15.1: Random Selection
Introduction
The usual treatments deal with a single random variable or a fixed, finite number of random variables, considered jointly. However, there are many common applications in which we select at random a member of a class of random variables and observe its value, or select a random number of random variables and obtain some function of those selected. This is formulated with the aid of a counting or selecting random variable \(N\), which is nonegative, integer valued. It may be independent of the class selected, or may be related in some sequential way to members of the class. We consider only the independent case. Many important problems require optional random variables, sometimes called Markov times . These involve more theory than we develop in this treatment.
Some common examples:
Total demand of \(N\) customers— \(N\) independent of the individual demands.
Total service time for \(N\) units— \(N\) independent of the individual service times.
Net gain in \(N\) plays of a game— \(N\) independent of the individual gains.
Extreme values of \(N\) random variables— \(N\) independent of the individual values.
Random sample of size \(N\)— \(N\) is usually determined by propereties of the sample observed.
Decide when to play on the basis of past results— \(N\) dependent on past
A useful model—random sums
As a basic model, we consider the sum of a random number of members of an iid class. In order to have a concrete interpretation to help visualize the formal patterns, we think of the demand of a random number of customers. We suppose the number of customers N is independent of the individual demands. We formulate a model to be used for a variety of applications.
A
basic sequence
\(\{X_n: 0 \le n\}\) [Demand of \(n\) customers]
An incremental sequence \(\{Y_n:0 \le n\}\) [Individual demands]
These are related as follows:
\(X_n = \sum_{k = 0}^{n} Y_k\) for \(n \ge 0\) and \(X_n = 0\) for \(n < 0\) \(Y_n = X_n - X_{n - 1}\) for all \(n\)
A counting random variable \(N\). If \(N = n\) then \(n\) of the \(Y_k\) are added to give the compound demand \(D\) (the random sum)
\(D = \sum_{k = 0}^{N} Y_k = \sum_{k = 0}^{\infty} I_{[N = k]} X_k = \sum_{k = 0}^{\infty} I_{\{k\}} (N) X_k\)
Note . In some applications the counting random variable may take on the idealized value \(\infty\). For example, in a game that is played until some specified result occurs, this may never happen, so that no finite value can be assigned to \(N\). In such a case, it is necessary to decide what value \(X_{\infty}\) is to be assigned. For \(N\) independent of the \(Y_n\) (hence of the \(X_n\)), we rarely need to consider this possibility.
Independent selection from an iid incremental sequence
We assume throughout
, unless specifically stated otherwise, that:
\(X_0 = Y_0 = 0\)
\(\{Y_k: 1 \le k\}\) is iid
\(\{N, Y_k: 0 \le k\}\) is an independent class
We utilize repeatedly two important propositions:
\(E[h(D)|N = n] = E[h(X_n)]\), \(n \ge 0\)
\(M_D (s) = g_N [M_Y (s)]\). If the \(Y_n\) are nonnegative integer valued, then so is \(D\) and \(g_D (s) = g_N[g_Y (s)]\)
DERIVATION
We utilize properties of generating functions, moment generating functions, and conditional expectation.
\(E[I_{\{n\}} (N) h(D)] = E[h(D)|N = n] P(N = n)\) by definition of conditional expectation, given an event, Now,
\(I_{\{n\}} (N) h(D) = I_{\{n\}} (N) h(X_n)\) and \(E[I_{\{n\}} (N) h(X_n)] = P(N = n) E[h(X_n)]\). Hence
\(E[h(D) |N = n] P(N = n) = P(N = n) E[h(X_n)]\). Division by \(P(N = n)\) gives the desired result.
By the law of total probability (CE1b), \(M_D(s)= E[e^{sD}] = E\{E[e^{sD} |N]\}\). By proposition 1 and the product rule for moment generating functions,
\(E[e^{sD}|N = n] = E[e^{sX_n}] = \prod_{k = 1}^{n} E[e^{sY_k}] = M_Y^n (s)\)
Hence
\(M_D(s) = \sum_{n = 0}^{\infty} M_Y^n (s) P(N = n) = g_N[M_Y (s)]\)
A parallel argument holds for \(g_D\)
— □
Remark . The result on \(M_D\) and \(g_D\) may be developed without use of conditional expectation.
in the integer-valued case.
\(M_D(s) = E[e^{sD}] = \sum_{k = 0}^{\infty} E[I_{\{N = n\}} e^{sX_n}] = \sum_{k = 0}^{\infty} P(N = n) E[e^{sX_n}]\)
\(= \sum_{k = 0}^{\infty} P(N = n) M_Y^n (s) = g_N [M_Y (s)]\)
— □
Example \(\PageIndex{1}\) A service shop
Suppose the number \(N\) of jobs brought to a service shop in a day is Poisson (8). One fourth of these are items under warranty for which no charge is made. Others fall in one of two categories. One half of the arriving jobs are charged for one hour of shop time; the remaining one fourth are charged for two hours of shop time. Thus, the individual shop hour charges \(Y_k\) have the common distribution
\(Y =\) [0 1 2] with probabilities \(PY =\) [1/4 1/2 1/4]
Make the basic assumptions of our model. Determine \(P(D \le 4)\).
Solution
\(g_N(s) = e^{8(s - 1)} g_Y (s) = \dfrac{1}{4} (1 + 2s + s^2)\)
According to the formula developed above,
\(g_D (s) = g_N [g_Y (s)] = \text{exp} ((8/4) (1 + 2s + s^2) - 8) = e^{4s} e^{2s^2} e^{-6}\)
Expand the exponentials in power series about the origin, multiply out to get enough terms. The result of straightforward but somewhat tedious calculations is
\(g_D (s) = e^{-6} ( 1 + 4s + 10s^2 + \dfrac{56}{3} s^3 + \dfrac{86}{3} s^4 + \cdot\cdot\cdot)\)
Taking the coefficients of the generating function, we get
\(P(D \le 4) \approx e^{-6} (1 + 4 + 10 + \dfrac{56}{3} + \dfrac{86}{3}) = e^{-6} \dfrac{187}{3} \approx 0.1545\)
Example \(\PageIndex{2}\) A result on Bernoulli trials
Suppose the counting random variable \(N\) ~ binomial \((n, p)\) and \(Y_i = I_{E_i}\), with \(P(E_i) = p_0\). Then
\(g_N = (q + ps)^n\) and \(g_Y (s) = q_0 + p_0 s\)
By the basic result on random selection, we have
\(g_D (s) = g_N [g_Y(s)] = [q + p(q_0 + p_0 s)]^n = [(1 - pp_0) + pp_0 s]^n\)
so that \(D\) ~ binomial \((n, pp_0)\).
In the next section we establish useful m-procedures for determining the generating function g D and the moment generating function \(M_D\) for the compound demand for simple random variables, hence for determining the complete distribution. Obviously, these will not work for all problems. It may helpful, if not entirely sufficient, in such cases to be able to determine the mean value \(E[D]\) and variance \(\text{Var} [D]\). To this end, we establish the following expressions for the mean and variance.
Example \(\PageIndex{3}\) Mean and variance of the compound demand
\(E[D] = E[N]E[Y]\) and \(\text{Var} [D] = E[N] \text{Var} [Y] + \text{Var} [N] E^2 [Y]\)
DERIVATION
\(E[D] = E[\sum_{n = 0}^{\infty} I_{\{N = n\}} X_n] = \sum_{n = 0}^{\infty} P(N = n) E[X_n]\)
\(= E[Y] \sum_{n = 0}^{\infty} n P(N = n) = E[Y] E[N]\)
\(E[D^2] = \sum_{n = 0}^{\infty} P(N = n) E[X_n^2] = \sum_{n = 0}^{\infty} P(N = n) \{\text{Var} [X_n] + E^2 [X_n]\}\)
\(= \sum_{n = 0}^{\infty} P(N = n) \{n \text{Var} [Y] = n^2 E^2 [Y]\} = E[N] \text{Var} [Y] + E[N^2] E^2[Y]\)
Hence
\(\text{Var} [D] = E[N] \text{Var} [Y] + E[N^2] E^2 [Y] - E[N]^2 E^2[Y] = E[N] \text{Var} [Y] + \text{Var} [N] E^2[Y]\)
Example \(\PageIndex{4}\) Mean and variance for Example 15.1.1
\(E[N] = \text{Var} [N] = 9\). By symmetry \(E[Y] = 1\). \(\text{Var} [Y] = 0.25(0 + 2 + 4) - 1 = 0.5\). Hence,
\(E[D] = 8 \cdot 1 = 8\), \(\text{Var} [D] = 8 \cdot 0.5 + 8 \cdot 1 = 12\)
Calculations for the compound demand
We have m-procedures for performing the calculations necessary to determine the distribution for a composite demand \(D\) when the counting random variable \(N\) and the individual demands \(Y_k\) are simple random variables with not too many values. In some cases, such as for a Poisson counting random variable, we are able to approximate by a simple random variable.
The procedure gend
If the \(Y_i\) are nonnegative, integer valued, then so is \(D\), and there is a generating function. We examine a strategy for computation which is implemented in the m-procedure gend . Suppose
\(g_N (s) = p_0 + p_1 s + p_2 s^2 + \cdot\cdot\cdot p_n s^n\)
\(g_Y (s) = \pi_0 + \pi_1 s + \pi_2 s^2 + \cdot\cdot\cdot \pi_m s^m\)
The coefficients of \(g_N\) and \(g_Y\) are the probabilities of the values of \(N\) and \(Y\), respectively. We enter these and calculate the coefficients for powers of \(g_Y\):
\(\begin{array} {lcr} {gN = [p_0\ p_1\ \cdot\cdot\cdot\ p_n]} & {1 \times (n + 1)} & {\text{Coefficients of } g_N} \\ {y = [\pi_0\ \pi_1\ \cdot\cdot\cdot\ \pi_n]} & {1 \times (m + 1)} & {\text{Coefficients of } g_Y} \\ {\ \ \ \ \ \cdot\cdot\cdot} & { } & { } \\ {y2 = \text{conv}(y,y)} & {1 \times (2m + 1)} & {\text{Coefficients of } g_Y^2} \\ {y3 = \text{conv}(y,y2)} & {1 \times (3m + 1)} & {\text{Coefficients of } g_Y^3} \\ {\ \ \ \ \ \cdot\cdot\cdot} & { } & { } \\ {yn = \text{conv}(y,y(n - 1))} & {1 \times (nm + 1)} & {\text{Coefficients of } g_Y^n}\end{array}\)
We wish to generate a matrix \(P\) whose rows contain the joint probabilities. The probabilities in the \(i\)th row consist of the coefficients for the appropriate power of \(g_Y\) multiplied by the probability \(N\) has that value. To achieve this, we need a matrix, each of whose \(n + 1\) rows has \(nm + 1\) elements, the length of \(yn\). We begin by “preallocating” zeros to the rows. That is, we set \(P = \text{zeros}(n + 1, n\ ^*\ m + 1)\). We then replace the appropriate elements of the successive rows. The replacement probabilities for the \(i\)th row are obtained by the convolution of \(g_Y\) and the power of \(g_Y\) for the previous row. When the matrix \(P\) is completed, we remove zero rows and columns, corresponding to missing values of \(N\) and \(D\) (i.e., values with zero probability). To orient the joint probabilities as on the plane, we rotate \(P\) ninety degrees counterclockwise. With the joint distribution, we may then calculate any desired quantities.
Example \(\PageIndex{5}\) A compound demand
The number of customers in a major appliance store is equally likely to be 1, 2, or 3. Each customer buys 0, 1, or 2 items with respective probabilities 0.5, 0.4, 0.1. Customers buy independently, regardless of the number of customers. First we determine the matrices representing \(g_N\) and \(g_Y\). The coefficients are the probabilities that each integer value is observed. Note that the zero coefficients for any missing powers must be included.
gN = (1/3)*[0 1 1 1]; % Note zero coefficient for missing zero power gY = 0.1*[5 4 1]; % All powers 0 thru 2 have positive coefficients gend Do not forget zero coefficients for missing powers Enter the gen fn COEFFICIENTS for gN gN % Coefficient matrix named gN Enter the gen fn COEFFICIENTS for gY gY % Coefficient matrix named gY Results are in N, PN, Y, PY, D, PD, P May use jcalc or jcalcf on N, D, P To view distribution for D, call for gD disp(gD) % Optional display of complete distribution 0 0.2917 1.0000 0.3667 2.0000 0.2250 3.0000 0.0880 4.0000 0.0243 5.0000 0.0040 6.0000 0.0003 EN = N*PN' EN = 2 EY = Y*PY' EY = 0.6000 ED = D*PD' ED = 1.2000 % Agrees with theoretical EN*EY P3 = (D>=3)*PD' P3 = 0.1167 [N,D,t,u,PN,PD,PL] = jcalcf(N,D,P); EDn = sum(u.*P)./sum(P); disp([N;EDn]') 1.0000 0.6000 % Agrees with theoretical E[D|N=n] = n*EY 2.0000 1.2000 3.0000 1.8000 VD = (D.^2)*PD' - ED^2 VD = 1.1200 % Agrees with theoretical EN*VY + VN*EY^2
Example \(\PageIndex{6}\) A numerical example
\(g_N (s) = \dfrac{1}{5} (1 + s + s^2 + s^3 + s^4)\) \(g_Y (s) = 0.1 (5s + 3s^2 + 2s^3\)
Note that the zero power is missing from \(gY\). corresponding to the fact that \(P(Y = 0) = 0\).
gN = 0.2*[1 1 1 1 1]; gY = 0.1*[0 5 3 2]; % Note the zero coefficient in the zero position gend Do not forget zero coefficients for missing powers Enter the gen fn COEFFICIENTS for gN gN Enter the gen fn COEFFICIENTS for gY gY Results are in N, PN, Y, PY, D, PD, P May use jcalc or jcalcf on N, D, P To view distribution for D, call for gD disp(gD) % Optional display of complete distribution 0 0.2000 1.0000 0.1000 2.0000 0.1100 3.0000 0.1250 4.0000 0.1155 5.0000 0.1110 6.0000 0.0964 7.0000 0.0696 8.0000 0.0424 9.0000 0.0203 10.0000 0.0075 11.0000 0.0019 12.0000 0.0003
p3 = (D == 3)*PD' % P(D=3) P3 = 0.1250 P4_12 = ((D >= 4)&(D <= 12))*PD' P4_12 = 0.4650 % P(4 <= D <= 12)
Example \(\PageIndex{7}\) Number of successes for random number \(N\) of trials.
We are interested in the number of successes in \(N\) trials for a general counting random variable. This is a generalization of the Bernoulli case in Example 15.1.2 . Suppose, as in Example 15.1.2 , the number of customers in a major appliance store is equally likely to be 1, 2, or 3, and each buys at least one item with probability \(p = 0.6\). Determine the distribution for the number \(D\) of buying customers.
Solution
We use \(gN\), \(gY\), and gend.
gN = (1/3)*[0 1 1 1]; % Note zero coefficient for missing zero power gY = [0.4 0.6]; % Generating function for the indicator function gend Do not forget zero coefficients for missing powers Enter gen fn COEFFICIENTS for gN gN Enter gen fn COEFFICIENTS for gY gY Results are in N, PN, Y, PY, D, PD, P May use jcalc or jcalcf on N, D, P To view distribution for D, call for gD disp(gD) 0 0.2080 1.0000 0.4560 2.0000 0.2640 3.0000 0.0720
The procedure gend is limited to simple \(N\) and \(Y_k\), with nonnegative integer values. Sometimes, a random variable with unbounded range may be approximated by a simple random variable. The solution in the following example utilizes such an approximation procedure for the counting random variable \(N\).
Example \(\PageIndex{8}\) Solution of the shop time Example 15.1.1
The number \(N\) of jobs brought to a service shop in a day is Poisson (8). The individual shop hour charges \(Y_k\) have the common distribution \(Y =\) [0 1 2] with probabilities \(PY =\) [1/4 1/2 1/4].
Under the basic assumptions of our model, determine \(P(D \le 4)\).
Solution
Since Poisson \(N\) is unbounded, we need to check for a sufficient number of terms in a simple approximation. Then we proceed as in the simple case.
pa = cpoisson(8,10:5:30) % Check for sufficient number of terms pa = 0.2834 0.0173 0.0003 0.0000 0.0000 p25 = cpoisson(8,25) % Check on choice of n = 25 p25 = 1.1722e-06 gN = ipoisson(8,0:25); % Approximate gN gY = 0.25*[1 2 1]; gend Do not forget zero coefficients for missing powers Enter gen fn COEFFICIENTS for gN gN Enter gen fn COEFFICIENTS for gY gY Results are in N, PN, Y, PY, D, PD, P May use jcalc or jcalcf on N, D, P To view distribution for D, call for gD disp(gD(D<=20,:)) % Calculated values to D = 50 0 0.0025 % Display for D <= 20 1.0000 0.0099 2.0000 0.0248 3.0000 0.0463 4.0000 0.0711 5.0000 0.0939 6.0000 0.1099 7.0000 0.1165 8.0000 0.1132 9.0000 0.1021 10.0000 0.0861 11.0000 0.0684 12.0000 0.0515 13.0000 0.0369 14.0000 0.0253 15.0000 0.0166 16.0000 0.0105 17.0000 0.0064 18.0000 0.0037 19.0000 0.0021 20.0000 0.0012 sum(PD) % Check on sufficiency of approximation ans = 1.0000 P4 = (D<=4)*PD' P4 = 0.1545 % Theoretical value (4 places) = 0.1545 ED = D*PD' ED = 8.0000 % Theoretical = 8 (Example 15.1.4) VD = (D.^2)*PD' - ED^2 VD = 11.9999 % Theoretical = 12 (Example 15.1.4)
The m-procedures mgd and jmgd
The next example shows a fundamental limitation of the gend procedure. The values for the individual demands are not limited to integers, and there are considerable gaps between the values. In this case, we need to implement the moment generating function \(M_D\) rather than the generating function \(g_D\).
In the generating function case, it is as easy to develop the joint distribution for \(\{N, D\}\) as to develop the marginal distribution for \(D\). For the moment generating function, the joint distribution requires considerably more computation. As a consequence, we find it convenient to have two m-procedures: mgd for the marginal distribution and jmgd for the joint distribution.
Instead of the convolution procedure used in gend to determine the distribution for the sums of the individual demands, the m-procedure mgd utilizes the m-function mgsum to obtain these distributions. The distributions for the various sums are concatenated into two row vectors, to which csort is applied to obtain the distribution for the compound demand. The procedure requires as input the generating function for \(N\) and the actual distribution, \(Y\) and \(PY\), for the individual demands. For \(gN\), it is necessary to treat the coefficients as in gend. However, the actual values and probabilities in the distribution for Y are put into a pair of row matrices. If \(Y\) is integer valued, there are no zeros in the probability matrix for missing values.
Example \(\PageIndex{9}\) Noninteger values
A service shop has three standard charges for a certain class of warranty services it performs: $10, $12.50, and $15. The number of jobs received in a normal work day can be considered a random variable \(N\) which takes on values 0, 1, 2, 3, 4 with equal probabilities 0.2. The job types for arrivals may be represented by an iid class \(\{Y_i: 1 \le i \le 4\}\), independent of the arrival process. The \(Y_i\) take on values 10, 12.5, 15 with respective probabilities 0.5, 0.3, 0.2. Let \(C\) be the total amount of services rendered in a day. Determine the distribution for \(C\).
Solution
gN = 0.2*[1 1 1 1 1]; % Enter data Y = [10 12.5 15]; PY = 0.1*[5 3 2]; mgd % Call for procedure Enter gen fn COEFFICIENTS for gN gN Enter VALUES for Y Y Enter PROBABILITIES for Y PY Values are in row matrix D; probabilities are in PD. To view the distribution, call for mD. disp(mD) % Optional display of distribution 0 0.2000 10.0000 0.1000 12.5000 0.0600 15.0000 0.0400 20.0000 0.0500 22.5000 0.0600 25.0000 0.0580 27.5000 0.0240 30.0000 0.0330 32.5000 0.0450 35.0000 0.0570 37.5000 0.0414 40.0000 0.0353 42.5000 0.0372 45.0000 0.0486 47.5000 0.0468 50.0000 0.0352 52.5000 0.0187 55.0000 0.0075 57.5000 0.0019 60.0000 0.0003
We next recalculate Example 15.1.6 , above, using mgd rather than gend.
Example \(\PageIndex{10}\) Recalculation of Example 15.1.6
In Example 15.1.6, we have
\(g_N (s) = \dfrac{1}{5} (1 + s + s^2 + s^3 + s^4)\) \(g_Y (s) = 0.1 (5s + 3s^2 + 2s^3)\)
The means that the distribution for \(Y\) is \(Y =\) [1 2 3] and \(PY =\) 0.1 * [5 3 2].
We use the same expression for \(gN\) as in Example 15.1.6.
gN = 0.2*ones(1,5); Y = 1:3; PY = 0.1*[5 3 2]; mgd Enter gen fn COEFFICIENTS for gN gN Enter VALUES for Y Y Enter PROBABILITIES for Y PY Values are in row matrix D; probabilities are in PD. To view the distribution, call for mD. disp(mD) 0 0.2000 1.0000 0.1000 2.0000 0.1100 3.0000 0.1250 4.0000 0.1155 5.0000 0.1110 6.0000 0.0964 7.0000 0.0696 8.0000 0.0424 9.0000 0.0203 10.0000 0.0075 11.0000 0.0019 12.0000 0.0003 P3 = (D==3)*PD' P3 = 0.1250 ED = D*PD' ED = 3.4000 P_4_12 = ((D>=4)&(D<=12))*PD' P_4_12 = 0.4650 P7 = (D>=7)*PD' P7 = 0.1421
As expected, the results are the same as those obtained with gend.
If it is desired to obtain the joint distribution for \(\{N, D\}\), we use a modification of mgd called jmgd . The complications come in placing the probabilities in the \(P\) matrix in the desired positions. This requires some calculations to determine the appropriate size of the matrices used as well as a procedure to put each probability in the position corresponding to its \(D\) value. Actual operation is quite similar to the operation of mgd, and requires the same data format.
A principle use of the joint distribution is to demonstrate features of the model, such as \(E[D|N = n] = nE[Y]\), etc. This, of course, is utilized in obtaining the expressions for \(M_D (s)\) in terms of \(g_N (s)\) and \(M_Y (s)\). This result guides the development of the computational procedures, but these do not depend upon this result. However, it is usually helpful to demonstrate the validity of the assumptions in typical examples.
Remark . In general, if the use of gend is appropriate, it is faster and more efficient than mgd (or jmgd). And it will handle somewhat larger problems. But both m-procedures work quite well for problems of moderate size, and are convenient tools for solving various “compound demand” type problems.


8.5: The Light-Dependent Reactions of Photosynthesis - Absorption of Light
- Differentiate between chlorophyll and carotenoids.
Absorption of Light
Light energy initiates the process of photosynthesis when pigments absorb the light. Organic pigments have a narrow range of energy levels that they can absorb. Energy levels lower than those represented by red light are insufficient to raise an orbital electron to an excited, or quantum, state. Energy levels higher than those in blue light will physically tear the molecules apart, a process called bleaching. For example, retinal pigments can only “see” (absorb) 700 nm to 400 nm light; this is visible light. For the same reasons, plant pigment molecules absorb only light in the wavelength range of 700 nm to 400 nm; plant physiologists refer to this range for plants as photosynthetically-active radiation.
The visible light seen by humans as the color white light actually exists in a rainbow of colors in the electromagnetic spectrum, with violet and blue having shorter wavelengths and, thus, higher energy. At the other end of the spectrum, toward red, the wavelengths are longer and have lower energy.
Understanding Pigments
Different kinds of pigments exist, each of which has evolved to absorb only certain wavelengths or colors of visible light. Pigments reflect or transmit the wavelengths they cannot absorb, making them appear in the corresponding color.
Chlorophylls and carotenoids are the two major classes of photosynthetic pigments found in plants and algae; each class has multiple types of pigment molecules. There are five major chlorophylls: a , b , c and d, along with a related molecule found in prokaryotes called bacteriochlorophyll.
With dozens of different forms, carotenoids are a much larger group of pigments. The carotenoids found in fruit, such as the red of tomato (lycopene), the yellow of corn seeds (zeaxanthin), or the orange of an orange peel (β-carotene), are used to attract seed-dispersing organisms. In photosynthesis, carotenoids function as photosynthetic pigments that are very efficient molecules for the disposal of excess energy. When a leaf is exposed to full sun, the light-dependent reactions are required to process an enormous amount of energy; if that energy is not handled properly, it can do significant damage. Therefore, many carotenoids are stored in the thylakoid membrane to absorb excess energy and safely release that energy as heat.
Each type of pigment can be identified by the specific pattern of wavelengths it absorbs from visible light, which is the absorption spectrum. Chlorophyll a absorbs light in the blue-violet region, while chlorophyll b absorbs red-blue light. Neither a or b absorb green light; because green is reflected or transmitted, chlorophyll appears green. Carotenoids absorb light in the blue-green and violet region and reflect the longer yellow, red, and orange wavelengths.
Many photosynthetic organisms have a mixture of pigments. In this way organisms can absorb energy from a wider range of wavelengths. Not all photosynthetic organisms have full access to sunlight. Some organisms grow underwater where light intensity and quality decrease and change with depth. Other organisms grow in competition for light. Plants on the rainforest floor must be able to absorb any light that comes through because the taller trees absorb most of the sunlight and scatter the remaining solar radiation
When studying a photosynthetic organism, scientists can determine the types of pigments present by using a spectrophotometer. These instruments can differentiate which wavelengths of light a substance can absorb. Spectrophotometers measure transmitted light and compute its absorption. By extracting pigments from leaves and placing these samples into a spectrophotometer, scientists can identify which wavelengths of light an organism can absorb.
Key Points
- Plant pigment molecules absorb only light in the wavelength range of 700 nm to 400 nm; this range is referred to as photosynthetically-active radiation.
- Violet and blue have the shortest wavelengths and the most energy, whereas red has the longest wavelengths and carries the least amount of energy.
- Pigments reflect or transmit the wavelengths they cannot absorb, making them appear in the corresponding color.
- Chorophylls and carotenoids are the major pigments in plants; while there are dozens of carotenoids, there are only five important chorophylls: a , b , c , d, and bacteriochlorophyll.
- Chlorophyll a absorbs light in the blue-violet region, chlorophyll b absorbs red-blue light, and both a and b reflect green light (which is why chlorophyll appears green).
- Carotenoids absorb light in the blue-green and violet region and reflect the longer yellow, red, and orange wavelengths; these pigments also dispose excess energy out of the cell.
Key Terms
- chlorophyll : Any of a group of green pigments that are found in the chloroplasts of plants and in other photosynthetic organisms such as cyanobacteria.
- carotenoid : Any of a class of yellow to red plant pigments including the carotenes and xanthophylls.
- spectrophotometer : An instrument used to measure the intensity of electromagnetic radiation at different wavelengths.


2.5: Exponents
Overview
- Exponential Notation
- Reading Exponential Notation
- The Order of Operations
Exponential Notation
In Section 2.4 we were reminded that multiplication is a description for repeated addition. A natural question is "Is there a description for repeated multiplication?" The answer is yes. The notation that describes repeated multiplication is exponential notation.
Factors
In multiplication, the numbers being multiplied together are called factors . In repeated multiplication, all the factors are the same. In nonrepeated multiplication, none of the factors are the same. For example,
\(18 \cdot 18 \cdot 18 \cdot 18\) Repeated Multiplication of 18. All four factors, 18 are the same.
\(x \cdot x \cdot x \cdot x \cdot x\) Repeated multiplication of \(x\). All five factors, \(x\), are the same.
\(3 \cdot 7 \cdot a\) Nonrepeated multiplication. None of the factors are the same.
Exponential notation is used to show repeated multiplication of the same factor. The notation consists of using a superscript on the factor that is repeated . The superscript is called an exponent .
Exponential Notation If \(x\) is any real number and \(n\) is a natural number, then
\(x^{n}=\underbrace{x \cdot x \cdot x \cdot \ldots \cdot x}_{n \text { factors of } x}\)
An exponent records the number of identical factors in multiplication.
Note that the definition for exponential notation only has meaning for natural number exponents. We will extend this notation to include other numbers as exponents later.
Sample Set A
\(7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 \cdot 7 = 7^6\)
The repeated factor is 7. The exponent 6 records the fact that 7 appears 6 times in the multiplication.
\(x \cdot x \cdot x \cdot x = x^4\)
The repeated factor is \(x\). The exponent 4 records the fact that \(x\) appears 4 times in the multiplication.
\((2y)(2y)(2y) = (2y)^3\)
The repeated factor is \(2y\). The exponent 3 records the fact that the factor \(2y\) appears 3 times in the multiplication.
\(2yyy = 2y^3\)
The repeated factor is \(y\). The exponent 3 records the fact that the factor \(y\) appears 3 times in the multiplication.
\((a + b)(a + b)(a - b)(a - b)(a - b) = (a + b)^2(a - b)^3\)
The repeated factors are \((a + b)\) and \((a - b)\), \((a + b)\) appearing 2 times and \((a - b)\) appearing 3 times.
Practice Set A
Write each of the following using exponents.
\(a \cdot a \cdot a \cdot a\)
- Answer
-
\(a^4\)
\((3b)(3b)(5c)(5c)(5c)(5c)\)
- Answer
-
\((3b)^2(5c)^4\)
\(2 \cdot 2 \cdot 7 \cdot 7 \cdot 7 \cdot (a - 4)(a - 4)\)
- Answer
-
\(2^2 \cdot 7^3(a - 4)^2\)
\(8xxxyzzzzz\)
- Answer
-
\(8x^3yz^5\)
Sample Set B
\(8x^3\) means \(8 \cdot xxx\) and not \(8x8x8x\). The exponent \(3\) applies only to the factor \(x\) since it is only to the factor \(x\) that the \(3\) is connected.
\((8x)^3\) means \((8x)(8x)(8x)\) since the parentheses indicate that the exponent \(3\) is directly connected to the factor \(8x\). Remember that the grouping symbols ( ) indicate that the quantities inside are to be considered as one single number.
\(34(a+1)^2\) means \(34 \cdot (a+1)(a+1)\) since the exponent \(2\) applies only to the factor \((a+1)\).
Practice Set B
Write each of the following without exponents.
\(4a^3\)
- Answer
-
\(4aaa\)
\((4a)^3\)
- Answer
-
\((4a)(4a)(4a)\)
Sample Set C
Select a number to show that \((2 x)^{2}\) is not always equal to \(2 x^{2}\).
Suppose we choose \(x\) to be 5 . Consider both \((2 x)^{2}\) and \(2 x^{2}\).
\(
\begin{array}{ll}
(2 x)^{2} & 2 x^{2} \\
(2 \cdot 5)^{2} & 2 \cdot 5^{2} \\
(10)^{2} & 2 \cdot 25 \\
100 & \neq 50
\end{array}
\)
Notice that \((2 x)^{2}=2 x^{2}\) only when \(x=0\).
Practice Set C
Select a number to show that \((5x)^2\) is not always equal to \(5x^2\)
- Answer
-
Select \(x=3\). Then \((5 \cdot 3)^2 = (15)^2 = 225\), but \(5 \cdot 3^2 = 5 \cdot 9 = 45\). \(225 \not = 45\)
Reading Exponential Notation
\(ln(x^n)\)
Base
\(x\) is the
base.
Exponent
\(n\) is the
exponent.
Power
The number represented by \(x^n\) is called a
power.
\(x\) to the \(n\)th Power
The term \(x^n\) is read as "\(x\) to the \(n\)th power," or more simply as "\(x\) to the \(n\)th."
\(x\) Squared and \(x\) Cubed
The symbol \(x^2\) is often read as "\(x\) squared," and \(x^3\) is often read as "\(x\) cubed." A natural question is "Why are geometric terms appearing in the exponent expression?" The answer for \(x^3\) is this: \(x^3\) means \(x \cdot x \cdot x\). In geometry, the volume of a rectangular box is found by multiplying the length by the width by the depth. A cube has the same length on each side. If we represent this length by the letter \(x\) then the volume of the cube is \(x \cdot x \cdot x\), which, of course, is described by \(x^3\). (Can you think of why \(x^2\) is read as \(x\) squared?)
Cube with
length = \(x\)
width = \(x\)
depth = \(x\)
Volume = \(xxx\) = \(x^3\)
The Order of Operations
In Section 4.2 we were introduced to the order of operations. It was noted that we would insert another operation before multiplication and division. We can do that now.
- Perform all operations inside grouping symbols beginning with the innermost set.
- Perform all exponential operations as you come to them, moving left-to-right.
- Perform all multiplications and divisions as you come to them, moving left-to-right.
- Perform all additions and subtractions as you come to them, moving left-to-right.
Sample Set D
Use the order of operations to simplify each of the following.
\(2^2 + 5 = 4 + 5 = 9\)
\(5^2 + 3^2 + 10 = 25 + 9 + 10 = 44\)
\(
\begin{aligned}
2^{2}+(5)(8)-1 &=4+(5)(8)-1 \\
&=4+40-1 \\
&=43
\end{aligned}
\)
\(
\begin{aligned}
7 \cdot 6-4^{2}+1^{5} &=7 \cdot 6-16+1 \\
&=42-16+1 \\
&=27
\end{aligned}
\)
\(
\begin{aligned}
(2+3)^{3}+7^{2}-3(4+1)^{2} &=(5)^{3}+7^{2}-3(5)^{2} \\
&=125+49-3(25) \\
&=125+49-75
\end{aligned}
\)
\(
\begin{aligned}
\left[4(6+2)^{3}\right]^{2} &=\left[4(8)^{3}\right]^{2} \\
&=[4(512)]^{2} \\
&=[2048]^{2} \\
&=4,194,304
\end{aligned}
\)
\(
\begin{aligned}
6\left(3^{2}+2^{2}\right)+4^{2} &=6(9+4)+4^{2} \\
&=6(13)+4^{2} \\
&=6(13)+16 \\
&=78+16 \\
&=94
\end{aligned}
\)
\(
\begin{aligned}
\dfrac{6^{2}+2^{2}}{4^{2}+6 \cdot 2^{2}}+\dfrac{1^{3}+8^{2}}{10^{2}-(19)(5)} &=\dfrac{36+4}{16+6 \cdot 4}+\dfrac{1+64}{100-95} \\
&=\dfrac{36+4}{16+24}+\dfrac{1+64}{100-95} \\
&=\dfrac{40}{40}+\dfrac{65}{5} \\
&=1+13 \\
&=14
\end{aligned}
\)
Practice Set D
Use the order of operations to simplify the following.
\(3^2 + 4 \cdot 5\)
- Answer
-
29
\(2^3 + 3^3 - 8 \cdot 4\)
- Answer
-
3
\(1^4 + (2^2 + 4)^2 \div 2^3\)
- Answer
-
9
\([6(10 - 2^3)]^2 - 10^2 - 6^2\)
- Answer
-
8
\(\dfrac{5^2 + 6^2 - 10}{1 + 4^2} + \dfrac{0^4 - 0^5}{7^2 - 6 \cdot 2^3}\)
- Answer
-
3
Exercises
For the following problems, write each of the quantities using exponential notation.
\(b\) to the fourth
- Answer
-
\(b^4\)
\(a\) squared
\(x\) to the eighth
- Answer
-
\(x^8\)
\((-3)\) cubed
\(5\) times \(s\) squared
- Answer
-
\(5s^2\)
\(3\) squared times \(y\) to the fifth
\(a\) cubed minus \((b + 7)\) squared
- Answer
-
\(a^3 - (b + 7)^2\)
\((21 - x)\) cubed plus \(x + 5\) to the seventh
\(xxxxx\)
- Answer
-
\(x^5\)
\((8)(8)xxxx\)
\(2 \cdot 3 \cdot 3 \cdot 3 \cdot 3xxyyyyy\)
- Answer
-
\(2(3^4)x^2y^5\)
\(2 \cdot 2 \cdot 5 \cdot 6 \cdot 6 \cdot 6xyyzzzwwww\)
\(7xx(a + 8)(a + 8)\)
- Answer
-
\(7x^2(a + 8)^2\)
\(10xyy(c + 5)(c + 5)(c + 5)\)
\(4x4x4x4x4x\)
- Answer
-
\((4x)^5\) or \(4^5x^5\)
\((9a)(9a)(9a)(9a)\)
\((-7)(-7)(-7)aabbba(-7)baab\)
- Answer
-
\((-7)^4a^5b^5\)
\((a - 10)(a - 10)(a + 10)\)
\((z + w)(z + w)(z + w)(z - w)(z - w)\)
- Answer
-
\((z + w)^3(z - w)^2\)
\((2y)(2y)2y2y\)
\(3xyxxy - (x + 1)(x + 1)(x + 1)\)
- Answer
-
\(3x^3y^2 - (x + 1)^3\)
For the following problems, expand the quantities so that no exponents appear.
\(4^3\)
\(6^2\)
- Answer
-
\(6 \cdot 6\)
\(7^3y^2\)
\(8x^3y^2\)
- Answer
-
\(8 \cdot x \cdot x \cdot x \cdot y \cdot y\)
\((18x^2y^4)^2\)
\((9a^3b^2)^3\)
- Answer
-
\((9aaabb)(9aaabb)(9aaabb)\) or \(9 \cdot 9 \cdot 9aaaaaaaaabbbbbb\)
\(5x^2(2y^3)^3\)
\(10a^3b^2(3c)^2\)
- Answer
-
\(10aaabb(3c)(3c)\) or \(10 \cdot 3 \cdot 3aaabbcc\)
\((a + 10)^2(a^2 + 10)^2\)
\((x^2 - y^2)(x^2 + y^2)\)
- Answer
-
\((xx - yy)(xx + yy)\)
For the following problems, select a number (or numbers) to show that
\((5x)^2\) is not generally equal to \(5x^2\).
\((7x)^2\) is not generally equal to \(7x^2\)
- Answer
-
Select \(x = 2\). Then, \(196 \not = 28\)
\((a + b)^2\) is not generally equal to \(a^2 + b^2\)
For what real number is \((6a)^2\) equal to \(6a^2\)
- Answer
-
\(0\)
For what real numbers, \(a\) and \(b\), is \((a + b)^2\) equal to \(a^2 + b^2\).
Use the order of operations to simplify the quantities for the following problems.
\(3^2 + 7\)
- Answer
-
\(16\)
\(4^3 - 18\)
\(5^2 + 2(40)\)
- Answer
-
\(105\)
\(8^2 + 3 + 5(2 + 7)\)
\(2^5 + 3(8 + 1)\)
- Answer
-
\(59\)
\(3^4 + 2^4(1 + 5)^3\)
\((6^2 - 4^2) \div 5\)
- Answer
-
\(4\)
\(2^2(10 - 2^3)\)
\((3^4 - 4^3) \div 17\)
- Answer
-
\(1\)
\((4 + 3)^2 + 1 \div (2 \cdot 5\)
\((2^4 + 2^5 - 2^3 \cdot 5)^2 \div 4^2\)
- Answer
-
\(4\)
\(1^6 + 0^8 + 5^2(2 + 8)^3\)
\((7)(16) - 9^2 + 4(1^1 + 3^2)\)
- Answer
-
\(71\)
\(\dfrac{2^3 - 7}{5^2}\)
\(\dfrac{(1 + 6)^2 + 2}{19}\)
- Answer
-
\(\dfrac{51}{19}\)
\(\dfrac{6^2 - 1}{5} + \dfrac{4^3 + (2)(3)}{10}\)
\(\dfrac{(2 + 1)^3 + 2^3 + 1^3}{6^2} - \dfrac{15^2 - [2(5)]^2}{5 \cdot 5^2}\)
\(\dfrac{6^3 - 2 \cdot 10^2}{2^2} + \dfrac{18(2^3 + 7^2)}{2(19) - 3^3}\)
- Answer
-
\(\dfrac{1070}{11}\) or \(97.27\)
Exercises for Review
Use algebraic notation to write the statement "a number divided by eight, plus five, is equal to ten."
Draw a number line that extends from −5 to 5 and place points at all real numbers that are strictly greater than −3 but less than or equal to 2.
- Answer
-
Is every integer a whole number?
Use the commutative property of multiplication to write a number equal to the number \(yx\).
- Answer
-
\(xy\)
Use the distributive property to expand \(3(x+6)\).


6.7: Anno Dom- 40724
Marvilous it may be to see and consider how some kind of wickednes did grow & breake forth here, in a land wher the same was so much witnesed against, and so narrowly looked unto, & severly punished when it was knowne; as in no place more, or so much, that I have known or heard of; insomuch as they have been somewhat censured, even by moderate and good men, for their severitie in punishments. And yet all this could not suppress y e breaking out of sundrie notorious sins, (as this year, besids other, gives us too many sad presidents and instances,) espetially drunkennes and unclainnes; not only incontinencie betweene persons unmaried, for which many both men & women have been punished sharply enough, but some maried persons allso. But that which is worse, even sodomie and bugerie, (things fearfull to name,) have broak forth in this land, oftener then once. I say it may justly be marveled at, and cause us to fear & tremble at the consideration of our corrupte natures, which are so hardly bridled, subdued, & mortified; nay, cannot by any other means but y e powerfull worke & grace of Gods spirite. But (besids this) one reason may be, that y e Divell may carrie a greater spite against the churches of Christ and y e gospell hear, by how much y e more they indeaour to preserve holynes and puritie amongst them, and strictly punisheth the contrary Home [460] when it ariseth either in church or comone wealth; that he might cast a [242] blemishe & staine upon them in y e eyes of [y e ] world, who use to be rash in judgmente. I would rather thinke thus, then that Satane hath more power in these heathen lands, as som have thought, then in more Christian nations, espetially over Gods servants in them.
2. An other reason may be, that it may be in this case as it is with waters when their streames are stopped or dam̅ed up, when they gett passage they flow with more violence, and make more noys and disturbance, then when they are suffered to rune quietly in their owne chanels. So wikednes being here more stopped by strict laws, and y e same more nerly looked unto, so as it cannot rune in a comone road of liberty as it would, and is inclined, it searches every wher, and at last breaks out wher it getts vente.
3. A third reason may be, hear (as I am verily perswaded) is not more evills in this kind, nor nothing nere so many by proportion, as in other places; but they are here more discoverd and seen, and made publick by due serch, inquisition, and due punishment; for y e churches looke narrowly to their members, and y e magistrats over all, more strictly then in other places. Besids, here the people are but few in comparison of other places, which are full & populous, and lye hid, as it were, in a wood or thickett, and many horrible evills by y t means are never seen nor Home [461] knowne; wheras hear, they are, as it were, brought into y e light, and set in y e plaine feeld, or rather on a hill, made conspicuous to y e veiw of all…
And after y e time of y e writīg of these things befell a very sadd accidente of the like foule nature in this govermente, this very year, which I shall now relate. Ther was a youth whose name was Thomas Granger; he was servant to an honest man of Duxbery, being aboute 16. or 17. years of age. (His father & mother lived at the same time at Sityate.) He was this year detected of buggery (and indicted for y e same) with a mare, a cowe, tow goats, five sheep, 2. calves, and a turkey. Horrible [249] it is to mention, but y e truth of y e historie requires it. He was first discovered by one y t accidentally Home [475] saw his lewd practise towards the mare. (I forbear perticulers.) Being upon it examined and com̅itted, in y e end he not only confest y e fact with that beast at that time, but sundrie times before, and at severall times with all y e rest of y e forenamed in his indictmente; and this his free-confession was not only in private to y e magistrats, (though at first he strived to deney it,) but to sundrie, both ministers & others, and afterwards, upon his indictmente, to y e whole court & jury; and confirmed it at his execution. And wheras some of y e sheep could not so well be knowne by his description of them, others with them were brought before him, and he declared which were they, and which were not. And accordingly he was cast by y e jury, and condemned, and after executed about y e 8. of Sept r , 1642. A very sade spectakle it was; for first the mare, and then y e cowe, and y e rest of y e lesser catle, were kild before his face, according to y e law, Levit: 20. 15. and then he him selfe was executed. The catle were all cast into a great & large pitte that was digged of purposs for them, and no use made of any part of them.
Upon y e examenation of this person, and also of a former that had made some sodomiticall attempts upon another, it being demanded of them how they came first to y e knowledge and practice of such wickednes, the one confessed he had long used it in old England; and this youth last spoaken of said he was taught it Home [476] by an other that had heard of such things from some in England when he was ther, and they kept catle togeather. By which it appears how one wicked person may infecte many; and what care all ought to have what servants they bring into their families.
But it may be demanded how came it to pass that so many wicked persons and profane people should so quickly come over into this land, & mixe them selves amongst them? seeing it was religious men y t begane y e work, and they came for religions sake. I confess this may be marveilled at, at least in time to come, when the reasons therof should not be knowne…


5.11: Ratios and Rate (Part 2)
Find Unit Rates
In the last example, we calculated that Bob was driving at a rate of \(\dfrac{175\; miles}{3\; hours}\). This tells us that every three hours, Bob will travel 175 miles. This is correct, but not very useful. We usually want the rate to reflect the number of miles in one hour. A rate that has a denominator of 1 unit is referred to as a unit rate .
A unit rate is a rate with denominator of 1 unit.
Unit rates are very common in our lives. For example, when we say that we are driving at a speed of 68 miles per hour we mean that we travel 68 miles in 1 hour. We would write this rate as 68 miles/hour (read 68 miles per hour). The common abbreviation for this is 68 mph. Note that when no number is written before a unit, it is assumed to be 1. So 68 miles/hour really means 68 miles/1 hour.
Two rates we often use when driving can be written in different forms, as shown:
| Example | Rate | Write | Abbreviate | Read |
|---|---|---|---|---|
| 68 miles in 1 hour | $$\dfrac{68\; miles}{1\; hour}$$ | 68 miles/hour | 68 mph | 68 miles per hour |
| 36 miles to 1 gallon | $$\dfrac{36\; miles}{1\; gallon}$$ | 36 miles/gallon | 36 mpg | 36 miles per gallon |
Another example of unit rate that you may already know about is hourly pay rate. It is usually expressed as the amount of money earned for one hour of work. For example, if you are paid $12.50 for each hour you work, you could write that your hourly (unit) pay rate is $12.50/hour (read $12.50 per hour.)
To convert a rate to a unit rate, we divide the numerator by the denominator. This gives us a denominator of 1.
Anita was paid $384 last week for working 32 hours. What is Anita’s hourly pay rate?
Solution
| Start with a rate of dollars to hours. Then divide. | $384 last week for 32 hours |
| Write as a rate. | $$\dfrac{$384}{32\; hours}$$ |
| Divide the numerator by the denominator. | $$\dfrac{$12}{1\; hour}$$ |
| Rewrite as a rate. | $12 / hour |
Anita’s hourly pay rate is $12 per hour.
Find the unit rate: $630 for 35 hours.
- Answer
-
$18 / hour
Find the unit rate: $684 for 36 hours.
- Answer
-
$19 / hour
Sven drives his car 455 miles, using 14 gallons of gasoline. How many miles per gallon does his car get?
Solution
Start with a rate of miles to gallons. Then divide.
| Write as a rate. | $$\dfrac{455\; miles}{14\; gallons}$$ |
| Divide 455 by 14 to get the unit rate. | $$\dfrac{32.5\; miles}{1\; gallon}$$ |
Sven’s car gets 32.5 miles/gallon, or 32.5 mpg.
Find the unit rate: 423 miles to 18 gallons of gas.
- Answer
-
23.5 mpg
Find the unit rate: 406 miles to 14.5 gallons of gas.
- Answer
-
28 mpg
Find Unit Price
Sometimes we buy common household items ‘in bulk’, where several items are packaged together and sold for one price. To compare the prices of different sized packages, we need to find the unit price. To find the unit price, divide the total price by the number of items. A unit price is a unit rate for one item.
A unit price is a unit rate that gives the price of one item.
The grocery store charges $3.99 for a case of 24 bottles of water. What is the unit price?
Solution
What are we asked to find? We are asked to find the unit price, which is the price per bottle.
| Write as a rate. | $$\dfrac{$3.99}{24\; bottles}$$ |
| Divide to find the unit price. | $$\dfrac{$0.16625}{1\; bottle}$$ |
| Round the result to the nearest penny. | $$\dfrac{$0.17}{1\; bottle}$$ |
The unit price is approximately $0.17 per bottle. Each bottle costs about $0.17.
Find the unit price. Round your answer to the nearest cent if necessary: 24-pack of juice boxes for $6.99
- Answer
-
\(\dfrac{$0.29}{1\; box}\)
Find the unit price. Round your answer to the nearest cent if necessary: 24-pack of bottles of ice tea for $12.72
- Answer
-
\(\dfrac{$0.53}{1\; bottle}\)
Unit prices are very useful if you comparison shop. The better buy is the item with the lower unit price. Most grocery stores list the unit price of each item on the shelves.
Paul is shopping for laundry detergent. At the grocery store, the liquid detergent is priced at $14.99 for 64 loads of laundry and the same brand of powder detergent is priced at $15.99 for 80 loads. Which is the better buy, the liquid or the powder detergent?
Solution
To compare the prices, we first find the unit price for each type of detergent.
| Liquid | Powder | |
| Write as a rate. | $$\dfrac{$14.99}{64\; loads}$$ | $$\dfrac{$15.99}{80\; loads}$$ |
| Find the unit price. | $$\dfrac{$0.234 \ldots}{1\; load}$$ | $$\dfrac{$0.199 \ldots}{1\; load}$$ |
| Round to the nearest cent. |
$0.23/load (23 cents per load.) |
$0.20/load (20 cents per load) |
Now we compare the unit prices. The unit price of the liquid detergent is about $0.23 per load and the unit price of the powder detergent is about $0.20 per load. The powder is the better buy.
Find each unit price and then determine the better buy. Round to the nearest cent if necessary.
Brand A Storage Bags, $4.59 for 40 count, or Brand B Storage Bags, $3.99 for 30 count
- Answer
-
Brand A costs $0.12 per bag. Brand B costs $0.13 per bag. Brand A is the better buy
Find each unit price and then determine the better buy. Round to the nearest cent if necessary.
Brand C Chicken Noodle Soup, $1.89 for 26 ounces, or Brand D Chicken Noodle Soup, $0.95 for 10.75 ounces
- Answer
-
Brand C costs $0.07 per ounce. Brand D costs $0.09 per ounce. Brand C is the better buy
Notice in Example \(\PageIndex{10}\) that we rounded the unit price to the nearest cent. Sometimes we may need to carry the division to one more place to see the difference between the unit prices.
Translate Phrases to Expressions with Fractions
Have you noticed that the examples in this section used the comparison words ratio of, to, per, in, for, on , and from ? When you translate phrases that include these words, you should think either ratio or rate. If the units measure the same quantity (length, time, etc.), you have a ratio. If the units are different, you have a rate. In both cases, you write a fraction.
Translate the word phrase into an algebraic expression: (a) 427 miles per h hours (b) x students to 3 teachers (c) y dollars for 18 hours
Solution
(a) 427 miles per h hours
| Write as a rate. | $$\dfrac{427\; miles}{h\; hours}$$ |
(b) x students to 3 teachers
| Write as a rate. | $$\dfrac{x\; students}{3\; teachers}$$ |
(c) y dollars for 18 hours
| Write as a rate. | $$\dfrac{$y}{18\; hours}$$ |
Translate the word phrase into an algebraic expression. (a) 689 miles per h hours (b) y parents to 22 students (c) d dollars for 9 minutes
- Answer a
-
\(\dfrac{689\; mi}{h\; hours}\)
- Answer b
-
\(\dfrac{y\; parents}{22\; students}\)
- Answer c
-
\(\dfrac{$d}{9\; min}\)
Translate the word phrase into an algebraic expression. (a) m miles per 9 hours (b) x students to 8 buses (c) y dollars for 40 hours
- Answer a
-
\(\dfrac{m\; mi}{9\; h}\)
- Answer b
-
\(\dfrac{x\; students}{8\; buses}\)
- Answer c
-
\(\dfrac{$y}{40\; h}\)
Ratios
Write Ratios as a Simplified Fractions Involving Decimals and Fractions
Write a Ratio as a Simplified Fraction
Rates and Unit Rates
Unit Rate for Cell Phone Plan
Practice Makes Perfect
Write a Ratio as a Fraction
In the following exercises, write each ratio as a fraction.
- 20 to 36
- 20 to 32
- 42 to 48
- 45 to 54
- 49 to 21
- 56 to 16
- 84 to 36
- 6.4 to 0.8
- 0.56 to 2.8
- 1.26 to 4.2
- \(1 \dfrac{2}{3}\) to \(2 \dfrac{5}{6}\)
- \(1 \dfrac{3}{4}\) to \(2 \dfrac{5}{8}\)
- \(4 \dfrac{1}{6}\) to \(3 \dfrac{1}{3}\)
- \(5 \dfrac{3}{5}\) to \(3 \dfrac{3}{5}\)
- $18 to $63
- $16 to $72
- $1.21 to $0.44
- $1.38 to $0.69
- 28 ounces to 84 ounces
- 32 ounces to 128 ounces
- 12 feet to 46 feet
- 15 feet to 57 feet
- 246 milligrams to 45 milligrams
- 304 milligrams to 48 milligrams
- total cholesterol of 175 to HDL cholesterol of 45
- total cholesterol of 215 to HDL cholesterol of 55
- 27 inches to 1 foot 430. 28 inches to 1 foot
Write a Rate as a Fraction
In the following exercises, write each rate as a fraction.
- 140 calories per 12 ounces
- 180 calories per 16 ounces
- 8.2 pounds per 3 square inches
- 9.5 pounds per 4 square inches
- 488 miles in 7 hours
- 527 miles in 9 hours
- $595 for 40 hours
- $798 for 40 hours
Find Unit Rates
In the following exercises, find the unit rate. Round to two decimal places, if necessary.
- 140 calories per 12 ounces
- 180 calories per 16 ounces
- 8.2 pounds per 3 square inches
- 9.5 pounds per 4 square inches
- 488 miles in 7 hours
- 527 miles in 9 hours
- $595 for 40 hours
- $798 for 40 hours
- 576 miles on 18 gallons of gas
- 435 miles on 15 gallons of gas
- 43 pounds in 16 weeks
- 57 pounds in 24 weeks
- 46 beats in 0.5 minute
- 54 beats in 0.5 minute
- The bindery at a printing plant assembles 96,000 magazines in 12 hours. How many magazines are assembled in one hour?
- The pressroom at a printing plant prints 540,000 sections in 12 hours. How many sections are printed per hour?
Find Unit Price
In the following exercises, find the unit price. Round to the nearest cent.
- Soap bars at 8 for $8.69
- Soap bars at 4 for $3.39
- Women’s sports socks at 6 pairs for $7.99
- Men’s dress socks at 3 pairs for $8.49
- Snack packs of cookies at 12 for $5.79
- Granola bars at 5 for $3.69
- CD-RW discs at 25 for $14.99
- CDs at 50 for $4.49
- The grocery store has a special on macaroni and cheese. The price is $3.87 for 3 boxes. How much does each box cost?
- The pet store has a special on cat food. The price is $4.32 for 12 cans. How much does each can cost?
In the following exercises, find each unit price and then identify the better buy. Round to three decimal places.
- Mouthwash, 50.7-ounce size for $6.99 or 33.8-ounce size for $4.79
- Toothpaste, 6 ounce size for $3.19 or 7.8-ounce size for $5.19
- Breakfast cereal, 18 ounces for $3.99 or 14 ounces for $3.29
- Breakfast Cereal, 10.7 ounces for $2.69 or 14.8 ounces for $3.69
- Ketchup, 40-ounce regular bottle for $2.99 or 64-ounce squeeze bottle for $4.39
- Mayonnaise, 15-ounce regular bottle for $3.49 or 22-ounce squeeze bottle for $4.99
- Cheese, $6.49 for 1 lb. block or $3.39 for \(\dfrac{1}{2}\) lb. block
- Candy, $10.99 for a 1 lb. bag or $2.89 for \(\dfrac{1}{4}\) lb. of loose candy
Translate Phrases to Expressions with Fractions
In the following exercises, translate the English phrase into an algebraic expression.
- 793 miles per p hours
- 78 feet per r seconds
- $3 for 0.5 lbs.
- j beats in 0.5 minutes
- 105 calories in x ounces
- 400 minutes for m dollars
- the ratio of y and 5x
- the ratio of 12x and y
Everyday Math
- One elementary school in Ohio has 684 students and 45 teachers. Write the student-to-teacher ratio as a unit rate.
- The average American produces about 1,600 pounds of paper trash per year (365 days). How many pounds of paper trash does the average American produce each day? (Round to the nearest tenth of a pound.)
- A popular fast food burger weighs 7.5 ounces and contains 540 calories, 29 grams of fat, 43 grams of carbohydrates, and 25 grams of protein. Find the unit rate of (a) calories per ounce (b) grams of fat per ounce (c) grams of carbohydrates per ounce (d) grams of protein per ounce. Round to two decimal places.
- A 16-ounce chocolate mocha coffee with whipped cream contains 470 calories, 18 grams of fat, 63 grams of carbohydrates, and 15 grams of protein. Find the unit rate of (a) calories per ounce (b) grams of fat per ounce (c) grams of carbohydrates per ounce (d) grams of protein per ounce.
Writing Exercises
- Would you prefer the ratio of your income to your friend’s income to be 3/1 or 1/3? Explain your reasoning.
- The parking lot at the airport charges $0.75 for every 15 minutes. (a) How much does it cost to park for 1 hour? (b) Explain how you got your answer to part (a). Was your reasoning based on the unit cost or did you use another method?
- Kathryn ate a 4-ounce cup of frozen yogurt and then went for a swim. The frozen yogurt had 115 calories. Swimming burns 422 calories per hour. For how many minutes should Kathryn swim to burn off the calories in the frozen yogurt? Explain your reasoning.
- Mollie had a 16-ounce cappuccino at her neighborhood coffee shop. The cappuccino had 110 calories. If Mollie walks for one hour, she burns 246 calories. For how many minutes must Mollie walk to burn off the calories in the cappuccino? Explain your reasoning.
Self Check
(a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
(b) After reviewing this checklist, what will you do to become confident for all objectives?
Contributors and Attributions
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Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


3.12: Procedural Law
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- Alison S. Burke, David Carter, Brian Fedorek, Tiffany Morey, Lore Rutz-Burri, & Shanell Sanchez
- Southern Oregon University via OpenOregon
As noted above, procedural law governs the process used to investigate and prosecute an individual who commits a crime. Procedural law also governs the ways a person convicted of a crime may challenge their convictions. The source of procedural law includes the same sources of law you have just read about which govern substantive criminal law: the constitution, cases law or judicial opinions, statutes, and common law. Whereas most substantive criminal law is now statutory, most procedural law is found in judicial opinions that interpret the Fourth, Fifth, Sixth, Eighth and Fourteenth Amendment to the U.S. Constitution, the U.S. Code, and the state constitutional and legislative counterparts. Generally, the federal and state constitutions set forth broad guarantees (for example, the right to a speedy trial), then statutes are enacted to provide more definite guidelines (for example, the Federal Speedy Trial Act) and then judges flesh out the meaning of those guarantees and statutes in their court opinions.
Phases of the Criminal Justice Process
The processing of a case through the criminal justice system can be broken down into five phases: investigative phase, the pre-trial phase, the trial phase, the sentencing phase, and the appellate or post-conviction phase.
Investigative Phase
The investigative phase is governed by laws covering searches and seizures (searches of persons and places, arrests and stops of individuals, seizures of belongings), interrogations and confessions, identification procedures (for example, line ups, showups, and photo arrays). This phase mostly involves what the police are doing to investigate a crime. However, when police apply for a search, seizure or arrest warrant, “neutral and detached” magistrates (i.e., judges) must decide whether probable cause exists to issue search warrants, arrest warrants, and warrants for the seizure of property and whether the scope of the proposed warrant is supported by the officer’s affidavit (sworn statement). When an individual is arrested without a warrant, judges will need to promptly review whether there is probable cause exists to hold them in custody before trial.
Pretrial Phase
The pretrial phase is governed by laws covering the initial appearance of the defendant before a judge or magistrate; the securing of defense counsel, the arraignment process (in which the defendant is informed of the charges which have been filed by the state); the process in which the court determines whether to release the defendant pre-trial either with some financial surety (posting bail) or on his or her own recognizance and with court-determined conditions imposed (for example, not having contact with the alleged victim); the selection and use of a grand jury or preliminary hearing processes (in which either a grand jury or a judge determines whether there is sufficient evidence that a felony has been committed); any pretrial motions such as motions to suppress evidence (for examples, asking the court not to let the government use evidence it may have obtained illegally through a search or getting a confession), motions to challenge a subpoena, motions to change venue (to move the trial), motions to join or sever cases (for example if two or more individuals are charged with the offense, should the trials be held together or separately). During the pretrial phase, prosecutors and defendants through their defense attorneys will engage in plea bargaining, and will generally resolve the case before a trial is held.
Trial Phase
The trial phase is governed by laws covering speedy trial guarantees, the selection and use of petit jurors (trial jurors); the rules of evidence (statutory and common law rules governing the admissibility of certain types of evidence such as hearsay or character evidence, the competency and impeachment of witnesses, the existence of any privilege, and the exclusion of witnesses during the testimony of other witnesses); the right of the defendant compulsory process (to secure favorable testimony and evidence); the right of the defendant to cross-examine any witnesses or evidence presented by the government against him; fair trials free of prejudicial adverse pre-trial or trial publicity; fair trials which are open to the public; and the continued right of the defendant to have the assistance of counsel and be present during his or her trial.
Sentencing Phase
The sentencing phase is governed by rules and laws concerning the substantive criminal laws on punishment (discussed above); time period in which a defendant must be sentenced; the defendant’s right of allocution (right to make a statement to the court before the judge imposes sentence); any victims’ rights to appear and make statements at sentencing; the defendant’s rights to present mitigation evidence and witnesses; and the defendant’s continued rights to the assistance of counsel at sentencing. In capital cases in which the state is seeking the death penalty, the trial will be bifurcated (a trial split into the “guilt/innocence phase” and the “penalty phase”) and the sentencing hearing will be more like a mini-trial.
Post-Conviction Phase (Appeals Phase)
The post-conviction phase is governed by rules and laws concerning the time period in which direct appeals must be taken; the defendant’s right to file an appeal of right (the initial appeal which must be reviewed by an appellate court) and right to file a discretionary appeal; the defendant’s right to have the assistance of counsel in helping to file either the appeal of right or a discretionary appeal. The post-conviction phase is also governed by rules and laws concerning the defendant’s ability to file a writ of habeas corpus (a civil suit against the entity who is currently holding the defendant in custody) or a post-conviction relief suit (a civil suit similar to a habeas corpus suit but one which can be filed by the defendant regardless if he or she is in custody). The post-conviction phase would also include any probation and parole revocation hearings.


2.13: The Degeneracy of an Isolated System and its Entropy
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In Section 20.10 , we observe that the entropy of an isolated equilibrium system can be defined as \(S=k{\ln W_{max}}\). In Section 20.12 , we see that the system-energy degeneracy is a sum of terms, one of which is \(W_{max}=W\left(N^{\textrm{⦁}}_i,g_i\right)\). That is, we have \[{\Omega}_E=W_{max}+\sum_{\left\{N_i\right\}\neq \left\{N^{\textrm{⦁}}_i\right\},E_{total}}{W\left(N_i,g_i\right)} \nonumber \]
where the last sum is taken over all energy-qualifying population sets other than the most-probable population set.
Let us now consider the relative magnitude of \(\Omega_E\) and \(W_{max}\). Clearly, \(\Omega_E\ge W_{max}\). If only one population set is consistent with the total-molecule and total-energy constraints of the isolated system, then \(\Omega_E=W_{max}\). In general, however, we must expect that there will be many, possibly an enormous number, of other population sets that meet the constraints. Ultimately, the relative magnitude of \({\Omega}_E\) and \(W_{max}\) depends on the energy levels available to the molecules and the number of molecules in the system and so could be almost anything. However, rather simple considerations lead us to expect that, for most macroscopic collections of molecules, the ratio \(\alpha ={\Omega_E}/W_{max}\) will be much less than \(W_{max}\). That is, although the value of \(\alpha\) may be very large, for macroscopic systems we expect to find \(\alpha \ll W_{max}\). If \(\Omega_E=W_{max}\), then \(\alpha =1\), and \({\ln \alpha }=0\).
Because \(W\) for any population set that contributes to \({\Omega}_E\) must be less than or equal to \(W_{max}\), the maximum value of \(\alpha\) must be less than the number of population sets which satisfy the system constraints. For macroscopic systems whose molecules have even a modest number of accessible energy levels, calculations show that \(W_{max}\) is a very large number indeed. Calculation of \(\alpha\) for even a small collection of molecules is intractable unless the number of accessible molecular energy levels is small. Numerical experimentation on small systems, with small numbers of energy levels, shows that the number of qualifying population sets increases much less rapidly than \(W_{max}\) as the total number of molecules increases. Moreover, the contribution that most qualifying population sets make to \({\Omega}_E\) is much less than \(W_{max}\).
For macroscopic systems, we can be confident that \(W_{max}\) is enormously greater than \(\alpha\). Hence \(\Omega_E\) is enormously greater than \(\alpha\). When we substitute for \(W_{max}\) in the isolated-system entropy equation, we find
\[ \begin{align*} S &=k \ln W_{max} \\[4pt] &=k \ln \left(\Omega_E/\alpha \right) \\[4pt] &=k \mathrm{l}\mathrm{n} \Omega_E -k \ln \alpha \\[4pt] &\approx k \ln \Omega_E \end{align*} \nonumber \]
where the last approximation is usually very good.
In many developments, the entropy of an isolated system is defined by the equation \(S=k{\ln {\Omega}_E}\) rather than the equation we introduced first, \(S=k{\ln W_{max}}\). From the considerations above, we expect the practical consequences to be the same. In Section 20.14 , we see that the approximate equality of \({\ln W_{max}}\) and \({\ln {\Omega}_E}\) is a mathematical consequence of our other assumptions and approximations.


31.1C: Essential Nutrients for Plants
- Distinguish among the essential nutrients for plants
Essential Nutrients
Plants require only light, water, and about 20 elements to support all their biochemical needs. These 20 elements are called essential nutrients. For an element to be regarded as essential, three criteria are required:
- a plant cannot complete its life cycle without the element
- no other element can perform the function of the element
- the element is directly involved in plant nutrition
Macronutrients and Micronutrients
The essential elements can be divided into macronutrients and micronutrients. Nutrients that plants require in larger amounts are called macronutrients. About half of the essential elements are considered macronutrients: carbon, hydrogen, oxygen, nitrogen, phosphorus, potassium, calcium, magnesium, and sulfur. The first of these macronutrients, carbon (C), is required to form carbohydrates, proteins, nucleic acids, and many other compounds; it is, therefore, present in all macromolecules. On average, the dry weight (excluding water) of a cell is 50 percent carbon, making it a key part of plant biomolecules.
The next-most-abundant element in plant cells is nitrogen (N); it is part of proteins and nucleic acids. Nitrogen is also used in the synthesis of some vitamins. Hydrogen and oxygen are macronutrients that are part of many organic compounds and also form water. Oxygen is necessary for cellular respiration; plants use oxygen to store energy in the form of ATP. Phosphorus (P), another macromolecule, is necessary to synthesize nucleic acids and phospholipids. As part of ATP, phosphorus enables food energy to be converted into chemical energy through oxidative phosphorylation. Light energy is converted into chemical energy during photophosphorylation in photosynthesis; and into chemical energy to be extracted during respiration. Sulfur is part of certain amino acids, such as cysteine and methionine, and is present in several coenzymes. Sulfur also plays a role in photosynthesis as part of the electron transport chain where hydrogen gradients are key in the conversion of light energy into ATP. Potassium (K) is important because of its role in regulating stomatal opening and closing. As the openings for gas exchange, stomata help maintain a healthy water balance; a potassium ion pump supports this process.
Magnesium (Mg) and calcium (Ca) are also important macronutrients. The role of calcium is twofold: to regulate nutrient transport and to support many enzyme functions. Magnesium is important to the photosynthetic process. These minerals, along with the micronutrients, also contribute to the plant’s ionic balance.
In addition to macronutrients, organisms require various elements in small amounts. These micronutrients, or trace elements, are present in very small quantities. The seven main micronutrients include boron, chlorine, manganese, iron, zinc, copper, and molybdenum. Boron (B) is believed to be involved in carbohydrate transport in plants; it also assists in metabolic regulation. Boron deficiency will often result in bud dieback. Chlorine (Cl) is necessary for osmosis and ionic balance; it also plays a role in photosynthesis. Copper (Cu) is a component of some enzymes. Symptoms of copper deficiency include browning of leaf tips and chlorosis (yellowing of the leaves). Iron (Fe) is essential for chlorophyll synthesis, which is why an iron deficiency results in chlorosis. Manganese (Mn) activates some important enzymes involved in chlorophyll formation. Manganese-deficient plants will develop chlorosis between the veins of its leaves. The availability of manganese is partially dependent on soil pH. Molybdenum (Mo) is essential to plant health as it is used by plants to reduce nitrates into usable forms. Some plants use it for nitrogen fixation; thus, it may need to be added to some soils before seeding legumes. Zinc (Zn) participates in chlorophyll formation and also activates many enzymes. Symptoms of zinc deficiency include chlorosis and stunted growth.
Deficiencies in any of these nutrients, particularly the macronutrients, can adversely affect plant growth. Depending on the specific nutrient, a lack can cause stunted growth, slow growth, or chlorosis. Extreme deficiencies may result in leaves showing signs of cell death.
Contributions and Attributions
- OpenStax College, Biology. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44712/latest...ol11448/latest . License : CC BY: Attribution
- OpenStax College, Biology. November 22, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest...ol11448/latest . License : CC BY: Attribution
- photosynthesis. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/photosynthesis . License : CC BY-SA: Attribution-ShareAlike
- nutrient. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/nutrient . License : CC BY-SA: Attribution-ShareAlike
- germinate. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/germinate . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Introduction. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44712/latest...31_00_01ab.jpg . License : CC BY: Attribution
- transpiration. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/transpiration . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Biology. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest...ol11448/latest . License : CC BY: Attribution
- xylem. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/xylem . License : CC BY-SA: Attribution-ShareAlike
- organic. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/organic . License : CC BY-SA: Attribution-ShareAlike
- inorganic. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/inorganic . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Introduction. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44712/latest...31_00_01ab.jpg . License : CC BY: Attribution
- OpenStax College, Nutritional Requirements of Plants. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest...e_31_01_01.jpg . License : CC BY: Attribution
- OpenStax College, Biology. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest...ol11448/latest . License : CC BY: Attribution
- Micronutrient. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Micronutrient . License : CC BY-SA: Attribution-ShareAlike
- chlorosis. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/chlorosis . License : CC BY-SA: Attribution-ShareAlike
- micronutrient. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/micronutrient . License : CC BY-SA: Attribution-ShareAlike
- macronutrient. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/macronutrient . License : CC BY-SA: Attribution-ShareAlike
- OpenStax College, Introduction. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44712/latest...31_00_01ab.jpg . License : CC BY: Attribution
- OpenStax College, Nutritional Requirements of Plants. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest...e_31_01_01.jpg . License : CC BY: Attribution
- OpenStax College, Biology. November 22, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest...ol11448/latest . License : CC BY: Attribution
- OpenStax College, Nutritional Requirements of Plants. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44714/latest/Figure_31_01_03abcd.jpg . License : CC BY: Attribution
Key Points
- An element is essential if a plant cannot complete its life cycle without it, if no other element can perform the same function, and if it is directly involved in nutrition.
- An essential nutrient required by the plant in large amounts is called a macronutrient, while one required in very small amounts is termed a micronutrient.
- Missing or inadequate supplies of nutrients adversely affect plant growth, leading to stunted growth, slow growth, chlorosis, or cell death.
- About half the essential nutrients are micronutrients such as boron, chlorine, manganese, iron, zinc, copper, molybdenum, nickel, silicon, and sodium.
Key Terms
- micronutrient : a mineral, vitamin, or other substance that is essential, even in very small quantities, for growth or metabolism
- chlorosis : a yellowing of plant tissue due to loss or absence of chlorophyll
- macronutrient : any of the elements required in large amounts by all living things


12.5: Affordability
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More people than ever before are partaking in commerce and communications thanks to digitisation lowering the barrier to entry. The traditional lengthy logistics chain to move products adds more cost. At each step along the chain the handling party requires a fee, which will be passed on to the customer by increasing the price of the product. By shortening the logistics chain and cutting out middlemen, manufacturers make cost savings. Although the cost of producing a product might stay the same, savings can be made when it comes to distributing, selling and marketing the product. These savings can be passed on to customers in the form of a lower price, with the manufacturer maintaining the same profit margin. This lower price can potentially attract customers who were previously barred by high prices. The digitisation of commerce can thus open up markets by making products more affordable.
The digital communications chain has been shortened in similar ways, with the same sort of cost benefits. However, the monetary cost of communications was never really high enough to pose a barrier to entry. The benefits of the digitisation of communications are not primarily price, but rather the lowering of the skills required to partake. Communicating via letters as outlined above requires the ability to both read and write. Until the spread of mass education in the twentieth century, these skills were limited to a relatively small subset of humanity. Now, since literacy levels are high in most developed states, digital communications have the power to make a difference for people with learning difficulties or in areas where education is limited. Courtesy of video messaging applications, real-time long-distance correspondence can be achieved via face-to-face communication. This bypasses any need to be able to read and write, requiring only the interpersonal communication skills every person has. It does of course require a device, such as a laptop or smartphone, on which to run the application. However, devices are becoming cheaper, and a single device can be shared and passed around. Shared ownership not only spreads the initial cost of purchasing the device, but is in itself a means for people to connect with one another. The ability of a family to gather around a laptop and video call with relatives on the other side of the world is a powerful way to maintain relationships otherwise challenged by distance and time.
Those previously separated by geographical distance and/or access to means of communication are now able to reconnect with lost acquaintances and even forge relationships with strangers on the other side of the globe. In this way, digital communications have the potential to increase humanity’s homogeneity. If everyone is connected, divisions between locations, races, nationalities, classes and wealth can be blurred. Rather than emphasise the things that have traditionally separated humanity, it is possible to concentrate on those things that unite us: the shared values that make us human.


8.5: EFA Interpretation
At the end of this section you should be able to answer the following questions:
- How do we assign a name or label to factors?
- How many variables or items can be recommended per factor?
Pattern Matrix & Structure/Factor Matrix
The next piece of information is the pattern and the structure matrixes. These provide useful information on the factor loadings of the items. The pattern matrix shows the unique contribution of a variable to a factor, and is generally simpler to interpret. The structure matrix shows the shared variance that is ignored in the pattern matrix, and is more complicated to interpret. Note: Jamovi automatically produces the pattern matrix.
PowerPoint: Matrices
Please have a look at the below output for a Pattern Matrix:
Number of Variables per Factor
The replicability and strength of a component are determined by the number of variables per factor/component. A good rule is that a minimum of 4 variables is recommended per factor. All items that load on to factor should have a score of greater than .40 on the pattern matrix. Just be aware that items can “crossload” (i.e. have scores great than .40 on more than one factor).
Pattern Matrix
As you can see on the output, the pattern matrix clearly shows the loadings of each item onto the two factors. The two factors show that there are siz items for each, with one factor measuring in-person support, and the other Facebook-based support. This makes sense, both statistically, and theoretically. It is important to note that this is an optimal factor loading as there are two distinct factors, with no poor loadings (i.e. < .40) and crossloadings (have scores great than .40 on more than one factor). If there were any poor loadings/crossloadings, these items would be removed for the next round of EFAs.
Interpretation of Factors/Components
When naming the factors found it is usual to characterise the factor by assigning a name or label related to the semantic topic the items measure. This involves not only the knowledge of the area (science) but is an artistic pursuit sometimes. For this example we would have one factor labeled in-person support, and the other Facebook-based support. And we can see that they have a moderate correlation with each other.


Learning Objectives
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Analyze the relationship between learning styles and academic performance to identify personal strengths and areas for improvement.
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Evaluate the impact of motivation and mindset on learning and academic success.
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Develop personalized strategies to enhance learning, including active listening and class preparation techniques.
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Apply self-regulation strategies to improve focus, time management, and readiness for class.
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Examine the concept of the hidden curriculum and its influence on student success in the academic environment.
Figure 3.1:
The School of Athens
by Raphael, a fresco in the Vatican, is thought to depict many of the greatest figures in Greek philosophy, including Plato and Aristotle as well as Nicomachus and Averroes. (Credit: Wikimedia Commons / Public Domain)
Student Survey
How do you feel about your learning abilities? Take this quick survey to figure it out, ranking questions on a scale of 1–4, 1 meaning “least like me” and 4 meaning “most like me.” These questions will help you determine how the chapter concepts relate to you right now. As you are introduced to new concepts and practices, it can be informative to reflect on how your understanding changes over time. We’ll revisit these questions at the end of the chapter to see whether your feelings have changed.
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Learning for me is easy. I don’t even have to think about it.
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I have a preferred learning style.
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If I can't learn something right away, I have difficulty staying with it.
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I think my teachers are the most significant aspect of my learning.
You can also take the
Chapter 2 survey(opens in new window)
anonymously online.
STUDENT PROFILE
“When I came to college, I was a great STEM student. I knew the best ways to study for understanding the complexity of cellular mechanisms, but I had no idea how to study for classes where I would need to draw upon political theory or even how to memorize vocabulary words for language classes. Since I am now a humanities student studying Russian, I learned the hard way that you cannot study for every class the same way.
“For my first Russian vocabulary quiz, I studied almost 14 hours because I could not remember the words no matter how hard I tried. I was studying the Russian textbook the same way that I would study for a Chemistry or Biology class: to simply read the chapter or vocabulary list over and over again. I knew that I could not afford to be this time-inefficient for the entire semester, so I asked my professor for some tips on how to study for her class. Now, I start studying three days before each quiz by making flash cards the first day, studying the words from Russian to English the second day, and then studying the words from English to Russian by writing them down the third day. This new method is not one that works well for every class, but that’s the beauty of it! I am a better learner because I have found ways to use a more diverse range of studying tactics.”
— Gabby Kennedy
, Baylor University
About This Chapter
In this chapter you will learn about the art of learning itself, as well as how to employ strategies that enable you to learn more efficiently.
After reading this chapter, you should be able to do the following:
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Discover the different types of learning and your learning practices.
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Make informed and effective learning choices in regards to personal engagement and motivation.
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Identify and apply the learning benefits of a growth mindset.
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Evaluate and make informed decisions about learning styles and learning skills.
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Recognize how personality type models influence learning and utilize that knowledge to improve your own learning.
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Identify the impact of outside circumstances on personal learning experiences and develop strategies to compensate for them.
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Recognize the presence of the “hidden curriculum” and how to navigate it.