Courses and Certificates Applied-Algebra Syllabus Exam Questions Answers

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:

▭ ( " A " )

The function f(n) gives the distance traveled by vehicle B.

The input n represents the distance traveled by vehicle A.

The problem says vehicle B travels 17 more than vehicle A. The phrase “17 more than” means we add 17 to the amount traveled by vehicle A.

So the relationship is:

f(n)=n+17

For example, if vehicle A traveled 50 miles, then vehicle B traveled:

f(50)=50+17=67

So the correct function is:

f(n)=n+17

Therefore, the correct answer is:

B

This is a polynomial function evaluation problem. Since I(m) represents daily income in dollars, the difference between I(7) and I(9) must also be in dollars. First evaluate I(7)=41(49)−490(7)+30=2009−3430+30=−1391. Then evaluate I(9)=41(81)−490(9)+30=3321−4410+30=−1059. The difference is |−1059−(−1391)|=332. Because the question asks for the difference between the two outputs, the correct unit is dollars, not months. So the difference between I(7) and I(9) is approximately $332. Therefore, the correct answer is C.

This question asks us to interpret unusually high values on a graph. The problem states that more customer accounts are opened during snowy weeks than during weeks without snow. Therefore, the likely snowy weeks correspond to the peaks on the graph, where the number of new customer accounts is noticeably higher than surrounding weeks. The graph shows clear spikes at approximately week 7 and week 14. These weeks have higher account openings compared with nearby weeks, making them the best evidence of snowy conditions. Week 1 and week 5 do not match the main peaks shown in the graph. Since the snowy weeks should correspond to the highest customer-account values, the correct answer is D.