Applied-Algebra WGU Applied Algebra FXO2 PFXP C957 Questions and Answers

The graph shows the queue size changing over time.

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Queue size "

The graph is a decreasing exponential curve. It starts near 750people and decreases quickly at first. Then the curve begins to flatten near:

y=100

A horizontal asymptote is the horizontal line that the graph approaches as time continues.

Here, the graph approaches the horizontal line:

y=100

This means that as time passes, the number of people in the virtual queue gets closer and closer to 100.

It does not mean the queue decreases to 0, because the graph levels off near 100, not near 0.

Therefore, the correct interpretation is:

" As time passes, the number of people in the queue approaches 100. "

The graph shows the weekly number of players on a gaming website.

The horizontal axis represents:

" years since launching "

The vertical axis represents:

" players in hundreds "

The graph is an upward-opening curve, so the minimum value occurs at the lowest point of the graph.

From the graph, the lowest point occurs at approximately:

t=4

This means the website reached its minimum weekly number of players approximately:

4 " years after it was launched "

Therefore, the correct answer is:

▭ ( " B " )

The graph shows customer visits by day.

To determine which days likely featured new releases, we should look for the days with unusually high customer counts.

From the graph:

Day 8 has about 40customers.

Day 11 also has about 40customers.

These are the highest values shown on the graph.

Since the problem says more customers show up on days when new releases are featured, the days with the highest customer counts are the most likely new-release days.

Therefore, the correct answer is:

▭ ( " D " )

The graph models the number of widgets the team can produce each day.

The horizontal axis represents:

" days "

The vertical axis represents:

" widgets "

The graph gives two labeled points:

(6│3.4)

and

(14│23)

The average rate of change is:

" change in widgets " / " change in days "

Substitute the values:

(23-3.4)/(14-6)

=19.6/8

=2.45

Rounded to the nearest tenth:

2.5

This means the team increased production by an average of approximately:

2.5 " widgets each day "

Therefore, the correct answer is:

▭ ( " B " )

The scatterplot shows the number of monthly train crossings decreasing over time. The decrease is not constant from one year to the next, so a linear model is not the best choice. Instead, the data decreases quickly at first and then begins to level off, which is characteristic of exponential decay. Exponential models are appropriate when a quantity changes by a repeated factor or percentage over equal time intervals. A logistic model would usually show an S-shaped pattern with a carrying capacity, and a polynomial model would be more appropriate for turning points or more complex curvature. Since the points show a smooth decreasing curve that flattens over time, an exponential function best models the data. Therefore, the correct answer is B.

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.

================

The model has:

r^2=0.42

This indicates a moderate fit, not a strong fit.

The prediction is being made at:

x=19.5

From the scatterplot, the observed data values appear to extend only to about:

x=16

So x=19.5is beyond the observed data range.

Because the model has only a moderate fit and the prediction is too far beyond the observed data, the prediction is not appropriate.

The correct interpretation is:

" No. The " r^2 " value indicates a moderate fit, but " x=19.5 " is more than " 25% " of the range beyond the maximum value. "

The horizontal asymptote describes the long-term behavior of the function. In this problem, the graph shows a decreasing exponential model for the number of licensed sellers. The graph begins near a higher value and then decreases, eventually leveling off near 300. This means the number of licensed sellers approaches 300 over time. It does not mean the number necessarily becomes exactly 300 at 21 months, and it does not mean the sellers increase back to the initial value. The initial value is separate from the horizontal asymptote. Since the graph levels off around 300, the correct interpretation is that in the long run, the number of licensed sellers nears 300. Therefore, the answer is C.

The regression model has r

2

=0.56, which indicates a moderate fit. A moderate fit is not as reliable as a strong fit, but it can still be used for limited prediction when the extrapolation is close to the observed data. The prediction is made at x=10.8 days. This value is slightly beyond the maximum observed x-value, so the prediction is an extrapolation. For a moderate fit, the x-value should remain within about 25% of the range beyond the maximum value. In this case, x=10.8 is within that acceptable range, so the prediction is appropriate. Therefore, the correct answer is B.

The graph represents:

p(t)= " number of active players "

where:

t= " hours after 11:00 a.m. "

The phrase “decreasing faster and faster” means two things are happening:

The graph is going downward, so the number of players is decreasing.

The graph is becoming steeper downward, so the rate of decrease is increasing.

This corresponds to a graph that is decreasing and concave down.

Looking at the graph, the number of active players reaches a maximum around:

t≈9

After that, the graph begins decreasing. Near the far right side of the graph, especially around:

t=11.2 " to " t=11.5

the curve is dropping more and more steeply.

That means the number of players is decreasing faster and faster on that interval.

The graph represents yearly profit as a function of time:

P(t)= " yearly profit "

where:

t= " years since opening "

The graph is a downward-opening curve, so its maximum value occurs at the highest point, also called the vertex.

From the graph, the highest point occurs at approximately:

t=10.5

So the maximum yearly profit happened approximately:

10.5 " years after opening "

The vertical axis measures yearly profit in millions of dollars. From the graph, the maximum value is approximately:

1.41

Since the units are millions of dollars:

1.41×1,000,000=1,410,000

So the maximum yearly profit was approximately:

$1,410,000

Therefore, the correct answer is:

▭ ( " C " )

The graph shows a linear relationship between the amount of red dye, r, and the amount of yellow dye, y.

The rate of change of a line is the slope:

" slope " =( " change in " y)/( " change in " r)

From the graph, the line goes through the origin and rises steeply. A clear point on the line is approximately:

(2│9)

This means when the amount of red dye is 2gallons, the amount of yellow dye is 9gallons.

So the rate of change is:

9/2

That means for every 2gallons of red dye, the factory uses 9gallons of yellow dye.

The relationship can be written as:

y=9/2 r

So the correct interpretation is:

" The amount of yellow dye must be " 9/2 " of the amount of red dye. "

Therefore, the correct answer is:

▭ ( " A " )

This situation has a constant rate of change because each sack weighs the same amount. After 5 sacks are removed, the total weight is 135 pounds. After 9 sacks are removed, the total weight is 87 pounds. From 5 to 9 sacks removed is 4 more sacks. Over the same interval, the total weight decreases by 135−87=48 pounds. Since 4 sacks account for a 48-pound decrease, each sack must weigh 48÷4=12 pounds. This uses the linear idea that equal changes in the input produce equal changes in the output. Therefore, the weight of each sack is 12 pounds, which is answer A.

================

The coefficient of determination, r

2

, measures how well a regression model fits the data. A value close to 1 indicates a strong fit, while a lower value indicates that the model does not explain the data as reliably. Here, r

2

=0.45, which is below 0.7. That means the model does not have a strong enough fit for reliable extrapolation in this question. Extrapolation means using the model to predict values outside the observed data range, which is riskier than interpolation. Because the model fit is not strong, choosing an extended range of x-values would not be appropriate. The correct reason is that extrapolation is not appropriate because r

2

< 0.7. Therefore, the answer is D.

The average rate of change compares the change in output to the change in input over an interval. From the graph, the two labeled points are approximately (20,58.5) and (26,91.2). The input is days, and the output is widgets produced each day. Use the formula

x

2

​

−x

1

​

y

2

​

−y

1

​

​

Substituting gives

26−20

91.2−58.5

​

=

6

32.7

​

=5.45, which rounds to approximately 5.5. Since this value is an average rate, it means the team’s daily production increased by about 5.5 widgets each day during the interval. It is not the total number of additional widgets over the whole interval. Therefore, the correct answer is A.

The graph shows the temperature of an object decreasing over time.

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Temperature in degrees Fahrenheit "

The curve decreases and approaches the horizontal axis as time increases.

A horizontal asymptote is a horizontal line that the graph approaches but may not reach. Horizontal lines have equations of the form:

y= " constant "

From the graph, the curve approaches:

y=0

So the horizontal asymptote is:

y=0

The options x=0and x=500are vertical lines, not horizontal asymptotes.

Therefore, the correct answer is:

▭ ( " B " )

A decreasing exponential function falls quickly at first and then levels off over time. The “slowest average decrease” occurs where the graph is flattest, because the output changes by the smallest amount per year. Earlier intervals, such as 2005 to 2008, occur when the graph is steepest, so those intervals have a faster average decrease. Later intervals occur after the exponential decay has slowed. From the graph, the interval from 2015 to 2025 is farthest to the right and has the least steep downward trend. Therefore, the number of registered voters decreases most slowly during that interval. This is why the correct answer is D.

The function A(t)represents the number of people expected to audition this year.

The input trepresents the number of people who auditioned last year.

The problem says this year’s number is expected to be 3 less than last year’s number. Therefore, the function rule is:

A(t)=t-3

We are told that 280 people auditioned last year, so:

t=280

Substitute 280into the function:

A(280)=280-3

A(280)=277

So, the notation that represents the number of people expected to audition this year is:

A(280)=277

This means if 280 people auditioned last year, then 277 people are expected to audition this year.

Therefore, the correct answer is:

▭ ( " A " )

This graph represents a logistic growth function, which has an S-shaped curve.

Key behavior of logistic functions:

Initial phase (early time) → growth is slow

Middle phase → growth speeds up (increasing rate)

Later phase → growth slows down as it approaches a maximum

Analyze the interval from day 1 to day 17:

This interval is in the early part of the graph.

The curve is increasing and getting steeper over this range.

That means the rate of increase is growing over time.

So, the number of comments is:

" increasing faster and faster "

We are given the function:

d(x)=14+65x

This is a linear function because it has the form:

d(x)=mx+b

where 65is the rate of change and 14is the starting value.

We need to find:

d(1.5)

That means substitute x=1.5into the function:

d(1.5)=14+65(1.5)

Now multiply:

65(1.5)=97.5

Then add:

d(1.5)=14+97.5

d(1.5)=111.5

So the distance from the tower to the object at 1.5seconds is:

111.5 " meters "

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.

The graph is a logistic growth function.

A logistic function often has an S-shape. It begins by increasing slowly, then increases faster and faster, and later levels off as it approaches a maximum carrying capacity.

From year 1 to year 9, the graph is rising and becoming steeper.

That means:

" the wild horse population is increasing "

and the slope is getting larger, so:

" the population is increasing faster and faster "

Therefore, the correct answer is:

▭ ( " A " )

We need to compare the two account values after:

t=3

First, find the value of Account 1:

A_1 (3)=6,500× 〖 1.05 〗 ^3

A_1 (3)=6,500×1.157625

A_1 (3)≈7,524.56

Now find the value of Account 2:

A_2 (3)=3,000× 〖 0.96 〗 ^3

A_2 (3)=3,000×0.884736

A_2 (3)≈2,654.21

Now subtract:

7,524.56-2,654.21=4,870.35

So Account 1 is about:

$4,870

greater than Account 2 after 3 years.

Therefore, the correct answer is:

▭ ( " C " )

This is a polynomial function evaluation problem. Since I(m) represents daily income in dollars, the difference between I(7) and I(9) must also be in dollars. First evaluate I(7)=41(49)−490(7)+30=2009−3430+30=−1391. Then evaluate I(9)=41(81)−490(9)+30=3321−4410+30=−1059. The difference is |−1059−(−1391)|=332. Because the question asks for the difference between the two outputs, the correct unit is dollars, not months. So the difference between I(7) and I(9) is approximately $332. Therefore, the correct answer is C.

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. "

In this function-notation problem, the input represents last year’s number of people and the output represents this year’s expected number. Because this year’s number is 7 more than last year’s, the rule is A(t)=t+7. The question says 440 people auditioned last year, so the correct input is 440. Substituting gives A(440)=440+7=447. This means if 440 people auditioned last year, then 447 are expected to audition this year. The notation must place 440 inside the parentheses because that is the input value, and 447 must be written as the output value. The other options reverse the input and output or subtract instead of adding. Therefore, option C is correct.

================

To decide which town is growing faster, compare the slopes of the two lines. In a linear population graph, the slope represents the growth rate, while the starting value represents the initial population. The graph shows that Town A has a slope of about 1.0 thousand people per year, while Town B has a slope of about 0.7 thousand people per year. Since 1.0 > 0.7, Town A is growing at a faster rate. The values 6.5 and 5.5 refer to population amounts, not growth rates, so they do not answer the question. The correct comparison must use the slopes. Therefore, the correct answer is D.

This problem requires reading a value from a graph. The horizontal axis gives time in minutes, and the vertical axis gives the number of viewers. To solve it, find 432 on the vertical axis, locate where the graph reaches that value, and then move down to the horizontal axis. From the exhibit, the graph reaches 432 viewers at minute 1. That means the ordered pair is approximately (1, 432), so the input is 1 minute and the output is 432 viewers. The other listed minutes correspond to different viewer counts on the graph. In graphical depictions, matching the correct input-output pair is the key step. Therefore, the correct answer is B.

================

The function is:

P(t)=24t+185

This is a linear function in the form:

P(t)=mt+b

where:

m= " rate of change "

and

b= " initial value "

In the function:

P(t)=24t+185

the coefficient of tis:

24

So the rate of change is:

24 " fish per month "

Because 24is positive, the fish population is increasing.

The number 185is not the rate of change. It represents the starting fish population at the beginning of the year, when t=0:

P(0)=24(0)+185=185

Therefore, the correct interpretation is:

" The number of fish in the lake is increasing at a constant rate of 24 fish per month. "

So the correct answer is:

▭ ( " B " )

From d=0to about d=2, the graph is decreasing, but it is flattening as it approaches a minimum. This means the number of raffle tickets sold is decreasing at a slower rate:

" decreases slower and slower "

After the minimum, from about d=2to d=4.4, the graph increases and becomes steeper. This means the number of raffle tickets sold is increasing at a faster rate:

" increases faster and faster "

So the correct description is:

" The number of raffle tickets decreases slower and slower and then increases faster and faster. "

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.

The value r

2

=0.99 indicates a very strong regression fit because it is close to 1. A strong fit means the model explains most of the variation in the data, so limited extrapolation can be appropriate. However, extrapolation should stay close to the observed data range. From the scatterplot, the data values appear to run from about x=5 to x=15. A reasonable extrapolation range extends slightly beyond those observed values, not too far outside them. The interval x=2.5 to x=17.5 stays close to the original data and is more appropriate than the wider interval x=0 to x=20. Therefore, the correct answer is B.

The graph compares the spread of two viruses over time.

The horizontal axis represents:

" time in months "

The vertical axis represents:

" percent of the population affected "

The graph shows:

Virus A as the solid blue curve.

Virus B as the dashed blue curve.

To determine which statement is correct, compare when each virus reaches the same population percentage.

At 20%, Virus A reaches that level earlier than Virus B. The solid curve is already at 20%before the dashed curve reaches 20%.

However, for higher percentages like 70%and 80%, Virus B reaches those levels before Virus A because the dashed curve becomes steeper and rises above the solid curve after about 50%.

So the correct conclusion is:

" Virus A affects " 20% " of the population before virus B does. "

Therefore, the correct answer is:

▭ ( " A " )

Because the vehicle is traveling away from the town at a fixed rate, the distance can be modeled by a linear function:

d(t)=mt+b

where:

m= " rate of change "

and

b= " initial distance from the town "

We are given two points:

(1│200)

and

(4│395)

These mean:

After 1hour, the vehicle is 200miles away.

After 4hours, the vehicle is 395miles away.

First, find the rate of change:

m=(395-200)/(4-1)

m=195/3

m=65

So the vehicle is moving away from the town at a rate of:

65 " miles per hour "

Now the function has the form:

d(t)=65t+b

Use the point (1│200)to find b:

200=65(1)+b

200=65+b

b=135

Therefore, the function is:

d(t)=65t+135

Check using t=4:

d(4)=65(4)+135

d(4)=260+135

d(4)=395

The graph represents the height of a launched ball over time. Such height functions are typically quadratic because gravity causes the ball to rise, slow down, reach a maximum height, and then fall. The graph shown is a downward-opening parabola. A downward-opening parabola is concave down for its entire domain. Concavity changes only if the curve switches from bending downward to bending upward, or from bending upward to bending downward. Since this graph bends downward the entire time, there is no point where the concavity changes from concave down to concave up. The vertex is a maximum point, but it is not a change in concavity. Therefore, the correct statement is that the graph is always concave down.

The graph shows the water volume decreasing over time.

The vertical axis represents:

" Water volume in gallons "

The horizontal axis represents:

" Time in minutes "

The graph is an exponential decay curve. It decreases quickly at first, then levels off.

From the graph, the curve approaches a horizontal value near:

y=1,000

This means the horizontal asymptote is approximately:

y=1,000

Since the graph approaches 1,000 gallons, it will not decrease to 500 gallons according to this model.

The water volume does not change by a constant rate because exponential functions do not have constant additive change. Also, the graph decreases fastest at the beginning, not later.

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 " )

The average rate of change tells us how much the output changes, on average, for each 1-unit increase in the input.

Here:

x= " hours after 7:00 a.m. "

and

y= " estimated wait time in minutes "

From the graph, point Ais approximately:

A=(5,88.3)

and point Bis approximately:

B=(9,65)

Now use the average rate of change formula:

" Average rate of change " =(y_2-y_1)/(x_2-x_1 )

Substitute the values:

(65-88.3)/(9-5)

=(-23.3)/4

=-5.825

Rounded to two decimal places:

-5.83

The negative sign means the estimated wait time is decreasing from point Ato point B.

So the average rate of change is approximately:

-5.83 " minutes per hour "

The maximum value of a polynomial graph is found at the highest point of the curve. In this problem, P(t) represents yearly profit in millions of dollars, and t represents years since opening. The graph reaches its highest point at approximately t=10.5, so the maximum yearly profit occurs about 10.5 years after opening. The vertical value at that point is about 2.25, but because the profit is measured in millions of dollars, this means 2.25 million dollars. Converting gives 2.25×1,000,000=2,250,000. Therefore, the correct interpretation must include both the correct time and the correct dollar amount. The correct answer is C.