WGU Applied Algebra FXO2 PFXP C957 Practice Tests 2026 | Pass Applied-Algebra with confidence!
Practice Courses and Certificates Applied-Algebra exam. Online Exam Practice Tests with detailed explanations!
NEW QUESTION # 34
The figure displays the graphs of two functions representing the heights, h_1and h_2, in feet, of two balls tseconds after being launched.
Which ball was lower 0.9seconds after being launched?
- A. Ball 1, because the height of ball 1 is less than the height of ball 2 at t=0.9.
- B. Ball 2, because the height of ball 2 is less than the height of ball 1 at t=0.9.
- C. Ball 1, because the height of ball 2 is less than the height of ball 1 at t=0.9.
- D. Ball 2, because the height of ball 1 is less than the height of ball 2 at t=0.9.
Answer: A
Explanation:
This question asks us to compare the heights of two balls at a specific time:
t=0.9
The graph shows:
Ball 1 as the solid blue curve.
Ball 2 as the dashed blue curve.
To determine which ball was lower at t=0.9, we look at the vertical positions of both curves when the time is
0.9seconds.
At approximately t=0.9:
" height of Ball 1 " < " height of Ball 2 "
This means Ball 1 is lower than Ball 2 at that time.
The correct reasoning must say both:
Ball 1 was lower, and
Ball 1's height was less than Ball 2's height at t=0.9.
That matches option A.
NEW QUESTION # 35
A person makes down quilts to sell.
The graph shows the functions that model the cost and revenue.
How many down quilts need to sell to break even/start making a profit?
- A. 0
- B. 1
- C. 2
- D. 3
Answer: B
Explanation:
This question asks us to interpret a graph showing cost and revenue.
In Applied Algebra, the break-even point is where:
" Revenue " = " Cost "
On a graph, this happens where the revenue line and the cost line intersect.
From the graph:
The solid blue line represents revenue.
The dashed blue line represents cost.
The break-even point occurs where these two lines cross.
Looking carefully at the graph, the two lines intersect at approximately:
x=6
The horizontal axis represents the number of units, meaning the number of down quilts sold.
So the person needs to sell about:
6 " quilts "
to break even.
After selling more than 6 quilts, the revenue line is above the cost line, meaning the person begins making a profit.
NEW QUESTION # 36
The scatterplot shows data on the usage of a computer ' s CPU over time. The graphed regression function has an r^2value of 0.25.
What is the appropriate range of x-values for extrapolation?
- A. x=3to x=15
- B. Extrapolation is not appropriate because r^2 < 0.7.
- C. Extrapolation is not appropriate because r^2 < 0.3.
- D. x=1to x=17
Answer: C
Explanation:
The value r^2measures how well a regression model fits the data.
Here:
r^2=0.25
This is a low r^2value, meaning the model explains only about:
25%
of the variation in the data.
That indicates a weak fit. When a model has a weak fit, using it for extrapolation is not reliable because predictions outside the data range may be very inaccurate.
Since:
0.25 < 0.3
the best conclusion is that extrapolation is not appropriate because the regression model does not fit the data well enough.
Therefore, the correct answer is:
# ( " C " )
NEW QUESTION # 37
The populations, in thousands, of two towns are shown in the graph, where the horizontal axis measures the time in years.
Which town's population is growing at a faster rate?
- A. Town B, because 8.0 > 6.5
- B. Town B, because 1.6 > 0.9
- C. Town A, because 1.6 > 0.9
- D. Town A, because 8.0 > 6.5
Answer: B
Explanation:
The graph compares the populations of two towns over time.
The horizontal axis represents:
" Time in years "
The vertical axis represents:
" Population in thousands "
The graph shows:
Town A as the solid blue line.
Town B as the dashed blue line.
To determine which town's population is growing faster, we compare the slopes of the two lines.
In Applied Algebra, the slope of a line represents the rate of change:
" slope " = " change in population " / " change in time "
From the graph:
" Town A grows at about " 0.9 " thousand people per year "
" Town B grows at about " 1.6 " thousand people per year "
Now compare the growth rates:
1.6 > 0.9
So Town B's population is growing at a faster rate.
The values 8.0and 6.5describe starting populations, not growth rates. Since the question asks about growing at a faster rate, we must compare the slopes.
NEW QUESTION # 38
The number of property sales in a region this year is expected to be 6 less than the number of property sales in the region last year. The function H(x)represents the number of property sales this year, where xrepresents the number of properties sold last year.
Which notation represents the number of property sales this year, given that the number of properties sold last year was 330?
- A. H(330)=324
- B. H(330)=336
- C. H(336)=330
- D. H(324)=330
Answer: A
Explanation:
We are told that this year's number of property sales is 6 less than last year's number of property sales.
Let:
x= " number of properties sold last year "
and
H(x)= " number of property sales this year "
Since this year's sales are 6 less than last year's sales, the function rule is:
H(x)=x-6
The question says that the number of properties sold last year was:
x=330
Substitute 330into the function:
H(330)=330-6
H(330)=324
So the notation that correctly represents the number of property sales this year is:
H(330)=324
This means: if 330 properties were sold last year, then 324 properties are expected to be sold this year.
NEW QUESTION # 39
The function
c(t)=-0.01t^4+0.25t^3-2.33t^2+9.02t+8
represents the number of active shopping carts, c, on a website thours after 8:00 a.m.
What is the difference between c(6)and c(8)?
- A. Approximately 2 active carts
- B. Approximately 2 hours
- C. Approximately 1 hour
- D. Approximately 1 active cart
Answer: D
Explanation:
The function c(t)gives the number of active shopping carts on a website.
Because c(t)represents active carts, the difference between c(6)and c(8)will be measured in:
" active carts "
not hours.
First evaluate c(6):
c(6)=-0.01(6)^4+0.25(6)^3-2.33(6)^2+9.02(6)+8
c(6)=19.28
Now evaluate c(8):
c(8)=-0.01(8)^4+0.25(8)^3-2.33(8)^2+9.02(8)+8
c(8)=18.08
Now find the difference:
19.28-18.08=1.20
This is approximately:
1 " active cart "
Therefore, the correct answer is:
# ( " B " )
NEW QUESTION # 40
The graph shows the daily practice duration for a musician, where the number of days since the beginning of the month is along the horizontal axis and the number of minutes practiced per day is along the vertical axis.
Based on the graph, what was the practice duration for day 8?
- A. 120 minutes
- B. 118 minutes
- C. 117 minutes
- D. 115 minutes
Answer: D
Explanation:
This question asks us to interpret a value from a graph.
The horizontal axis represents:
Days
The vertical axis represents:
Practice duration in minutes
We need to find the practice duration on day 8.
To do this:
Locate 8 on the horizontal axis.
Move vertically until you reach the blue graph.
Read the corresponding value on the vertical axis.
From the graph, when:
x=8
the blue graph is at:
y=115
So the musician practiced for:
115 minutes
NEW QUESTION # 41
The graph shows the number of user accounts on a social media website over time.
When did the number of user accounts reach 4,500?
- A. After 3.2months
- B. After 2.0months
- C. After 7.0months
- D. After 8.8months
Answer: B
Explanation:
The graph represents the number of user accounts over time.
The horizontal axis represents:
" Time in months "
The vertical axis represents:
" Number of user accounts "
We need to find when the number of user accounts reached:
4,500
On the vertical axis, 4,500is halfway between:
3,000
and
6,000
So we look for the point where the blue curve reaches that height.
From the graph, the curve reaches approximately 4,500user accounts at:
x#2.0
That means the website had about 4,500user accounts after approximately:
2.0 " months "
NEW QUESTION # 42
The population of bison in a preserve can be modeled using the logistic function f(x), where x represents the number of years since the preserve was established and f(x) represents the population. The graph of f(x) is shown.
How does the bison population change as time progresses from year 7 to year 10?
- A. The bison population decreases faster and faster.
- B. The bison population decreases slower and slower.
- C. The bison population increases faster and faster.
- D. The bison population increases slower and slower.
Answer: D
Explanation:
A logistic function often increases quickly in the middle and then begins to level off as it approaches a maximum carrying capacity. From year 7 to year 10, the graph is still increasing, so the bison population is going up. However, the curve is becoming flatter during this interval. A flatter curve means the slope is getting smaller, so the population is increasing at a slower rate. This behavior is described as "increases slower and slower." It is not decreasing, because the graph continues to move upward. It is also not increasing faster and faster, because the slope is not becoming steeper in this interval. Therefore, the correct interpretation is answer B.
NEW QUESTION # 43
The logistic function f(x), whose graph is shown, models the number of registrants for an academic conference, where x represents the number of weeks since registration opened and f(x) represents the number of registrants.
How does the number of registrants change as time progresses from week 1 to week 7?
- A. The number of registrants increases slower and slower.
- B. The number of registrants decreases faster and faster.
- C. The number of registrants increases faster and faster.
- D. The number of registrants decreases slower and slower.
Answer: C
Explanation:
The graph shows a logistic growth function for conference registration. From week 1 to week 7, the number of registrants is increasing because the graph rises as time moves forward. The curve also becomes steeper throughout this interval, meaning the number of registrants is increasing at a faster rate each week. In Applied Algebra, when a graph rises and its slope increases, the correct interpretation is "increases faster and faster." The graph is not decreasing, so both decreasing options are incorrect. The graph is also not flattening during this interval, so "increases slower and slower" does not match. Therefore, the correct answer is C.
NEW QUESTION # 44
The function P(t)represents the daily profit, in hundreds of dollars, for a museum since opening. The graph of P(t)is shown.
What is the correct interpretation of the maximum value?
- A. Approximately 20 years after opening, a maximum daily profit of approximately $1,650was earned.
- B. Approximately 20 years after opening, a maximum daily profit of approximately $299was earned.
- C. Approximately 9.5 years after opening, a maximum daily profit of approximately $299was earned.
- D. Approximately 9.5 years after opening, a maximum daily profit of approximately $1,650was earned.
Answer: D
Explanation:
The graph shows a curved, downward-opening function. This type of graph is commonly associated with a quadratic polynomial function.
The maximum value of a downward-opening parabola occurs at its highest point, called the vertex.
From the graph, the highest point occurs at approximately:
t=9.5
This means the museum reaches its maximum daily profit approximately:
9.5 " years after opening "
The vertical axis represents daily profit in hundreds of dollars. From the graph, the maximum P(t)-value is approximately:
16.5
Since the profit is measured in hundreds of dollars:
16.5×100=1650
So the maximum daily profit is approximately:
$1,650
Therefore, the correct interpretation is:
" Approximately 9.5 years after opening, a maximum daily profit of approximately " $1,650 " was earned. "
NEW QUESTION # 45
The scatterplot shows data on the population of rabbits in a nature preserve. The graphed regression function has an r^2value of 0.94.
What is the appropriate range of x-values for extrapolation?
- A. x=-4to x=28
- B. Extrapolation is not appropriate because r^2 < 1.
- C. Extrapolation is not appropriate because r^2 > 0.7.
- D. x=0to x=24
Answer: D
Explanation:
The graph shows a logistic regression model for rabbit population data.
The given value is:
r^2=0.94
An r^2value close to 1means the model fits the data well. Since 0.94is close to 1, the regression model is a strong fit.
However, extrapolation should still be done carefully. Extrapolation means using the model to predict values outside the observed data range.
From the scatterplot, the data values appear to run roughly from about:
x=4 " to " x=20
A reasonable extrapolation range should stay close to the observed data. The interval:
x=0 " to " x=24
extends slightly beyond the data on both sides, so it is more reasonable than the much wider interval:
x=-4 " to " x=28
The statements saying extrapolation is not appropriate because r^2 < 1or because r^2 > 0.7are not valid interpretations.
NEW QUESTION # 46
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?
- A. (8#16)
- B. (0#8)
- C. (0#12)
- D. (0#16)
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 # 47
The number of letters processed daily at a mail center is modeled by the decreasing exponential function shown in the graph.
What is the long-term trend in the number of letters processed per day, based on the equation of the horizontal asymptote?
- A. 1,000
- B. 2,250
- C. 4,500
- D. 0
Answer: A
Explanation:
The graph shows a decreasing exponential function.
In Applied Algebra, a decreasing exponential function may approach a fixed value over time. This fixed value is called the horizontal asymptote.
The horizontal asymptote represents the long-term value that the function gets closer and closer to, but does not necessarily cross or reach exactly.
From the graph, the curve decreases quickly at first, then begins to level off near:
y=1,000
This means that as time continues, the number of letters processed per day approaches:
1,000
So the long-term trend is that the mail center will process about:
1,000 " letters per day "
NEW QUESTION # 48
The function f(z) represents the relationship between the number of units in two inventories, where z is the number of units in inventory A and f(z) is the number of units in inventory B. The number of units in inventory B is 2 more than the number of units in inventory A. Which function represents this situation?
- A. f(z)=z+2
- B. f(z)=z#2
- C. f(z)=2z
- D. f(z)=z/2
Answer: A
Explanation:
This problem translates a verbal statement into function notation. The variable z represents the number of units in inventory A, and f(z) represents the number of units in inventory B. The phrase "2 more than" tells us to add 2 to the number of units in inventory A. That gives the rule f(z)=z+2. For example, if inventory A has
10 units, then inventory B has f(10)=12 units. Option A subtracts 2, option C multiplies by 2, and option D divides by 2, so those do not match the stated relationship. Because inventory B is always exactly 2 more than inventory A, the correct function is f(z)=z+2. Therefore, the answer is B.
NEW QUESTION # 49
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