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

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.

The graph represents:

p(t)= " number of people at the gym "

where:

t= " hours after 7:00 a.m. "

The question asks about the concavity of the graph between:

t=1.3

and

t=3.4

On this interval, the graph is still going upward, so the number of people is increasing.

However, the graph is beginning to flatten as it approaches its highest point. That means the slope is positive, but the slope is getting smaller.

In other words:

" The number of people is increasing, but at a slower rate. "

This is called concave down behavior. For a concave down graph that is increasing, the correct interpretation is:

" increasing slower and slower "

Therefore, the correct answer is:

▭ ( " C " )

The value r

2

=0.99 indicates a very strong regression fit because it is close to 1. A strong fit means the model explains most of the variation in the data, so limited extrapolation can be appropriate. However, extrapolation should stay close to the observed data range. From the scatterplot, the data values appear to run from about x=5 to x=15. A reasonable extrapolation range extends slightly beyond those observed values, not too far outside them. The interval x=2.5 to x=17.5 stays close to the original data and is more appropriate than the wider interval x=0 to x=20. Therefore, the correct answer is B.

Since xrepresents the number of years since 2003:

2003 ⇒ x=0

2007 ⇒ x=4

The average rate of change from 2003 to 2007 is:

(f(4)-f(0))/(4-0)

Find f(0):

f(0)=280× 〖 1.08 〗 ^0

f(0)=280

Find f(4):

f(4)=280× 〖 1.08 〗 ^4

f(4)≈380.94

Now calculate:

(380.94-280)/4

=100.94/4

≈25.23

Rounded to the nearest whole number:

25

Since the value is positive, the number of subscribers is increasing.

So the correct interpretation is:

" The number of subscribers increases at an average rate of 25 subscribers per year. "

Therefore, the correct answer is:

▭ ( " C " )

The function c(t)gives the number of active shopping carts on a website.

Because c(t)represents active carts, the difference between c(6)and c(8)will be measured in:

" active carts "

not hours.

First evaluate c(6):

c(6)=-0.01(6)^4+0.25(6)^3-2.33(6)^2+9.02(6)+8

c(6)=19.28

Now evaluate c(8):

c(8)=-0.01(8)^4+0.25(8)^3-2.33(8)^2+9.02(8)+8

c(8)=18.08

Now find the difference:

19.28-18.08=1.20

This is approximately:

1 " active cart "

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. "

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 graph represents a logistic growth model for viral infections. From day 9 to day 12, the curve is moving upward, so the number of infected people is increasing. During this interval, the graph is also becoming steeper, which means the rate of increase is getting larger. When a function is increasing and its slope is increasing, the correct description is that the quantity increases faster and faster. This interval occurs before the logistic graph begins to level off near its carrying capacity. The choices involving decrease are incorrect because the graph is not going downward. The choice “increases slower and slower” would apply to a later interval where the graph flattens. Therefore, the correct answer is A.

The function A(t)represents the number of people expected to audition this year.

The input trepresents the number of people who auditioned last year.

The problem says this year’s number is expected to be 3 less than last year’s number. Therefore, the function rule is:

A(t)=t-3

We are told that 280 people auditioned last year, so:

t=280

Substitute 280into the function:

A(280)=280-3

A(280)=277

So, the notation that represents the number of people expected to audition this year is:

A(280)=277

This means if 280 people auditioned last year, then 277 people are expected to audition this year.

Therefore, the correct answer is:

▭ ( " A " )

The function is:

f(x)=10× 〖 1.09 〗 ^x

This is an exponential growth function because:

1.09 > 1

For an exponential growth function, the rate of increase becomes larger as time passes. That means later intervals usually have faster average increases than earlier intervals.

Since xrepresents years since 2001:

2001 ⇒ x=0

2003 ⇒ x=2

2005 ⇒ x=4

2013 ⇒ x=12

2016 ⇒ x=15

2019 ⇒ x=18

2021 ⇒ x=20

The latest interval is:

2013 " to " 2021

Because the graph is exponential growth and is steepest later in time, this interval has the fastest average increase.