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

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 " )

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 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:

▭ ( " A " )

We are told that this year’s number of property sales is 6 less than last year’s number of property sales.

Let:

x= " number of properties sold last year "

and

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

Since this year’s sales are 6 less than last year’s sales, the function rule is:

H(x)=x-6

The question says that the number of properties sold last year was:

x=330

Substitute 330into the function:

H(330)=330-6

H(330)=324

So the notation that correctly represents the number of property sales this year is:

H(330)=324

This means: if 330 properties were sold last year, then 324 properties are expected to be sold this year.

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 function is:

f(x)=190× 〖 0.99 〗 ^x

Since xrepresents the number of years since 2003:

2012 ⇒ x=9

2015 ⇒ x=12

The average rate of change from 2012 to 2015 is:

(f(12)-f(9))/(12-9)

First, find f(9):

f(9)=190× 〖 0.99 〗 ^9

f(9)≈173.59

Now find f(12):

f(12)=190× 〖 0.99 〗 ^12

f(12)≈168.37

Now calculate the average rate of change:

(168.37-173.59)/(12-9)

=(-5.22)/3

≈-1.74

Rounded to the nearest whole number:

-2

The negative sign means the number of subscribers is decreasing.

So the correct interpretation is:

" The number of subscribers decreases at an average rate of about 2 subscribers per year. "

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 graph shows the temperature of an object changing over time.

The horizontal axis represents:

" Time in minutes "

The vertical axis represents:

" Temperature in degrees Fahrenheit "

The curve is decreasing quickly at first and then begins to level off. This is the shape of an exponential decay function.

A horizontal asymptote is a horizontal line that the graph approaches as time increases.

Because a horizontal asymptote is a horizontal line, its equation must have the form:

y= " constant "

From the graph, the temperature approaches about:

25^ ∘ F

So the horizontal asymptote is:

y=25

This means the object’s temperature gets closer and closer to 25^ ∘ Fover time.

The graph shows a decreasing exponential function.

In Applied Algebra, a decreasing exponential function may approach a fixed value over time. This fixed value is called the horizontal asymptote.

The horizontal asymptote represents the long-term value that the function gets closer and closer to, but does not necessarily cross or reach exactly.

From the graph, the curve decreases quickly at first, then begins to level off near:

y=1,000

This means that as time continues, the number of letters processed per day approaches:

1,000

So the long-term trend is that the mail center will process about:

1,000 " letters per day "

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 "