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

This question asks us to interpret a graph showing cost and revenue.

In Applied Algebra, the break-even point is where:

" Revenue " = " Cost "

On a graph, this happens where the revenue line and the cost line intersect.

From the graph:

The solid blue line represents revenue.

The dashed blue line represents cost.

The break-even point occurs where these two lines cross.

Looking carefully at the graph, the two lines intersect at approximately:

x=6

The horizontal axis represents the number of units, meaning the number of down quilts sold.

So the person needs to sell about:

6 " quilts "

to break even.

After selling more than 6 quilts, the revenue line is above the cost line, meaning the person begins making a profit.

The graph shows:

P(t)= " weekly profit in thousands of dollars "

where:

t= " years since opening "

The graph is a downward-opening curve, so it has a maximum value at its highest point. In Applied Algebra, the highest point of this kind of polynomial graph is called the vertex.

To answer the question, we need the t-value of the vertex because the question asks when the store had its maximum weekly profit.

From the graph, the highest point occurs between:

t=5

and

t=6

More specifically, it is approximately:

t=5.5

So the store had its maximum weekly profit approximately:

5.5 " years after opening "

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.

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:

â–­ ( " A " )

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.

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:

â–­ ( " D " )

The maximum value of a graph is the highest output value the function reaches. Since P(t) represents yearly profit, the maximum value occurs where the store earns its greatest yearly profit. The graph is a downward-opening curve, so its highest point is the vertex. To answer the question, we need the t-value of the vertex, not the profit amount. From the graph, the highest point occurs between 4 and 5 years after opening, closest to 4.5 years. This means the virtual store reached its maximum yearly profit approximately 4.5 years after it opened. The later values decrease after the vertex, so options involving 7.5 or 14 years are not correct. Therefore, the correct answer is B.

The function F(n) gives the number of animals in enclosure B.

The input n represents the number of animals in enclosure A.

The problem says enclosure B has 20 more animals than enclosure A. The phrase “20 more than” means we add 20:

F(n)=n+20

For example, if enclosure A had 30 animals, then enclosure B would have:

F(30)=30+20=50

So the function that represents the situation is:

F(n)=n+20