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

The graph shows a linear relationship between:

w= " water in cups "

and

l= " lemon juice in cups "

The line passes through the origin, so the relationship has the form:

l=mw

where mis the rate of change.

From the graph, a clear point on the line is approximately:

(8│5)

This means when there are 8 cups of water, there are 5 cups of lemon juice.

The rate of change is:

m= " change in lemon juice " / " change in water "

m=5/8

So the relationship is:

l=5/8 w

This means the amount of lemon juice must be:

5/8

of the amount of water.

A logistic function often increases quickly in the middle and then begins to level off as it approaches a maximum carrying capacity. From year 7 to year 10, the graph is still increasing, so the bison population is going up. However, the curve is becoming flatter during this interval. A flatter curve means the slope is getting smaller, so the population is increasing at a slower rate. This behavior is described as “increases slower and slower.” It is not decreasing, because the graph continues to move upward. It is also not increasing faster and faster, because the slope is not becoming steeper in this interval. Therefore, the correct interpretation is answer B.

The model has:

r^2=0.42

This indicates a moderate fit, not a strong fit.

The prediction is being made at:

x=19.5

From the scatterplot, the observed data values appear to extend only to about:

x=16

So x=19.5is beyond the observed data range.

Because the model has only a moderate fit and the prediction is too far beyond the observed data, the prediction is not appropriate.

The correct interpretation is:

" No. The " r^2 " value indicates a moderate fit, but " x=19.5 " is more than " 25% " of the range beyond the maximum value. "

The scatterplot shows a decreasing pattern.

At first, the number of monthly crossings decreases quickly. Then the values begin to level off.

This type of pattern is characteristic of an exponential decay model.

A linear model would show points decreasing at a constant rate, forming an approximately straight line. Here, the decrease is not constant; it is steep at first and then slows down.

A logistic model usually has an S-shaped pattern, which is not shown here.

A polynomial model may curve, but the long-term leveling behavior shown in the scatterplot is best represented by an exponential decay function.

Therefore, the correct answer is:

▭ ( " D " )

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 value r

2

=0.89 indicates a strong fit because it is close to 1. A strong fit means the regression function follows the scatterplot data closely enough to support reasonable prediction. The question asks whether a prediction at x=18.6 months is appropriate. This value is outside the observed data range, so the prediction involves extrapolation. For a strong regression model, extrapolation can be acceptable if the new x-value is not too far beyond the maximum observed value. Here, x=18.6 is within 50% of the range beyond the maximum value, so the prediction is considered appropriate. Therefore, the correct answer is C.

This question asks us to read and interpret a value from a graph.

The horizontal axis represents:

" Days since the beginning of the month "

The vertical axis represents:

" Practice duration in minutes "

We need to find the practice duration for:

" day " 9

To do this, locate 9on the horizontal axis, move upward to the blue graph, and then read the corresponding value on the vertical axis.

From the graph, the point at day 9is approximately:

127

So the musician practiced for:

127 " minutes "

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 " )

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 u(t) gives the number of units in the inventory warehouse t hours after 9:00 a.m. The expression u(8) means the value of the inventory function 8 hours after 9:00 a.m. Since 8 hours after 9:00 a.m. is 5:00 p.m., the time must be 5:00 p.m., not 10:00 a.m. Now evaluate the function: u(8)=0.21(8)

3

−4.09(8)

2

+18.89(8)+69.5. This gives approximately 66.38, which rounds to about 66 units. Therefore, the correct interpretation is that there are approximately 66 units in the inventory at 5:00 p.m.