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| Section | Objectives |
|---|---|
| Logistic Functions | - Interpret asymptotes for logistic functions - Interpret inputs and outputs for logistic functions - Interpret rates of change for logistic functions |
| Functions and Algebra of Functions | - Derive conclusions based on notation - Derive conclusions based on data - Derive conclusions based on graphs |
| Polynomial Functions | - Interpret concavity for polynomial functions - Interpret rates of change for polynomial functions - Interpret inputs and outputs for polynomial functions |
| Graphical Depictions | - Interpret inputs and outputs for situations - Interpret maximum and minimum for situations - Interpret concavity for situations - Interpret rates of change for situations - Interpret asymptotes for situations |
| Exponential Functions | - Interpret concavity for exponential functions - Interpret inputs and outputs for exponential functions - Interpret rates of change for exponential functions |
| Validity of Models | - Determine validity - Determine fit - Examine utility |
| Linear Functions | - Interpret inputs and outputs of linear functions - Interpret rates of change for linear functions |
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NEW QUESTION # 112
In the graph showing the number of new customer accounts, the horizontal axis shows the number of weeks since the beginning of winter and the vertical axis shows the number of new customer accounts. More customer accounts are opened during snowy weeks than during weeks without snow.
When was it likely snowy?
Answer: A
Explanation:
This question asks us to interpret unusually high values on a graph. The problem states that more customer accounts are opened during snowy weeks than during weeks without snow. Therefore, the likely snowy weeks correspond to the peaks on the graph, where the number of new customer accounts is noticeably higher than surrounding weeks. The graph shows clear spikes at approximately week 7 and week 14. These weeks have higher account openings compared with nearby weeks, making them the best evidence of snowy conditions.
Week 1 and week 5 do not match the main peaks shown in the graph. Since the snowy weeks should correspond to the highest customer-account values, the correct answer is D.
NEW QUESTION # 113
The graph shows the value, O, in dollars, of an investment account over time.
What represents the value of the account after 8 years?
Answer: D
Explanation:
The graph shows the total value of an investment account as time increases.
The horizontal axis represents:
" Time in years "
The vertical axis represents:
" Total value in dollars "
We need to find the value of the account after:
8 " years "
From the graph, when t=8, the curve is slightly above $2,200. The closest answer choice is:
$2,221.12
So the value of the account after 8 years is approximately:
$2,221.12
Therefore, the correct answer is:
# ( " D " )
NEW QUESTION # 114
A new company just launched and is using the function M(t) to predict its market share after t years. The graph of M(t) is shown.
When should the company expect to have a market share of 60%?
Answer: C
Explanation:
This question uses a logistic function, which often models growth that starts slowly, increases rapidly, and then levels off near a maximum value. The input t represents years, and M(t) represents market share as a percent. To find when the company reaches a 60% market share, locate 60 on the vertical axis and identify where the graph reaches that height. Then read the corresponding value on the horizontal axis. From the graph, the curve reaches 60% slightly after 6 years, approximately at t=6.6. The value after 6 years is close but not yet at 60%. Therefore, the company should expect to reach 60% market share after about 6.6 years.
The correct answer is D.
NEW QUESTION # 115
The function P(t)represents the weekly profit, in thousands of dollars, for a rental store since opening. The graph of P(t)is shown.
When did the store have its maximum weekly profit?
Answer: A
Explanation:
The graph shows:
P(t)= " weekly profit in thousands of dollars "
where:
t= " years since opening "
The graph is a downward-opening curve, so it has a maximum value at its highest point. In Applied Algebra, the highest point of this kind of polynomial graph is called the vertex.
To answer the question, we need the t-value of the vertex because the question asks when the store had its maximum weekly profit.
From the graph, the highest point occurs between:
t=5
and
t=6
More specifically, it is approximately:
t=5.5
So the store had its maximum weekly profit approximately:
5.5 " years after opening "
NEW QUESTION # 116
Based on collected data, the value of a painting can be modeled using the exponential function f(x)=240(1.12) x
. In this case, x represents the number of years since 2005, and f(x) represents the value of the painting in dollars.
Which value represents the average yearly rate of change of the painting's value from 2013 to 2015?
Answer: B
Explanation:
This problem asks for an average yearly rate of change over a time interval. Since x represents years since
2005, the year 2013 corresponds to x=8, and the year 2015 corresponds to x=10. Use the average rate of change formula:
10#8
f(10)#f(8)
Evaluating the model gives f(8)=240(1.12)
8
#594.23 and f(10)=240(1.12)
10
#745.40. The total change is approximately 151.17 dollars over 2 years. Dividing by 2 gives about 75.59 dollars per year. Since the question asks for the average yearly rate, not the total increase, the correct answer is B.
NEW QUESTION # 117
......
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