WGU Applied-Algebra - WGU Applied Algebra FXO2 PFXP C957

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 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.

This question asks us to interpret unusually high values on a graph. The problem states that more customer accounts are opened during snowy weeks than during weeks without snow. Therefore, the likely snowy weeks correspond to the peaks on the graph, where the number of new customer accounts is noticeably higher than surrounding weeks. The graph shows clear spikes at approximately week 7 and week 14. These weeks have higher account openings compared with nearby weeks, making them the best evidence of snowy conditions. Week 1 and week 5 do not match the main peaks shown in the graph. Since the snowy weeks should correspond to the highest customer-account values, the correct answer is D.

The 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:

â–­ ( " B " )

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 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 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. "

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 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.

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 " )