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

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.

The average rate of change measures how much the output changes per one unit of input. Here, the input is time in hours after 6:00 a.m., and the output is the number of vehicles stopped at the toll booth. The two given points are A=(1,49.4) and B=(4,45.8). Use the average rate of change formula: (y

2

​

−y

1

​

)/(x

2

​

−x

1

​

). Substituting the coordinates gives (45.8−49.4)/(4−1). The numerator is −3.6, and the denominator is 3, so the average rate of change is −1.2. The negative sign means the number of vehicles decreased over the interval.

The function is:

D(t)=48t+23

This is a linear function because it has the form:

D(t)=mt+b

where:

48

is the rate of change, and:

23

is the starting distance from the rest stop.

We need to find:

D(1.7)

That means substitute:

t=1.7

into the function.

D(1.7)=48(1.7)+23

First multiply:

48(1.7)=81.6

Now add 23:

D(1.7)=81.6+23

D(1.7)=104.6

So the vehicle is:

104.6 miles

from the rest stop after 1.7 hours.

The graph shows a decreasing exponential function.

The vertical axis represents:

" Letters per day "

The horizontal axis represents:

" Time in years "

For an exponential decay graph, the value decreases quickly at first and then begins to level off. The value it approaches in the long run is called the horizontal asymptote.

From the graph, the curve levels off near:

y=2,500

So the number of letters processed per day is trending toward:

2,500

This does not mean the mail center immediately processes exactly 2,500 letters per day. It means that as time continues, the number gets closer and closer to 2,500.

Therefore, the correct answer is:

â–­ ( " B " )

The graph represents a logistic growth model for viral infections. From day 9 to day 12, the curve is moving upward, so the number of infected people is increasing. During this interval, the graph is also becoming steeper, which means the rate of increase is getting larger. When a function is increasing and its slope is increasing, the correct description is that the quantity increases faster and faster. This interval occurs before the logistic graph begins to level off near its carrying capacity. The choices involving decrease are incorrect because the graph is not going downward. The choice “increases slower and slower” would apply to a later interval where the graph flattens. Therefore, the correct answer is A.

This problem compares two logistic growth curves. Each curve shows the percentage of the population affected by a disease over time. To determine which conclusion is valid, compare when each disease reaches the same percentage level. From the graph, Disease B reaches the 70% population level before Disease A reaches 70%. This means Disease B affects 70% of the population earlier in time. For other listed percentages, the graph does not support the same timing relationship as clearly or correctly. Logistic curves can cross or change relative position, so it is important to compare the same output level on both curves. Based on the graph, the valid conclusion is that Disease B affects 70% of the population before Disease A does. 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:

â–­ ( " B " )

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 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:

â–­ ( " A " )

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

Slow growth at the beginning (near point A)

Rapid growth in the middle (near point B and C)

Slowing growth at the end (near point D)

Key Concept:

Instantaneous rate of change = slope of the tangent line at a single point

Average rate of change = slope between two points

Analyze the graph:

At point A → slope is small (slow increase)

From A to C → increasing but not maximum

At point C → curve is steepest → maximum slope

From C to D → curve flattens → smaller slope

At point D → slope is very small (almost flat)

Conclusion:

The greatest rate of change occurs where the graph is steepest, which is at:

" Point C "

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:

â–­ ( " C " )

This question asks which town’s population is growing at a faster rate.

Since both population graphs are straight lines, we compare their slopes.

The slope of a line represents the rate of change:

" slope " = " change in population " / " change in time "

From the graph:

Town A starts at about 8.0thousand people and increases at a rate of about 1.0thousand people per year.

Town B starts at about 5.5thousand people and increases at a rate of about 1.3thousand people per year.

Even though Town A starts with a larger population, the question asks about the growth rate, not the starting population.

Compare the rates:

1.3 > 1.0

So Town B is growing faster than Town A.

The function f(n) gives the distance traveled by vehicle B.

The input n represents the distance traveled by vehicle A.

The problem says vehicle B travels 17 more than vehicle A. The phrase “17 more than” means we add 17 to the amount traveled by vehicle A.

So the relationship is:

f(n)=n+17

For example, if vehicle A traveled 50 miles, then vehicle B traveled:

f(50)=50+17=67

So the correct function is:

f(n)=n+17

Therefore, the correct answer is:

B

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.

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

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.

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 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.

This situation can be modeled with a linear relationship because each box has the same weight.

We are given:

4 " boxes " →116 " pounds "

and

9 " boxes " →161 " pounds "

The weight increases because more boxes are added.

First, find the change in total weight:

161-116=45

Then find the change in the number of boxes:

9-4=5

Since 5 additional boxes increased the total weight by 45 pounds, the weight of each box is:

45/5=9

So each box weighs:

9 " pounds "

Check:

If each box weighs 9 pounds, then adding 5 more boxes adds:

5(9)=45

116+45=161

This matches the given information.

The function F(n) gives the number of animals in enclosure B.

The input n represents the number of animals in enclosure A.

The problem says enclosure B has 20 more animals than enclosure A. The phrase “20 more than” means we add 20:

F(n)=n+20

For example, if enclosure A had 30 animals, then enclosure B would have:

F(30)=30+20=50

So the function that represents the situation is:

F(n)=n+20

We are told that computer B completes 12 more tasks than computer A.

Let:

x= " number of tasks completed by computer A "

and

F(x)= " number of tasks completed by computer B "

The phrase “12 more than” means add 12 to the original amount:

F(x)=x+12

So, if computer A completes xtasks, computer B completes:

x+12

tasks.

For example, if computer A completed 50 tasks, then computer B completed:

F(50)=50+12=62

Therefore, the correct answer is:

â–­ ( " B " )

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:

â–­ ( " B " )

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 graph shows population in thousands, and the horizontal axis represents:

" years since 1995 "

Point Pis located at approximately:

x=12

Since xrepresents years since 1995:

1995+12=2007

So point Pcorresponds to the year:

2007

The instantaneous rate of change at point Pis:

3.74

Because the vertical axis is measured in thousands, this means:

3.74 " thousand animals per year "

Convert thousands to actual animals:

3.74×1000=3740

So in 2007, the population was increasing by approximately:

3,740 " animals per year "

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 "

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:

â–­ ( " A " )

The function V(t)represents the number of travelers in the travel depot.

The input trepresents:

" hours after 7:00 a.m. "

The output V(t)represents:

" number of travelers "

So V(2)means the number of travelers in the depot:

2 " hours after 7:00 a.m. "

From the graph, when:

t=2

the value of V(t)is approximately:

79.2

Since V(t)measures travelers, not hours, V(2)represents:

79.2 " travelers "

Therefore, the correct answer is:

â–­ ( " D " )

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.

This is an exponential growth problem because the base 1.13 is greater than 1. The function starts with 2,900 bacteria and grows by a factor of 1.13 each hour, which represents a 13% hourly increase. To find the population after 18 hours, substitute t=18 into the function: f(18)=2900(1.13)

18

Evaluating the exponent and multiplying by 2900 gives approximately 26,170.38. Since bacteria are counted as whole organisms, this rounds to about 26,170 bacteria. The other answer choices do not match the exponential model at t=18. Therefore, the correct answer is C.

In an exponential decay model, the horizontal asymptote represents the long-term value that the function approaches as time increases. The graph shows the number of letters processed per day decreasing quickly at first and then leveling off. Since the graph approaches the horizontal line y=1000, the long-term trend is 1,000 letters per day. This does not mean the mail center immediately processes exactly 1,000 letters, but it means the model predicts the daily number will get closer and closer to 1,000 over time. The other choices either represent values too low or values from earlier parts of the graph, not the asymptote. Therefore, the correct answer is B.

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 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 "

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.

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 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 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 problem states that the greatest number of trees are sold when the temperature is over 70°F. Therefore, we need to identify the days on the graph when tree sales are highest. The graph shows two major peaks in the number of trees sold per day. These peaks occur around day 2 and day 10. Since higher tree sales are linked to temperatures over 70°F, those two days are the most likely days when the temperature exceeded 70°F. Day 6 appears near a low point, so it is not likely associated with high temperature based on the given relationship. Day 12 also does not match the main peaks. Therefore, the correct answer is C.

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. "

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