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

The function w(t)gives the number of workers at the job site thours after 6:00 a.m.

We need to find the difference between:

w(4)

and

w(8)

Because w(t)represents workers, the difference will be measured in workers, not hours.

First, evaluate w(4):

w(4)=0.03(4)^4-0.81(4)^3+7.55(4)^2-26.62(4)+44

w(4)=0.03(256)-0.81(64)+7.55(16)-106.48+44

w(4)=7.68-51.84+120.8-106.48+44

w(4)=14.16

Now evaluate w(8):

w(8)=0.03(8)^4-0.81(8)^3+7.55(8)^2-26.62(8)+44

w(8)=0.03(4096)-0.81(512)+7.55(64)-212.96+44

w(8)=122.88-414.72+483.2-212.96+44

w(8)=22.4

Now find the difference:

22.4-14.16=8.24

This is approximately:

8 " workers "

The table gives the percent of customers with fiber internet for each year since 2020.

Since 2023 is:

2023−2020=3

we use the row where:

Years since 2020=3

From the table, the percent of customers with fiber internet in 2023 is:

57%

The company has approximately:

1,050

customers each year.

Now find 57% of 1,050:

0.57×1,050=598.5

The closest answer choice is:

598

So approximately 598 customers had fiber internet in 2023.

Therefore, the correct answer is:

C

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.

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 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 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.

This question asks for correct function notation. The input x represents the number of properties sold last year, and the output H(x) represents the expected number of property sales this year. Since this year’s number is expected to be 9 more than last year’s number, the function rule is H(x)=x+9. The problem states that 140 properties were sold last year, so the correct input is x=140. Substituting gives H(140)=140+9=149. This means that when last year’s sales were 140, this year’s expected sales are 149. The input must be inside the parentheses, and the output must follow the equals sign. Therefore, the correct answer is B.

A horizontal asymptote is a horizontal line that a graph approaches as the input increases. Horizontal lines always have equations in the form y=constant, not x=constant. In this graph, the number of client accounts decreases over time and levels off near 125 accounts. Since the graph approaches 125 on the vertical axis, the horizontal asymptote is y=125. Choices x=125 and x=550 are vertical lines, so they cannot represent horizontal asymptotes. Choice y=550 represents a higher value near the beginning of the graph, not the long-term level. Therefore, the correct equation of the horizontal asymptote is y=125, making the answer B.

The graph shows an increasing exponential curve.

The number of subscribers is going up as time passes, so the function is increasing.

Also, the graph becomes steeper as time increases. This means the rate of increase is getting larger.

So the number of subscribers is:

" increasing faster and faster "

This is a key feature of exponential growth.

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 "

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 "

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. "

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.

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.

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

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 shows an exponential decay model for traffic flow. A horizontal asymptote represents the long-term value the function approaches. In this case, the graph decreases but levels off above zero. Because the horizontal asymptote is above the horizontal axis, the model predicts that the traffic flow approaches a positive number rather than decreasing all the way to zero. Exponential functions do not change by a constant additive rate; that is a property of linear functions. Instead, exponential models change by a constant factor. Therefore, statements about constant rate are not valid here. The valid conclusion based on the asymptote is that the traffic flow will not decrease to 0 vehicles per hour. The correct answer is B.

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 regression function has:

r^2=0.54

This indicates a moderate fit, not a strong fit. The model explains some of the variation in the data, but not as strongly as a model with an r^2value closer to 1.

Now consider the prediction at:

x=20.6

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

x=18

The prediction at x=20.6is beyond the observed data, so it is an extrapolation. However, it is not very far beyond the data range.

Because x=20.6is within a reasonable extrapolation range and the model has a moderate fit, the prediction is appropriate.

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 " )

The function is:

G(t)=1,600× 〖 1.31 〗 ^t

This is an exponential growth function because the base is greater than 1:

1.31 > 1

We need to find the population after 25 minutes, so substitute:

t=25

into the function:

G(25)=1,600× 〖 1.31 〗 ^25

Evaluate the exponent and multiply:

G(25)≈1,367,421.41

Since the population is a number of bacteria cells, round to the nearest whole number:

G(25)≈1,367,421

To describe “decreasing slower and slower,” the graph must be moving downward while becoming less steep. This means the function values are decreasing, but the rate of decrease is becoming smaller in magnitude. On the graph of x(t), the number of users first increases, then decreases after reaching a local maximum. The interval from t=6.8 to t=7.4 lies on a decreasing part of the curve where the graph is beginning to flatten compared with the steeper earlier decrease. That behavior means the number of users is still going down, but not as quickly as before. The other intervals either occur while the graph is increasing or include behavior that does not consistently match decreasing slower and slower. Therefore, the correct answer is C.

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 " )

The regression model has r

2

=0.42, which indicates a moderate fit. This means the model captures some of the pattern in the data, but it is not strong enough to support distant predictions. The proposed prediction is for x=12.4 days, which lies beyond the largest observed x-value in the scatterplot. Since the model is only moderate, acceptable extrapolation must remain close to the original data range. The value x=12.4 is more than 25% of the range beyond the maximum observed value, so it is too far outside the data for a reliable prediction. Therefore, the prediction is not appropriate. The correct answer is B.

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. "

The function is:

W(x)=740× 〖 0.87 〗 ^x

This is an exponential decay function because the base 0.87is between 0 and 1:

0 < 0.87 < 1

That means the sample’s weight decreases over time.

We need to find the weight when:

x=3

Substitute 3into the function:

W(3)=740× 〖 0.87 〗 ^3

First calculate the exponent:

〖 0.87 〗 ^3=0.87×0.87×0.87

〖 0.87 〗 ^3=0.658503

Now multiply by 740:

W(3)=740×0.658503

W(3)=487.29222

Rounded to two decimal places:

W(3)≈487.29

So the weight of the radioactive sample after 3 years is approximately:

487.29 " oz. "

The graph represents:

p(t)= " number of people at the gym "

where:

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

The question asks about the concavity of the graph between:

t=1.3

and

t=3.4

On this interval, the graph is still going upward, so the number of people is increasing.

However, the graph is beginning to flatten as it approaches its highest point. That means the slope is positive, but the slope is getting smaller.

In other words:

" The number of people is increasing, but at a slower rate. "

This is called concave down behavior. For a concave down graph that is increasing, the correct interpretation is:

" increasing slower and slower "

Therefore, the correct answer is:

▭ ( " C " )

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.

The average rate of change tells how much the output changes per one unit of input.

Here:

t= " hours after 5:00 a.m. "

and

n(t)= " number of animals "

The two given points are:

A=(1,10.6)

and

B=(8,6.59)

Use the average rate of change formula:

(y_2-y_1)/(x_2-x_1 )

Substitute the coordinates:

(6.59-10.6)/(8-1)

=(-4.01)/7

≈-0.573

The negative sign means the number of animals is decreasing.

So the correct interpretation is:

" The number of animals at the station decreases at an average rate of " 0.573 " animals per hour. "

Therefore, the correct answer is:

▭ ( " B " )

The value r

2

=0.96 indicates a strong regression fit because it is very close to 1. A strong fit supports prediction better than a weak or moderate fit. However, prediction also depends on whether the x-value is within a reasonable extrapolation range. The prediction is made at x=18.1, which is beyond the maximum observed x-value in the scatterplot. Even though the model fits the data well, the prediction is too far beyond the observed data range. For a strong model, extrapolation should not go more than about 50% of the range beyond the maximum value. Since x=18.1 exceeds that limit, the prediction is not appropriate. Therefore, the correct answer is B.

The function M(t)represents the company’s predicted market share after tyears.

The horizontal axis represents:

t= " time in years "

The vertical axis represents:

M(t)= " market share percentage "

We need to find when the company’s market share reaches:

60%

To answer this from the graph:

Locate 60on the vertical axis.

Move horizontally until reaching the blue curve.

Move downward to the horizontal axis to read the corresponding time.

From the graph, the blue curve reaches 60%a little after 6years, approximately at:

t=6.6

So the company should expect to have a market share of 60%after about:

6.6 " years "

The function a(x)represents the total value of an investment account.

The horizontal axis represents:

x= " time in years "

The vertical axis represents:

a(x)= " total account value in dollars "

We are asked to find the total value when:

x=4

To answer this from the graph:

Locate x=4on the horizontal axis.

Move vertically upward until reaching the blue graph.

Read the corresponding value on the vertical axis.

From the graph, when x=4, the value is slightly above $1,800, approximately:

a(4)≈1,830.59

So the total value of the investment account after 4 years is approximately:

$1,830.59

The graph shows a decreasing exponential model.

The vertical axis represents:

" Licensed sellers "

The horizontal axis represents:

" Time in years "

The graph starts near 700 licensed sellers and decreases over time. However, it does not appear to decrease all the way to 0. Instead, it levels off near:

y=100

That horizontal line is the horizontal asymptote.

A horizontal asymptote represents the long-term value the function approaches as time continues.

So the correct interpretation is:

" In the long run, the number of licensed sellers nears 100. "

Therefore, the correct answer is:

▭ ( " C " )

The scatterplot shows a decreasing pattern.

At first, the number of monthly crossings decreases quickly. Then the values begin to level off.

This type of pattern is characteristic of an exponential decay model.

A linear model would show points decreasing at a constant rate, forming an approximately straight line. Here, the decrease is not constant; it is steep at first and then slows down.

A logistic model usually has an S-shaped pattern, which is not shown here.

A polynomial model may curve, but the long-term leveling behavior shown in the scatterplot is best represented by an exponential decay function.

Therefore, the correct answer is:

▭ ( " D " )

The graph is a logistic growth curve.

A logistic growth curve has two main concavity regions:

" concave up before the inflection point "

and

" concave down after the inflection point "

From the graph, the function increases faster and faster until about:

x=6

After x=6, the graph continues increasing, but it begins to flatten out. That means the number of members is increasing slower and slower.

This is concave down behavior.

Therefore, one interval where the graph is concave down is:

(6│11)

So the correct answer is:

▭ ( " D " )

The graph compares the spread of two viruses over time.

The horizontal axis represents:

t= " time in months "

The vertical axis represents:

" percentage 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, we compare when each virus reaches the same population percentage.

At 20%:

Virus B reaches 20%earlier, around t≈5months.

Virus A reaches 20%later, around t≈6.5months.

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

For higher percentages such as 60%, 70%, and 80%, Virus A rises more quickly after about 50%, so Virus A reaches those percentages before Virus B.

Therefore, the correct conclusion is:

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

The function is:

p(m)=5m+5

This is a linear function in the form:

p(m)=mx+b

For a linear function, the average rate of change over any interval is the same as the slope.

The slope is the coefficient of m:

5

So the price increases at a constant rate of:

$5 " per month "

We can also verify using the average rate of change formula:

(p(10)-p(1))/(10-1)

Find p(10):

p(10)=5(10)+5=55

Find p(1):

p(1)=5(1)+5=10

Now calculate:

(55-10)/(10-1)

=45/9

=5

This question asks us to read and interpret a value from a graph.

The horizontal axis represents:

" Days since the beginning of the month "

The vertical axis represents:

" Practice duration in minutes "

We need to find the practice duration for:

" day " 9

To do this, locate 9on the horizontal axis, move upward to the blue graph, and then read the corresponding value on the vertical axis.

From the graph, the point at day 9is approximately:

127

So the musician practiced for:

127 " minutes "

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.