Applied-Algebra Exam Dumps : WGU Applied Algebra FXO2 PFXP C957

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 average rate of change measures how much the output changes per one unit of input.

Here:

t= " hours after 8:00 a.m. "

and

n(t)= " number of people at the park "

The two given points are:

A=(2,110)

and

B=(6,182)

This means:

At t=2hours after 8:00 a.m., there were 110people at the park.

At t=6hours after 8:00 a.m., there were 182people at the park.

Now calculate the average rate of change:

" Average rate of change " =(182-110)/(6-2)

=72/4

=18

Because the value is positive, the number of people is increasing.

So the correct interpretation is:

" The number of people at the park increases at an average rate of 18 people per hour. "

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 " )