# Find exponential models y_1=Ce^{k_1t} and y_2=C(2)^{k_2t} That pass through the two given points. Compare the values of k_1 and k_2. (If you round your answer, round to four decimal places.) (0,16),(60,frac{1}{4}) y_1=Ce^{k_1t} where C=? and k_1=? y_2=C(2)^{k_2t} where C=? and k_2=?

Question
Exponential models
Find exponential models
$$\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}}$$
and
$$\displaystyle{y}_{{2}}={C}{\left({2}\right)}^{{{k}_{{2}}{t}}}$$
That pass through the two given points. Compare the values of $$\displaystyle{k}_{{1}}$$ and $$\displaystyle{k}_{{2}}$$. (If you round your answer, round to four decimal places.)
$$\displaystyle{\left({0},{16}\right)},{\left({60},{\frac{{{1}}}{{{4}}}}\right)}$$
$$\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}}$$ where C=?
and $$\displaystyle{k}_{{1}}=?$$
$$\displaystyle{y}_{{2}}={C}{\left({2}\right)}^{{{k}_{{2}}{t}}}$$ where C=?
and $$\displaystyle{k}_{{2}}=?$$

2021-03-07
Given equations are $$\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}}$$
$$\displaystyle{y}_{{2}}={C}{\left({2}^{{{k}_{{2}}{t}}}\right)}$$
They pass through points (0, 16) and $$\displaystyle{\left({60},{\frac{{{1}}}{{{4}}}}\right)}$$
now $$\displaystyle{y}_{{1}}={C}{e}^{{{k}_{{1}}{t}}},{\left({0},{16}\right)}$$
$$\displaystyle{16}={C}{e}^{{0}}={C}$$
$$\displaystyle{C}={16}$$
$$\displaystyle{y}_{{1}}={16}{e}^{{{k}_{{1}}{t}}},{\left({60},{\frac{{{1}}}{{{4}}}}\right)}$$
$$\displaystyle{\frac{{{1}}}{{{4}}}}={16}{e}^{{{60}{k}_{{1}}}}$$
$$\displaystyle{e}^{{{60}{k}_{{1}}}}={\frac{{{1}}}{{{64}}}}$$
$$\displaystyle{60}{k}_{{1}}={\ln{{\frac{{{1}}}{{{64}}}}}}=-{4.158883}$$
$$\displaystyle{k}_{{1}}={\frac{{-{4.158883}}}{{{60}}}}=-{0.0693}$$
$$\displaystyle{k}_{{1}}=-{0.0693}$$
Again
$$\displaystyle{y}_{{1}}={C}{2}^{{{k}_{{2}}{t}}},$$
$$\displaystyle{16}={C}{2}^{{0}}={C}$$
$$\displaystyle{C}={16}$$
$$\displaystyle{y}_{{1}}={16}\times{2}^{{{k}_{{2}}{t}}}$$
$$\displaystyle{\frac{{{1}}}{{{4}}}}={16}\times{2}^{{{60}{k}_{{2}}}}$$
$$\displaystyle{2}^{{{60}{k}_{{2}}}}={\frac{{{1}}}{{{64}}}}={2}^{{-{6}}}$$
$$\displaystyle{60}{k}_{{2}}=-{6}$$
$$\displaystyle{k}_{{2}}=-{\frac{{{6}}}{{{60}}}}=-{0.1}$$
$$\displaystyle{k}_{{2}}=-{0.1}$$
$$\displaystyle{C}={16}$$
$$\displaystyle{k}_{{1}}=-{0.0693}$$
$$\displaystyle{k}_{{2}}=-{0.1}$$

### Relevant Questions

Scientists are working with a sample of cobalt-56 in their laboratory. They begin with a sample that has 60 mg of cobalt-56, and they measure that after 31 days, the mass of cobalt-56 sample is 45.43 mg. Recall that the differential equation which models exponential decay is $$\frac{dm}{dt}=-km$$ and the solution of that differential equation if $$m(t)=m_0e^{-kt}$$, where $$m_0$$ is the initial mass and k is the relative decay rate.
a) Use the information provided to compute the relative decay rate k. Show your calculation (do not just cit a formula).
b) Use the information provided to determine the half-life of cobalt-56. Give your answer in days and round to the second decimal place. Show your calculation (do not just cite a formula).
c) To the nearest day, how many days will it take for the initial sample of 60mg of cobalt-56 to decay to just 10mg of cobalt-56?
d) What will be the rate at which the mass is decaying when the sample has 50mg of cobalt-56? Make sure to indicate the appropriate units and round your answer to three decimal places.
Scientists are working with a sample of cobalt-56 in their laboratory. They begin with a sample that has 60 mg of cobalt-56, and they measure that after 31 days, the mass of cobalt-56 sample is 45.43 mg. Recall that the differential equation which models exponential decay is $$\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}=-{k}{m}$$ and the solution of that differential equation if $$\displaystyle{m}{\left({t}\right)}={m}_{{0}}{e}^{{-{k}{t}}}$$, where $$\displaystyle{m}_{{0}}$$ is the initial mass and k is the relative decay rate.
a) Use the information provided to compute the relative decay rate k. Show your calculation (do not just cit a formula).
b) Use the information provided to determine the half-life of cobalt-56. Give your answer in days and round to the second decimal place. Show your calculation (do not just cite a formula).
c) To the nearest day, how many days will it take for the initial sample of 60mg of cobalt-56 to decay to just 10mg of cobalt-56?
d) What will be the rate at which the mass is decaying when the sample has 50mg of cobalt-56? Make sure to indicate the appropriate units and round your answer to three decimal places.
Several models have been proposed to explain the diversification of life during geological periods. According to Benton (1997), The diversification of marine families in the past 600 million years (Myr) appears to have followed two or three logistic curves, with equilibrium levels that lasted for up to 200 Myr. In contrast, continental organisms clearly show an exponential pattern of diversification, and although it is not clear whether the empirical diversification patterns are real or are artifacts of a poor fossil record, the latter explanation seems unlikely. In this problem, we will investigate three models fordiversification. They are analogous to models for populationgrowth, however, the quantities involved have a differentinterpretation. We denote by N(t) the diversification function,which counts the number of taxa as a function of time, and by rthe intrinsic rate of diversification.
(a) (Exponential Model) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{e}}}{N}\ {\left({8.86}\right)}.$$ Solve (8.86) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{e}}}$$ can be estimated from $$\displaystyle{r}_{{{e}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ {\left({8.87}\right)}$$
(b) (Logistic Growth) This model is described by $$\displaystyle{\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}}={r}_{{{l}}}{N}\ {\left({1}\ -\ {\frac{{{N}}}{{{K}}}}\right)}\ {\left({8.88}\right)}$$ where K is the equilibrium value. Solve (8.88) with the initial condition N(0) at time 0, and show that $$\displaystyle{r}_{{{l}}}$$ can be estimated from $$\displaystyle{r}_{{{l}}}={\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{K}\ -\ {N}{\left({0}\right)}}}{{{N}{\left({0}\right)}}}}\right]}\ +\ {\frac{{{1}}}{{{t}}}}\ {\ln{\ }}{\left[{\frac{{{N}{\left({t}\right)}}}{{{K}\ -\ {N}{\left({t}\right)}}}}\right]}\ {\left({8.89}\right)}$$ for $$\displaystyle{N}{\left({t}\right)}\ {<}\ {K}.$$
(c) Assume that $$\displaystyle{N}{\left({0}\right)}={1}$$ and $$\displaystyle{N}{\left({10}\right)}={1000}.$$ Estimate $$\displaystyle{r}_{{{e}}}$$ and $$\displaystyle{r}_{{{l}}}$$ for both $$\displaystyle{K}={1001}$$ and $$\displaystyle{K}={10000}.$$
(d) Use your answer in (c) to explain the following quote from Stanley (1979): There must be a general tendency for calculated values of $$\displaystyle{\left[{r}\right]}$$ to represent underestimates of exponential rates,because some radiation will have followed distinctly sigmoid paths during the interval evaluated.
(e) Explain why the exponential model is a good approximation to the logistic model when $$\displaystyle\frac{{N}}{{K}}$$ is small compared with 1.
factor in determining the usefulness of an examination as a measure of demonstrated ability is the amount of spread that occurs in the grades. If the spread or variation of examination scores is very small, it usually means that the examination was either too hard or too easy. However, if the variance of scores is moderately large, then there is a definite difference in scores between "better," "average," and "poorer" students. A group of attorneys in a Midwest state has been given the task of making up this year's bar examination for the state. The examination has 500 total possible points, and from the history of past examinations, it is known that a standard deviation of around 60 points is desirable. Of course, too large or too small a standard deviation is not good. The attorneys want to test their examination to see how good it is. A preliminary version of the examination (with slight modifications to protect the integrity of the real examination) is given to a random sample of 20 newly graduated law students. Their scores give a sample standard deviation of 70 points. Using a 0.01 level of significance, test the claim that the population standard deviation for the new examination is 60 against the claim that the population standard deviation is different from 60.
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H_{0}:\sigma=60,\ H_{1}:\sigma\ <\ 60H_{0}:\sigma\ >\ 60,\ H_{1}:\sigma=60H_{0}:\sigma=60,\ H_{1}:\sigma\ >\ 60H_{0}:\sigma=60,\ H_{1}:\sigma\ \neq\ 60$$
(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)
What are the degrees of freedom?
What assumptions are you making about the original distribution?
We assume a binomial population distribution.We assume a exponential population distribution. We assume a normal population distribution.We assume a uniform population distribution.
For the following exercises, use a graphing utility to create a scatter diagram of the data given in the table. Observe the shape of the scatter diagram to determine whether the data is best described by an exponential, logarithmic, or logistic model. Then use the appropriate regression feature to find an equation that models the data. When necessary, round values to five decimal places.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}{x}&{1}&{2}&{3}&{4}&{5}&{6}&{7}&{8}&{9}&{10}\backslash{h}{l}\in{e}{f{{\left({x}\right)}}}&{409.4}&{260.7}&{170.4}&{110.6}&{74}&{44.7}&{32.4}&{19.5}&{12.7}&{8.1}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
Suspect was Armed:
Black - 543
White - 1176
Hispanic - 378
Total - 2097
Suspect was unarmed:
Black - 60
White - 67
Hispanic - 38
Total - 165
Total:
Black - 603
White - 1243
Hispanic - 416
Total - 2262
Give your answer as a decimal to at least three decimal places.
a) What percent are Black?
b) What percent are Unarmed?
c) In order for two variables to be Independent of each other, the P $$(A and B) = P(A) \cdot P(B) P(A and B) = P(A) \cdot P(B).$$
This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
d) Now let's get the real percent that are Black and Unarmed by using the table?
If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
e) What percent are White and Unarmed?
f) What percent are Hispanic and Unarmed?
If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
Why is that?
This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
g) What percent of blacks shot and killed by police were unarmed?
h) What percent of whites shot and killed by police were unarmed?
i) What percent of Hispanics shot and killed by police were unarmed?
You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
j) Why do you believe this is happening?
Do a search on the internet for reasons why blacks are more likely to be killed by police. Read a few articles on the topic. Write your response using the articles as references. Give the websites used in your response. Your answer should be several sentences long with at least one website listed. This part of this problem will be graded after the due date.
Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
In this exercise, you will use the correlation and regression applet to create scatter plots with 10 points that have a correlation close to 0.7. The lesson here is that many models may have the same correlation. Always compile your data before trusting correlations. (a) Stop after adding the first two points. What is the value of correlation?
(Enter your answer, rounded to four decimal places).
r=?
Why does correlation matter? Two is the minimum number of data points required to calculate the correlation. This value is the default correlation.
Because two points define a line, correlation always matters.
The mean of these two values always has this value.
$$\begin{array}{|l|l|l|}\hline t(\text{in years})&m(\text{amoun of radioactive material})\\\hline0&\\\hline5730\\\hline11460\\\hline17190\\\hline\end{array}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{l}\right|}{l}{\left|{l}\right|}\right\rbrace}{h}{l}\in{e}{t}{\left(\text{in years}\right)}&{m}{\left(\text{amoun of radioactive material}\right)}\backslash{h}{l}\in{e}{0}&\backslash{h}{l}\in{e}{5730}\backslash{h}{l}\in{e}{11460}\backslash{h}{l}\in{e}{17190}\backslash{h}{l}\in{e}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$