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| Section | Objectives |
|---|---|
| Foundations of Algebra | - Algebraic Expressions
|
| Linear Functions and Systems | - Linear Equations
|
| Exponential and Logarithmic Concepts | - Logarithmic Basics
|
| Polynomial and Rational Expressions | - Polynomials
|
| Functions and Graphs | - Function Concepts
|
>> Latest Real Applied-Algebra Exam <<
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NEW QUESTION # 15
A coach is placing an order for team shirts. The graph shows the total cost based on the number of shirts.
What is the cost of each additional shirt?
Answer: C
Explanation:
This question asks for the cost of each additional shirt, which means we need to find the rate of change of the linear function.
On a graph, the rate of change is the slope:
" slope " = " change in total cost " / " change in number of shirts "
From the graph, two labeled points are:
(0#33)
and
(10#173)
This means:
When 0shirts are ordered, the cost is $33.
When 10shirts are ordered, the cost is $173.
Now find the slope:
m=(173-33)/(10-0)
m=140/10
m=14
So the cost increases by:
$14
for each additional shirt.
The $33represents a fixed starting cost, not the cost per shirt.
NEW QUESTION # 16
The population of fish in a lake is changing according to the function P(t)=31t+438, where t is the number of months since the beginning of the year and P(t) is the fish population at time t. Which interpretation of the rate of change is correct?
Answer: C
Explanation:
The function P(t)=31t+438 is linear and follows the form P(t)=mt+b. In this form, m is the rate of change and b is the initial value. Here, the coefficient of t is 31, so the fish population changes by 31 fish per month.
Because 31 is positive, the population is increasing rather than decreasing. The number 438 is the initial fish population at the start of the year, not the rate of change. Therefore, the correct interpretation is that the number of fish in the lake is increasing at a constant rate of 31 fish per month. That matches option A.
NEW QUESTION # 17
Through a local marketplace, a person makes livestock feeders to sell. The graph shows the functions modeling the cost and revenue.
What is the minimum number of livestock feeders that must be sold in order for the revenue to cover the costs?
Answer: B
Explanation:
This question asks for the break-even point, which is the point where:
Revenue=Cost
On the graph:
The solid blue line represents revenue.
The dashed blue line represents cost.
The revenue line starts below the cost line, meaning the person is not yet covering costs. The revenue line eventually crosses the cost line at a little more than 4 units.
Since the question asks for the minimum number of livestock feeders that must be sold, we need a whole number of feeders.
The graph shows the intersection occurs between:
4 and 5
So selling 4 feeders is not enough, but selling 5 feeders is enough for revenue to cover the costs.
Therefore, the correct answer is:
C
NEW QUESTION # 18
The logistic function f(x), whose graph is shown, models the number of people who have created a website account, where xrepresents the number of days since the website started and f(x)represents the number of people who have created an account.
What is one range of values for which the graph is concave down?
Answer: A
Explanation:
For a logistic graph, concavity changes at the inflection point.
Before the inflection point, the graph is usually concave up. This means the function is increasing faster and faster.
After the inflection point, the graph is concave down. This means the function is still increasing, but it is increasing slower and slower.
From the graph, the curve changes concavity around:
x=8
After x=8, the graph begins to level off as it approaches its upper limit.
So one interval where the graph is concave down is:
(8#16)
Therefore, the correct answer is:
# ( " A " )
NEW QUESTION # 19
The scatterplot shows data on the number of shirts sold at a gift shop per week over time. The graphed regression function has an r
2
value of 0.09.
Which range of x-values is appropriate for extrapolation?
Answer: C
Explanation:
The r
2
value tells how closely the regression model fits the scatterplot data. In this problem, r
2
=0.09, which is extremely low. This means the regression model explains only a very small portion of the variation in the shirt-sales data. When r
2
is this weak, the model is not reliable for predictions outside the data range. Extrapolation is already less certain than interpolation because it predicts beyond the observed data. With such a poor fit, extending the model beyond the data would not be appropriate. Since 0.09 < 0.3, the best justification is that extrapolation is not appropriate because r
2
< 0.3. Therefore, the correct answer is C.
NEW QUESTION # 20
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