Below is data collected from the growth of two different trees over time. Each tree was planted in 1960, and the tree's height has been collected every ten years. What type of function is the growth of Tree A? How can you tell? What type of function is the growth of Tree B? How can you tell? Write an equation for each function of the trees' growth over time. For time, you may use x=0 for 1960. Compare the growth rate for each of the trees. Compare the starting heights of each of the trees. When will Tree A's height exceed Tree B's height?

Question
Below is data collected from the growth of two different trees over time. Each tree was planted in 1960, and the tree's height has been collected every ten years.
What type of function is the growth of Tree A? How can you tell?
What type of function is the growth of Tree B? How can you tell?
Write an equation for each function of the trees' growth over time. For time, you may use x=0 for 1960.
Compare the growth rate for each of the trees.
Compare the starting heights of each of the trees.
When will Tree A's height exceed Tree B's height?

Answers (1)

2020-11-25
Tree A is curved which means it could be an exponential function. The yy-coordinates are 10, 20, 40, and 80. Since the yy-coordinates are doubling every 10 years, then the function is exponential since exponential functions have a constant factor (the number you multiply by).
2. Unlike the yy-coordinates for Tree A, the yy-coordinates for Tree B are not increasing by a constant factor. Since 37−12=25, 62−37=25, and 87−62=25, then the yy-coordinates for Tree B are increasing by a constant amount of 25. This means the function has a constant rate of change of 25 feet per 10 years. The function is then linear since linear functions have constant rates of change.
Exponential functions are of the form y=a⋅b^(x/c​) where aa is the amount when x=0, b is the growth factor, and c is how often the amount changes by the growth factor. Since the y-coordinate is 10 when x=0 (the year 1960), then a=10. Since the y-coordinates are doubling every 10 years, then b=2 and c=10. The function for Tree A is then y=10⋅2^(x/10).
Linear functions are of the form y=mx+b where mm is the constant rate of change and bb is the y-intercept. Since the rate of change is 25 feet per 10 years, then m=25/10=2.5m. Since the height is 12 feet when x=0 (the year 1960), then b=12. The function for Tree B is then y=2.5x+12.
From the previous problems, Tree A is growing by a constant factor of 2 every 10 years and Tree B is growing by a constant amount of 25 feet every 10 years. In the long term, Tree A has a greater growth rate than Tree B since growing by a constant factor will give greater increases than growing by a constant amount as time increases.
5. From the graph, the initial height of Tree A was 10 feet in 1960. From the table, the initial height of Tree B was 12 feet in 1960. Therefore, Tree B had a greater initial height.
6. From the graph and table, in 1990 Tree A had a height of 80 feet and Tree B had a height of 87 feet. Tree A's height will then exceed Tree B's height sometime after 1990. Since 1990 is 30 years after 1960, then x=30x=30 corresponds to 1990. Make a table finding the heights of the trees after 1990:
The height of Tree A will then exceed the height of Tree B after x=33x=33 years which corresponds to 1993 1993 ​ .
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Relevant Questions

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asked 2020-10-23
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White - 67
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White - 1243
Hispanic - 416
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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).
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Remember, the previous answer is only correct if the variables are Independent.
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Let's compare the percentage of unarmed shot for each race.
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f) What percent are Hispanic and Unarmed?
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Why is that?
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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.
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j) Why do you believe this is happening?
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asked 2021-01-28
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asked 2020-10-23
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\(\displaystyle{X}_{{1}}:\) Rate of hay fever per 1000 population for people under 25
\(\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}\)
A random sample of \(\displaystyle{n}_{{2}}={14}\) regions in western Kansas gave the following information for people over 50 years old.
\(\displaystyle{X}_{{2}}:\) Rate of hay fever per 1000 population for people over 50
\(\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}\)
(i) Use a calculator to calculate \(\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.\) (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use \(\displaystyle\alpha={0.05}.\)
(a) What is the level of significance?
State the null and alternate hypotheses.
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}\)
\(\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}\)
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference \(\displaystyle\mu_{{1}}-\mu_{{2}}\). Round your answer to three decimal places.)
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P-value \(\displaystyle>{0.250}\)
\(\displaystyle{0.125}<{P}-\text{value}<{0},{250}\)
\(\displaystyle{0},{050}<{P}-\text{value}<{0},{125}\)
\(\displaystyle{0},{025}<{P}-\text{value}<{0},{050}\)
\(\displaystyle{0},{005}<{P}-\text{value}<{0},{025}\)
P-value \(\displaystyle<{0.005}\)
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
asked 2021-01-31
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.
asked 2020-10-23
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Time t (in hours)012345678 Population P(t)501002004008001,6003,2006,40012,800
a. Complete another row of the table with values log (population) and identify the familiar function pattern illustrated by values in that row.
b. Use your calculator to find log 2 and see how that value relates to the pattern you found in the log P(t) row of the data table.
c. Suppose that you had a different set of experimental data that you suspected was an example of exponential growth or decay, and you produced a similar “third row” with values equal to the logarithms of the population data.
How could you use the pattern in that “third row” to figure out the actual rule for the exponential growth or decay model?
asked 2021-02-12
Michelle is studying the relationship between the hours worked (per week) and time spent reading (per day) and has collected the data shown in the table. The line of best fit for the data is \(y=−0.79x+98.8\).
Hours Worked (per week) 30405060 Minutes Reading (per day) 75685852
(a) According to the line of best fit, the predicted number of minutes spent reading for a person who works 27 hours (per week) is 77.47.
(b) Is it reasonable to use this line of best fit to make the above prediction?
Select the correct answer below:
1.The estimate, a predicted time of 77.47 minutes, is unreliable but reasonable.
2.The estimate, a predicted time of 77.47 minutes, is reliable but unreasonable.
3.The estimate, a predicted time of 77.47 minutes, is both unreliable and unreasonable.
4.The estimate, a predicted time of 77.47 minutes, is both reliable and reasonable.
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