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

Because the vehicle is traveling away from the town at a fixed rate, the distance can be modeled by a linear function:

d(t)=mt+b

where:

m= " rate of change "

and

b= " initial distance from the town "

We are given two points:

(1│200)

and

(4│395)

These mean:

After 1hour, the vehicle is 200miles away.

After 4hours, the vehicle is 395miles away.

First, find the rate of change:

m=(395-200)/(4-1)

m=195/3

m=65

So the vehicle is moving away from the town at a rate of:

65 " miles per hour "

Now the function has the form:

d(t)=65t+b

Use the point (1│200)to find b:

200=65(1)+b

200=65+b

b=135

Therefore, the function is:

d(t)=65t+135

Check using t=4:

d(4)=65(4)+135

d(4)=260+135

d(4)=395

From d=0to about d=2, the graph is decreasing, but it is flattening as it approaches a minimum. This means the number of raffle tickets sold is decreasing at a slower rate:

" decreases slower and slower "

After the minimum, from about d=2to d=4.4, the graph increases and becomes steeper. This means the number of raffle tickets sold is increasing at a faster rate:

" increases faster and faster "

So the correct description is:

" The number of raffle tickets decreases slower and slower and then increases faster and faster. "

The regression model has:

r^2=0.99

This means the model is a very strong fit for the data because 0.99is close to 1.

However, a strong r^2value does not automatically make every prediction appropriate. We also have to check whether the x-value is within a reasonable extrapolation range.

The data shown on the graph appear to extend to about:

x=11

The prediction is for:

x=16.4

This is far beyond the observed data range. Even though the model fits the known data very well, predicting too far beyond the data can be unreliable.

The correct statement is that the r^2value indicates a strong fit, but x=16.4is more than 50%of the range beyond the maximum observed value.

Therefore, the correct answer is:

â–­ ( " A " )

The graph shows an increasing exponential curve.

The number of subscribers is going up as time passes, so the function is increasing.

Also, the graph becomes steeper as time increases. This means the rate of increase is getting larger.

So the number of subscribers is:

" increasing faster and faster "

This is a key feature of exponential growth.

The r

2

value tells how closely the regression model fits the scatterplot data. In this problem, r

2

=0.09, which is extremely low. This means the regression model explains only a very small portion of the variation in the shirt-sales data. When r

2

is this weak, the model is not reliable for predictions outside the data range. Extrapolation is already less certain than interpolation because it predicts beyond the observed data. With such a poor fit, extending the model beyond the data would not be appropriate. Since 0.09 < 0.3, the best justification is that extrapolation is not appropriate because r

2

< 0.3. Therefore, the correct answer is C.

The graph models the number of widgets the team can produce each day.

The horizontal axis represents:

" days "

The vertical axis represents:

" widgets "

The graph gives two labeled points:

(6│3.4)

and

(14│23)

The average rate of change is:

" change in widgets " / " change in days "

Substitute the values:

(23-3.4)/(14-6)

=19.6/8

=2.45

Rounded to the nearest tenth:

2.5

This means the team increased production by an average of approximately:

2.5 " widgets each day "

Therefore, the correct answer is:

â–­ ( " B " )

This problem asks for an average yearly rate of change over a time interval. Since x represents years since 2005, the year 2013 corresponds to x=8, and the year 2015 corresponds to x=10. Use the average rate of change formula:

10−8

f(10)−f(8)

​

Evaluating the model gives f(8)=240(1.12)

8

≈594.23 and f(10)=240(1.12)

10

≈745.40. The total change is approximately 151.17 dollars over 2 years. Dividing by 2 gives about 75.59 dollars per year. Since the question asks for the average yearly rate, not the total increase, the correct answer is B.

The graph shows weekly profit as a function of time.

The horizontal axis represents:

" years since opening "

The vertical axis represents:

" weekly profit in hundreds of dollars "

The graph is an upward-opening curve, so its minimum value occurs at the lowest point, called the vertex.

From the graph, the lowest point occurs at approximately:

t=6

So the minimum weekly profit happened approximately:

6 " years after opening "

The vertical value at this point is approximately:

9.9

Since the profit is measured in hundreds of dollars:

9.9×100=990

So the minimum weekly profit was approximately:

$990

Therefore, the correct answer is:

â–­ ( " A " )