Find the probability that all three are left-handed.
- Find the probability that none of the three are left-handed.
- Find the probability that only the first one selected is left-handed (and the others are right-handed.)
- Answer
-
- 0.003375 = 0.3375%
- 0.614125 = 61.4125%
- 0.108375 = 10.8375%
'At Least Once' Rule for Independent Events
Many times we need to calculate the probability that an event will happen "at least once" in many trials. The calculation can get quite complicated if there are more than a couple of trials. Using the complement to calculate the probability can simplify the problem considerably. That is, we will find the probability of what we don't want to happen first, and then subtract that probability from 1. The following example will help you understand the formula.
The probability that a child forgets her homework on a given day is 0.15. What is the probability that she will forget her homework at least once in the next five days?
Solution
Assume that whether she forgets or not one day has no effect on whether she forgets or not the second day. The events are independent.
If \(P(\text{forgets}) = 0.15\), then \(P(\text{not forget}) = 0.85\).
\(\begin{aligned} P(\text{forgets at least once in 5 tries}) &= P(\text{forgets 1, 2, 3, 4 or 5 times in 5 tries}) \\[4pt] & = 1 - P(\text{forgets 0 times in 5 tries}) \\[4pt] &= 1 - P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \\[4pt] &= 1 - (0.85)(0.85)(0.85)(0.85)(0.85) \\[4pt] & = 1 - (0.85)^{5} \\[4pt] &\approx0.5563 \end{aligned}\)
The probability that the child will forget her homework at least one day in the next five days is 0.5563 or 55.63%
The idea in the previous example can be generalized as the At Least Once Rule .
If an experiment is repeated \(n\) times where the \(n\) trials are independent and the probability of event \(A\) occurring one time is \(P(A)\),
\(P(A \text{ occurs at least once}) = 1 - P(\text{not } A)^{n}\).
The probability of seeing a falcon near the lake during a day of bird watching is 0.21. What is the probability that a birdwatcher will see a falcon at least once in eight trips to the lake?
Solution
If \(P(\text{sees falcon}) = 0.21\), then \(P(\text{doesn't see a falcon}) = 1- 0.21 =0.79\).
\(\begin{aligned} P(\text{sees falcon at least once in 8 trips}) &= 1 - P(\text{doesn't see a falcon})^8 \\[4pt] &= 1 - (0.79)^{8} \\[4pt] &\approx 0.8483 \end{aligned}\)
The probability of seeing a falcon at least once in eight trips to the lake is approximately 0.8483 or 84.83%.
A multiple choice test consists of six questions. Each question has four choices for answers (A, B, C, or D), only one of which is correct. A student guesses on all six questions. What is the probability that he gets at least one answer correct?
Solution
Let \(A\) be the event that the answer to a question is correct. Since each question has four choices and only one correct choice, \(P(\text{correct}) = \dfrac{1}{4}\). That means \(P(\text{not correct}) =1 - \dfrac{1}{4} = \dfrac{3}{4}\).
\(\begin{aligned} P(\text{at least one correct in six trials}) &= 1 - P(\text{not correct})^{6} \\[4pt] &= 1 - \left(\dfrac{3}{4}\right)^{6} \\[4pt] &\approx 0.8220 \end{aligned}\)
The probability that he gets at least one answer correct is 0.8220 or 82.20%.
Mary is in the archery club. The probability that she hits the target on an attempt is 65%. She will make 4 attempts to hit the target. What's the probability that Mary hits the target at least once on her attempts?
- Answer
-
0.9850 or 98.50%


7.6: Dilution of Concentrated Solutions
In the laboratory, a chemist will often prepare solutions of known concentration beginning with a standard stock solution. A stock solution is generally concentrated and, of course, the molar concentration of the solute must be known. To perform a reaction, a measured amount of this stock solution will be withdrawn and added to another reactant, or will be diluted into a larger volume for some other use. The calculations involved in these dilutions are trivial and simply involve calculating the number of moles transferred and dividing this by the final volume. For example, 15.0 mL of a stock solution of 1.00 M hydrochloric acid (HCl) is withdrawn and diluted into 75 mL of distilled water; what is the final concentration of hydrochloric acid?
First, the number of moles of HCl is calculated from the volume added and the concentration of the stock solution:
\[0.0150L\times \left ( \frac{1.00\: moles}{1\: L} \right )=0.0150\; moles\; HCl \nonumber \]
We have diluted this number of moles into (15.0 + 75.0) = 90.0 mL, therefore the final concentration of HCl is given by:
\[\left ( \frac{0.0150\; moles\; HCl}{0.0900\: L} \right )=(0.167\; moles\; HCl/L)\; or\; 0.167\; M \nonumber \]
An even simpler way to approach these problems is to multiply the initial concentration of the stock solution by the ratio of the aliquot (the amount withdrawn from the stock solution) to the final volume, using the equation below:
\[(stock\; concentration)\times \left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=final\; concentration \nonumber \]
Using this method on the previous problem,
\[(1.00\; M)\times \left ( \frac{15.0\; ml}{90.0\; ml} \right )=0.167\; M \nonumber \]
Note that we did not have to convert our volumes (15.0 and 90.0 mL) into L when we use this approach because the units of volume cancel in the equation. If the units that are given for the aliquot and the final volume are different, a metric conversion ratio may be required. For example, 10.0 µL of a 1.76 M solution of HNO 3 (nitric acid) are diluted into 10.0 mL of distilled water; what is the final concentration of nitric acid?
In this problem, we need to convert µL and mL into a common unit. We can do this using the ratios,
\[\left ( \frac{10^{-6}L}{1\mu L} \right )\; and\; \left ( \frac{10^{-3}L}{1\: mL} \right ) \nonumber \]
We need to multiply each of our volumes by the appropriate factor to get our volumes in terms of liters, and then simply multiply by the initial concentration. Thus,
\[1.76\: M\times \left \{ \frac{10.0\mu L\left ( \frac{10^{-6}L}{1\mu L}\right )}{10.0\: mL\left ( \frac{10^{-3}L}{1\: mL}\right )} \right \}=1.76\times 10^{-3}M \nonumber \]
The final volume in this problem is actually (1.00×10 -2 L) + (1.00×10 -5 L) = 1.001×10 -2 L, but because our calculation is only accurate to three significant figures, the volume of the aliquot is not significant and the final volume has been rounded.
The standard method we have used here can also be adapted to the type of problem in which you need to find the volume of a stock solution that must be diluted to a certain volume in order to produce a solution of a given concentration. For example, what volume of 0.029 M CaCl 2 must be diluted to exactly 0.500 L in order to give a solution that is 50.0 µM?
In order to solve this problem in the simplest of terms, we should re-examine the above equation:
\[(stock\; concentration)\times \left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=final\; concentration \nonumber \]
This equation can be re-written as:
\[\left ( \frac{volume\; of\; the\; aliquot}{final\; volume} \right )=\left ( \frac{final\; concentration}{stock\; concentration} \right ) \nonumber \]
Or
\[\left ( \frac{V}{V_{f}} \right )=\left ( \frac{C_{f}}{C_{i}} \right ) \nonumber \]
where C i and C f are the stock and final concentrations, respectively, V is the volume of the aliquot and V f is the final volume of the solution. Stated another way, this is simply a set of ratios; aliquot to final volume , and final concentration to initial concentration (operationally, these ratios will always be “ small value/larger value ”). Working with this set of ratios, we can directly solve this type of problem as follows:
First, we need to convert our final concentration (50.0 µM) into M, to match the units of our stock solution. The metric multiplier for µ is 10 -6 , making our final concentration 50.0×10 -6 M, or more properly, 5.00×10 -5 M. Our equation is therefore:
\[\left ( \frac{V}{0.500L} \right )=\left ( \frac{5.00\times 10^{-5}M}{0.029M} \right ) \nonumber \]
\[V=\left ( \frac{5.00\times 10^{-5}M}{0.029M} \right )\times 0.500L \nonumber \]
The volume of the aliquot, V , is 8.62×10 -4 L, or using the conversion factor
\[\left ( \frac{10^{3}mL}{1L} \right ) \nonumber \]
the required volume is 0.86 mL (there are only two significant figures in the concentration of the stock solution, 0.029 M).
Dilution problems can be solved directly using the above equation, or, as you become more comfortable with the math, using the initial and final ratios like we did in this problem (remember, the numbers in the two ratios are “smaller/larger” ).
- A 1.50 mL aliquot of a 0.177 M solution of sulfuric acid (H 2 SO 4 ) is diluted into 10.0 mL of distilled water, to give solution A . A 10.0 mL aliquot of A is then diluted into 50.0 mL of distilled water, to give solution B . Finally, 10.0 mL of B is diluted into 900.0 mL of distilled water to give solution C . Additional distilled water is then added to C to give a final volume of 1.0000 L. What is the final concentration of sulfuric acid in solution C ?
- A solution was prepared by mixing 250 mL of 0.547 M NaOH with 50.0 mL of 1.62 M NaOH and then diluting to a final volume of 1.50 L. What is the molarity of Na + in this solution? To what final volume should 75.00 mL of 0.889 M HCl(aq) be diluted to prepare 0.800 M HCl(aq)?


Skip to main content
4.3.S: Culture and Meaning (Summary)
-
-
Last updated
-
-
Save as PDF
-
Key Terms and Concepts
|
Beliefs
|
Native Anthropologists
|
|
Conflict Theory
|
Non-Material Culture
|
|
Cultural Sociology
|
Norms
|
|
Culture
|
Practices
|
|
Enculturation
|
Qualitative Data
|
|
Environmental Theory
|
Quantitative Data
|
|
Ethnographers
|
Sapir-Worf Hypothesis
|
|
Exchange Theory
|
Social Location
|
|
Feminism
|
Sociological Imagination
|
|
Functionalism
|
Symbolic Interactionism
|
|
Macrosociology
|
Symbols
|
|
Material Culture
|
Theoretical paradigms
|
|
Microsociology
|
Values
|
References
-
Alexander, Jeffrey C. 2003. “The Strong Program in Cultural
Sociology: Elements of a Structural Hermeneutic (with Philip
Smith).” Pp. 11-26 in
The Meanings of Social Life: A Cultural
Sociology
. New York: Oxford University Press.
-
Back, Les, Andy Bennett, Laura Desfor Edles, Margaret Gibson,
David Inglis, Ronald Joacobs, and Ian Woodward. 2012.
Cultural
Sociology: An Introduction
. United Kingdom:
Wiley-Blackwell.
-
Best, Stephen. 2007. “Cultural Turn.”
Blackwell
Encyclopedia of Sociology
. Hoboken, NJ: John Wiley & Sons,
Ltd.
-
Carl, John D. 2013.
Think Social Problems
. 2nd ed.
Boston, MA: Pearson Education, Inc.
-
Chaney, D. 1996.
Lifestyles
. London: Routledge.
-
Durkeim, Emile. 1965.
The Elementary Forms of the Religious
Life
. New York: Free Press.
-
Giddens, Anthony. 1991.
Modernity and Self-Identity: Self
and Society in the Late Modern Age
. Cambridge: Polity.
-
Griffiths, Heather, Nathan Keirns, Eric Strayer, Suasn
Cody-Rydzewsk, Gail Scaramuzzo, Tommy Sadler, Sally Vyain, Jeff
Byer, and Faye Jones. 2015.
Introduction to Sociology 2e
.
Houston, TX: OpenStax College.
-
Griswold, Wendy. 2013.
Cultures and Societies in a Changing
World
. 4th ed. Thousand Oaks, CA: Sage Publications, Inc.
-
Henslin, James M. 2011.
Essentials of Sociology: A
Down-to-Earth Approach
. 11th ed. Upper Saddle River, NJ:
Pearson.
-
Human Rights Watch. 2015. “Unequal and Unprotected: Women’s
Rights under Lebanese Personal Status Laws.” Retrieved January 9,
2018 (www.hrw.org/report/2015/01/1...womens-rights-
under-lebanese-personal-status-laws).
-
Kennedy, Vera, Romney Norwood, and Matthew Jendian. 2017.
Critical Thinking about Social Problems
. Dubuque, IA:
Kendall Hunt Publishing Company.
-
Kottak, Conrad Phillip and Kathryn A. Kozaitis. 2012.
On
Being Different: Diversity and Multiculturalism in the North
American Mainstream
. 4th ed. New York: McGraw-Hill
-
Companies, Inc.
-
Marx, Karl. 1977. “Preface to a Contribution to the Critique of
Political Economy.” Pp. 388-392 in
Karl Marx: Selected
Writings
, edited by David McLellan. Oxford, UK: Oxford
University Press.
-
Mills, C. Wright. 2000 [1959].
The Sociological
Imagination
. New York: Oxford University Press.
-
Tyler, Edward. B. 1958 [1871].
Primitive Culture:
Researches into the Development of Mythology, Philosophy, Religion,
Art, and Custom
. Gloucester, MA: Smith.
-
Weber, Max. 1946. From
Max Weber: Essays in Sociology
,
edited by H.H. Gerth and C. Wright Mills. New York: Oxford
University Press.


9.3: Team Building Exercises
The following is a list of exercises designed to build a strong and supportive learning environment. Feel free to add any additional activities you have discovered.
High Fives
Exercise
- Begin by having the class move throughout the space for about thirty seconds to one minute so that the group feels comfortable navigating the space without collisions.
- Next, have each member of the class hold up one hand and continue moving throughout the space, but each time they make eye contact with someone in the group they must give each other a high five and say “hey!” in an enthusiastic way.
Purpose
- This activity forces the group to acknowledge each other and interact in a positive way. When someone looks at you and gets excited and you are met with an enthusiastic high five you immediately feel accepted and you get a small confidence boost. You will notice within thirty seconds how the energy in the room shifts and how forced smiles that occured in the beginning of the activity will turn into genuine smiles of happiness.
Name Game
Exercise
- Have everyone in the group gather into a circle.
- Have each member of the group introduce themselves by saying his/her name and offer a few interesting facts about themselves.
- Create a movement/dance move based on one interesting fact and perform this move while saying the student's name.
- Have everyone in the group repeat the movement and name several times.
- Repeat with each member of the group.
Purpose
- This activity allows the group to learn each other's names quickly by connecting the name to a physical movement while also providing information that can help students connect through shared interests or experiences.
Ground Rules
Exercise
- Break the class up into pairs of two and have each pair create four rules that they feel the class should follow throughout the semester/camp. Some examples include: Treat others with respect, no favoritism, etc.
- Once each pair has agreed on the rules have them merge with another group and form a group of four. Each group must share the rules each have created and select only four to agree on together. This will narrow the rules from eight to four.
- Once each group of four has agreed on which four rules they feel should be adopted by the group, have two groups of four merge into a group of eight. In this new group each previous group must share the rules each have created and select only four to agree on together. This will narrow the rules from eight to four.
- Once each group of eight has agreed on which four rules they feel should be adopted by the group, have two groups of eight merge into a group of sixteen. In this new group each previous group must share the rules each have created and select only four to agree on together. This will narrow the rules from eight to four.
- Repeat this process until the class is back together forming one large group. Decide which four rules will be adopted by the group and then write the rules down.
Purpose
- This exercise provides many wonderful outcomes. Each student is allowed to contribute rules and ideas that address fears they carried into the group. Each student also must listen and collaborate with other members of the group to agree on rules that address both students’ needs. This allows the student to feel confident speaking up for his/her needs, understand compromise and think about the overall needs of the group. This also secretly has the students create their own culture of respect and trust while verbally creating the culture of respect and trust.
Spotlight
Exercise
- Place a chair in the front of the room.
- Have each member of the class or group sit and face the chair.
- One at a time, have each member of the class come up to the chair and sit down and let the audience look at them for one minute.
- In order to pass the time the student is allowed to answer one of the following questions. This question is selected by the instructor and the student receives only one question. The student is not required to talk, the question is there to help the student take their mind off of being stared at for a period of time. The point of the exercise is to gain experience being themselves in front of a group and allowing the group to see them, not to answer a question or entertain the group.
Questions
- What are your dreams?
- What is the most embarrassing thing that has ever happened to you?
- Who is your hero and why?
- If you could go back in time and relive any day of your life to experience it again, what would it be? Why?
- What gives you strength and what is your greatest fear?
- What was the happiest day of your life? What was the worst day of your life?
- What was an obstacle you were forced to overcome?
- What was the time you were your most vulnerable?
- What was a time you helped someone and a time that you betrayed someone?
- If you could go back in time and change one thing in your life what would it be?
- What was a time you felt helpless and a time you were on top of the world?
- What would your best friend say about you? What would your worst enemy say?
- What do you hope to be and what do you hope you never become?
Once the minute has passed the student is thanked and the next member of the class goes to the spotlight. Once everyone in the group has participated, gather the group into a circle to discuss the experience.
Purpose
- This is without question one of, if not the best team building exercise I have ever witnessed. The activity not only provides the experience of being the center of attention, it forces the student to confront the most terrifying aspects of stage fright by having them be the center of attention while also being unprepared. The student is then offered the ability to escape the uncomfortable situation by sharing an experience in his/her life that is true. Great acting requires truth and you will never be able to connect truthfully to a character or another human being if you cannot be comfortable being vulnerable in front of a group and connecting to a true part of yourself.  Once a fear is confronted and conquered, acceleration can occur.
- There is another magical quality about this exercise. All friendships and meaningful relationships occur when we allow ourselves to be vulnerable with another person and share something true about ourselves that asks the other individual not to judge us. When the other person listens, does not judge, and shares something in return we have a strong foundation on which to build the relationship.
- In the discussion you will find that people feel more connected with each other and that people will share how they now feel bonded with others who share the same struggles.


12.1: Module Introduction
I like the night. Without the dark, we’d never see the stars.
Stephenie Meyer
Twilight
“I think that we sare like stars. Something happens to burst us open; but when we burst open and think we are dying; we’re actually turning into a supernova. And then when we look at ourselves again, we see that we’re suddenly more beautiful than we ever were before!”
C. JoyBell C.
This module looks at how stars form and develop over time, including the less-massive stars, like the Sun, and more-massive stars.
Objectives
Upon completion of this module, the student will be able to:Upon completion of this module, the student will be able to:
- Define the biology-like terms astronomers use when describing stars.
- Identify the characteristics of low-mass stars.
- Describe how white dwarfs, novae, and Type 1a supernovae are formed in low-mass stars.
- Identify the characteristics of high-mass stars.
- Describe how neutron stars, supernovae, and black holes are formed in high-mass stars.
- Define GRBs.
- Describe GRB characteristics


6.21: Introduction to Continuous Probability Distribution
What you’ll learn to do: Use a probability distribution for a continuous random variable to estimate probabilities and identify unusual events.
In the last section, we studied discrete (listable) random variables and their distributions. Now we explore continuous (decimal valued) random variables that can take on values anywhere in an interval. For example, a person’s exact weight without rounding is a continuous random variable. If rounded to the nearest pound, weight is a discrete random variable. Decimal valued numbers arise often in real life, often in measuring things such as weight or length. To best study real life data that has values lying all over an interval, we need to build a solid foundation in continuous probability distributions.
Contributors and Attributions
CC licensed content, Shared previously
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution