Applied-Algebra Exam Dumps - WGU Courses and Certificates Questions and Answers

The graph represents:

p(t)= " number of active players "

where:

t= " hours after 11:00 a.m. "

We need to interpret the concavity between:

t=0.4

and

t=3.6

On this interval, the graph is rising, so the number of players is increasing.

But as the graph rises, it begins to level off near the top. This means the rate of increase is getting smaller.

So the number of players is not increasing faster and faster. Instead, it is increasing, but more slowly as time passes.

That means the graph is concave down on this interval.

The correct interpretation is:

" The number of players is increasing slower and slower. "

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 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 "

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

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 " )

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:

▭ ( " A " )

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.

This graph represents a logistic growth function, which has an S-shaped curve:

Slow growth at the beginning (near point A)

Rapid growth in the middle (near point B and C)

Slowing growth at the end (near point D)

Key Concept:

Instantaneous rate of change = slope of the tangent line at a single point

Average rate of change = slope between two points

Analyze the graph:

At point A → slope is small (slow increase)

From A to C → increasing but not maximum

At point C → curve is steepest → maximum slope

From C to D → curve flattens → smaller slope

At point D → slope is very small (almost flat)

Conclusion:

The greatest rate of change occurs where the graph is steepest, which is at:

" Point C "

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 " )

This question asks for the cost of each additional shirt, which means we need to find the rate of change of the linear function.

On a graph, the rate of change is the slope:

" slope " = " change in total cost " / " change in number of shirts "

From the graph, two labeled points are:

(0│33)

and

(10│173)

This means:

When 0shirts are ordered, the cost is $33.

When 10shirts are ordered, the cost is $173.

Now find the slope:

m=(173-33)/(10-0)

m=140/10

m=14

So the cost increases by:

$14

for each additional shirt.

The $33represents a fixed starting cost, not the cost per shirt.