Applied-Algebra Sample Questions Answers

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=0.99

This means the model is a very strong fit for the data because 0.99is close to 1.

However, a strong r^2value does not automatically make every prediction appropriate. We also have to check whether the x-value is within a reasonable extrapolation range.

The data shown on the graph appear to extend to about:

x=11

The prediction is for:

x=16.4

This is far beyond the observed data range. Even though the model fits the known data very well, predicting too far beyond the data can be unreliable.

The correct statement is that the r^2value indicates a strong fit, but x=16.4is more than 50%of the range beyond the maximum observed value.

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.

The graph shows a decreasing exponential function.

The horizontal axis represents:

" Time in months "

The vertical axis represents:

" New user accounts per day "

The graph starts near 7,000new user accounts per day and decreases over time. However, it does not decrease toward 0. Instead, the graph levels off near a horizontal value around:

y=2,500

This horizontal value is the horizontal asymptote.

A horizontal asymptote shows the long-term value the function approaches. Since the asymptote is above 0, the model predicts that the number of daily new user accounts approaches a positive value, not zero.

So the justified statement is:

" The number of daily new user accounts will not decrease to 0. "

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 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 " )

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.

This question uses a logistic function, which often models growth that starts slowly, increases rapidly, and then levels off near a maximum value. The input t represents years, and M(t) represents market share as a percent. To find when the company reaches a 60% market share, locate 60 on the vertical axis and identify where the graph reaches that height. Then read the corresponding value on the horizontal axis. From the graph, the curve reaches 60% slightly after 6 years, approximately at t=6.6. The value after 6 years is close but not yet at 60%. Therefore, the company should expect to reach 60% market share after about 6.6 years. The correct answer is D.

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 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 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

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.

Since xrepresents the number of years since 2003:

2003 ⇒ x=0

2007 ⇒ x=4

The average rate of change from 2003 to 2007 is:

(f(4)-f(0))/(4-0)

Find f(0):

f(0)=280× 〖 1.08 〗 ^0

f(0)=280

Find f(4):

f(4)=280× 〖 1.08 〗 ^4

f(4)≈380.94

Now calculate:

(380.94-280)/4

=100.94/4

≈25.23

Rounded to the nearest whole number:

25

Since the value is positive, the number of subscribers is increasing.

So the correct interpretation is:

" The number of subscribers increases at an average rate of 25 subscribers per year. "

Therefore, the correct answer is:

▭ ( " C " )

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.

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 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 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. "

This situation can be modeled using a linear relationship because each sack has the same weight.

We are given:

After 3sacks are removed, the total weight is 116pounds.

After 6sacks are removed, the total weight is 101pounds.

From 3 sacks removed to 6 sacks removed, the number of removed sacks increases by:

6-3=3

During that time, the total weight decreases from 116pounds to 101pounds:

116-101=15

So removing 3 additional sacks decreases the total weight by 15 pounds.

Now divide to find the weight of one sack:

15/3=5

So each sack weighs:

5 " pounds "

Check:

If 3 more sacks are removed and each sack weighs 5 pounds, the total weight should decrease by:

3(5)=15

116-15=101

This matches the given information.

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. "

Since xrepresents the number of years since 2004:

2013-2004=9

so:

2013 ⇒ x=9

Also:

2018-2004=14

so:

2018 ⇒ x=14

The average yearly rate of change from 2013 to 2018 is:

(f(14)-f(9))/(14-9)

Now evaluate the function:

f(9)=330× 〖 1.04 〗 ^9

f(9)≈469.37

f(14)=330× 〖 1.04 〗 ^14

f(14)≈571.13

Now calculate the average rate of change:

(571.13-469.37)/5

=101.76/5

=20.35

So the painting’s value increased by an average of about:

$20.35 " 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 "

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 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 function u(t) gives the number of units in the inventory warehouse t hours after 9:00 a.m. The expression u(8) means the value of the inventory function 8 hours after 9:00 a.m. Since 8 hours after 9:00 a.m. is 5:00 p.m., the time must be 5:00 p.m., not 10:00 a.m. Now evaluate the function: u(8)=0.21(8)

3

−4.09(8)

2

+18.89(8)+69.5. This gives approximately 66.38, which rounds to about 66 units. Therefore, the correct interpretation is that there are approximately 66 units in the inventory at 5:00 p.m.

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 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 " )

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 " )

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

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 " )

We are told that the number of property sales this year is expected to be 8 times the number of property sales last year.

Let:

x= " number of properties sold last year "

and

H(x)= " number of property sales this year "

Since this year’s sales are 8 times last year’s sales, the function rule is:

H(x)=8x

The question says that last year’s property sales were:

x=360

Substitute 360into the function:

H(360)=8(360)

H(360)=2,880

So the correct notation is:

H(360)=2,880

This means: if 360 properties were sold last year, then 2,880 property sales are expected this year.

Therefore, the correct answer is:

▭ ( " C " )

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

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 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 question asks us to identify a correct conclusion from the graph.

The vertical axis represents:

" number of daily homework problems "

The horizontal axis represents:

" school days since the beginning of the term "

From the graph, the lowest point occurs at the far right, around:

" day " 22

The corresponding number of homework problems is approximately:

18

So the minimum number of daily homework problems is 18, and it occurs at day 22.

Therefore, the correct answer is:

▭ ( " D " )

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.

This is a linear function evaluation problem. The function G(x)=24+47x gives the drone’s distance from the operator after x minutes. The constant 24 is the starting distance, and 47 is the rate of change in yards per minute. To evaluate G(1.9), substitute 1.9 for x. Then compute G(1.9)=24+47(1.9). Multiplying first gives 47×1.9=89.3. Adding 24 gives 113.3. Because the function measures distance in yards, the result means the drone is 113.3 yards from the operator after 1.9 minutes. The other options do not result from correctly substituting 1.9 into the function. Therefore, the correct answer is D.

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

The graph shows a linear relationship between:

w= " water in cups "

and

l= " lemon juice in cups "

The line passes through the origin, so the relationship has the form:

l=mw

where mis the rate of change.

From the graph, a clear point on the line is approximately:

(8│5)

This means when there are 8 cups of water, there are 5 cups of lemon juice.

The rate of change is:

m= " change in lemon juice " / " change in water "

m=5/8

So the relationship is:

l=5/8 w

This means the amount of lemon juice must be:

5/8

of the amount of water.