Pass the WGU Courses and Certificates Applied-Algebra Questions and answers with CertsForce

The function is:

f(x)=1,300× 〖 0.92 〗 ^x

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

0 < 0.92 < 1

So the number of registered voters is decreasing over time.

Since xrepresents the number of years since 2001:

2001 ⇒ x=0

2007 ⇒ x=6

2008 ⇒ x=7

2009 ⇒ x=8

2012 ⇒ x=11

2021 ⇒ x=20

To determine the fastest average decrease, compare the average rates of change. A faster decrease means the average rate of change is more negative.

For 2001 to 2007:

(f(6)-f(0))/(6-0)

f(0)=1300

f(6)=1300(0.92)^6≈788.31

(788.31-1300)/6≈-85.28

So from 2001 to 2007, the number of registered voters decreased by about:

85.28 " voters per year "

For 2007 to 2008:

(f(7)-f(6))/(7-6)≈-63.06

For 2008 to 2021:

(f(20)-f(7))/(20-7)≈-42.58

For 2009 to 2012:

(f(11)-f(8))/(11-8)≈-45.13

Now compare the decreases:

-85.28, -63.06, -42.58, -45.13

The most negative rate is:

-85.28

That means the fastest average decrease happened from:

2001 " to " 2007

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 " )

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.

We need to compare the two account values after:

t=3

First, find the value of Account 1:

A_1 (3)=6,500× 〖 1.05 〗 ^3

A_1 (3)=6,500×1.157625

A_1 (3)≈7,524.56

Now find the value of Account 2:

A_2 (3)=3,000× 〖 0.96 〗 ^3

A_2 (3)=3,000×0.884736

A_2 (3)≈2,654.21

Now subtract:

7,524.56-2,654.21=4,870.35

So Account 1 is about:

$4,870

greater than Account 2 after 3 years.

Therefore, the correct answer is:

▭ ( " C " )

The graph shows the value of the coin increasing over time. Since the curve rises from left to right, the value is not decreasing. The graph also becomes steeper as time increases, which means the rate of increase is getting larger. In Applied Algebra, when a graph is increasing and its slope is increasing, the quantity is described as increasing faster and faster. This is typical of exponential growth, where values grow by a repeated percent or factor rather than by a constant amount. The choices involving decreasing are incorrect because the graph moves upward. The choice “increasing slower and slower” would require the graph to flatten, which it does not. Therefore, the correct answer is B.

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.

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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 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 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 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 "