Applied-Algebra Questions and Answers

The value

r^2=0.96

indicates a strong fit because it is very close to 1. This means the regression model explains much of the variation in the traffic-frequency data.

However, we also need to check whether the prediction at

x=16.2

is a reasonable extrapolation.

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

x=14

The predicted value is slightly beyond the observed data range, but not extremely far beyond it.

Since x=16.2is still within a reasonable extrapolation range, and the model has a strong r^2value, the prediction is appropriate.

Therefore, the correct answer is:

▭ ( " C " )

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

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

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

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 function M(t)represents the company’s 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 market share reaches:

40%

From the graph, locate 40on the vertical axis. Then move horizontally to the blue curve and read the corresponding value on the horizontal axis.

The curve reaches 40%at approximately:

t=7.4

So the company should expect to have a market share of 40%after about:

7.4 " years "

Therefore, the correct answer is:

▭ ( " C " )

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.

This is also a logistic growth curve.

Analyze the interval from month 2 to month 6:

The graph is increasing during this time.

The curve is becoming steeper, meaning the slope is increasing.

This indicates the rate of change is getting larger over time.

So, the number of affected acres is:

" increasing faster and faster "

This corresponds to the early-to-middle growth phase of a logistic function.

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 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 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:

▭ ( " D " )

From t=6.2to about t=8.5, the graph is increasing, but it is becoming less steep as it approaches a local maximum.

That means the number of customers is:

" increasing slower and slower "

After the local maximum, from about t=8.5to t=10.1, the graph begins to decrease and becomes steeper downward.

That means the number of customers is:

" decreasing faster and faster "

So the full description is:

" The number of customers increases slower and slower and then decreases faster and faster. "

Therefore, the correct answer is:

▭ ( " D " )

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 " )

This question asks for a description of concavity and rate of change on a specific interval. From d=10 to part of the interval, the elevation is increasing, but the graph becomes less steep as it approaches a local maximum. That means the elevation increases slower and slower. After the peak, the graph begins to decrease, and the downward slope becomes steeper. This means the elevation decreases faster and faster. The correct statement must describe both parts of the interval in order: first increasing at a slowing rate, then decreasing at an increasing rate. The other answer choices either reverse the behavior or describe the wrong direction of change. Therefore, the correct answer is A.

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.

We are told that the number of people expected to audition this year is 4 times the number of people who auditioned last year.

Let:

t=number of people who auditioned last year

and

A(t)=number of people expected to audition this year

Since this year’s number is 4 times last year’s number, the function is:

A(t)=4t

The question says that 330 people auditioned last year, so:

t=330

Now substitute 330 into the function:

A(330)=4(330)

A(330)=1320

So, the correct function notation is:

A(330)=1,320

This means: when 330 people auditioned last year, 1,320 people are expected to audition this year.

Therefore, the correct answer is:

B

The graph represents the value of an investment account over time. The horizontal axis gives time in years, and the vertical axis gives the total account value in dollars. To find the value after 4 years, locate x=4 on the horizontal axis and read the corresponding G(x)-value on the graph. The plotted exponential growth curve is slightly above $1,000 at 4 years, closest to $1,052.87. Because the graph shows an increasing exponential trend, values such as $480.52, $584.62, and $711.28 are too low for the account value at x=4. The correct interpretation is that after 4 years, the investment account is worth approximately $1,052.87. Therefore, the correct answer is D.

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

Here:

t= " hours after 8:00 a.m. "

and

n(t)= " number of people at the park "

The two given points are:

A=(2,110)

and

B=(6,182)

This means:

At t=2hours after 8:00 a.m., there were 110people at the park.

At t=6hours after 8:00 a.m., there were 182people at the park.

Now calculate the average rate of change:

" Average rate of change " =(182-110)/(6-2)

=72/4

=18

Because the value is positive, the number of people is increasing.

So the correct interpretation is:

" The number of people at the park increases at an average rate of 18 people per hour. "

The average daily change is the average rate of change over the interval from day 16 to day 20. From the graph, the performance score at day 16 is approximately 30.1, and the score at day 20 is approximately 57.6. Use the formula

x

2

​

−x

1

​

y

2

​

−y

1

​

​

Substituting gives

20−16

57.6−30.1

​

The numerator is 27.5, and the denominator is 4, so the average daily change is 27.5/4=6.875, which rounds to 6.88. This means the performance score increased by an average of about 6.88 points per day over the interval. Therefore, the correct answer is B.

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 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 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 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

The graph shows a linear relationship between:

" Number of boards "

and

" Total cost "

The cost of each additional board is the rate of change, or slope, of the line.

From the graph, two labeled points are:

(0│27)

and

(10│177)

This means:

When 0 boards are ordered, the total cost is $27.

When 10 boards are ordered, the total cost is $177.

Now calculate the slope:

m=(177-27)/(10-0)

m=150/10

m=15

So each additional board costs:

$15

Therefore, the correct answer is:

▭ ( " B " )

The function is:

f(x)=330× 〖 1.15 〗 ^x

Because xrepresents the number of years since 2005:

2006 ⇒ x=1

2011 ⇒ x=6

The average yearly rate of change from 2006 to 2011 is:

(f(6)-f(1))/(6-1)

First, find f(1):

f(1)=330× 〖 1.15 〗 ^1

f(1)=330×1.15

f(1)=379.50

Now find f(6):

f(6)=330× 〖 1.15 〗 ^6

f(6)≈330×2.31306

f(6)≈763.31

Now calculate the average rate of change:

(763.31-379.50)/(6-1)

=383.81/5

=76.762

Rounded to two decimal places:

76.76

So the average yearly rate of change of the artifact’s value from 2006 to 2011 is approximately:

$76.76 " per year "

Therefore, the correct answer is:

▭ ( " A " )

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.

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 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 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 "

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 for the correct interpretation of a labeled point on a graph. In function notation, N(x)=y means when the input is x weeks, the output is y books checked out. The exhibit shows several plotted points. Point B is located at approximately x=2.8 and y=177, so Point B represents N(2.8)=177. To verify a graph statement, both the horizontal coordinate and vertical coordinate must match the actual point on the graph. If either coordinate is wrong, then the statement is not accurate. Since Point B matches the stated coordinates and output value, it is the valid conclusion. Therefore, the correct answer is A.

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

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.

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

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 "

From week 1 to week 7, the graph is increasing and becoming steeper.

In Applied Algebra, when a graph is increasing and its slope is getting larger, the quantity is said to be:

" increasing faster and faster "

This interval is on the early increasing part of the logistic curve, before the graph begins to level off.

Therefore, the correct answer is:

▭ ( " 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.

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 graph shows the total value of an investment account as time increases.

The horizontal axis represents:

" Time in years "

The vertical axis represents:

" Total value in dollars "

We need to find the value of the account after:

8 " years "

From the graph, when t=8, the curve is slightly above $2,200. The closest answer choice is:

$2,221.12

So the value of the account after 8 years is approximately:

$2,221.12

Therefore, the correct answer is:

▭ ( " D " )