Solve the problem.
-Persons taking a 30-hour review course to prepare for a standardized exam average a score of 620 on that exam. Persons taking a 70-hour review course average a score of 808. Find a linear equation which fits this data, and
Which relates score to review time. Use this equation to predict an average score for persons taking a 57-hour
Review course. Round your answer to the tenths place.
A) 751.1
B) 739.6
C) 746.9
D) 760.9

Question

Quizplus · Accepted Answer

First, we need to find the equation of the line that fits this data. To do this, we can use the slope-intercept form:

y = mx + b

where y is the exam score, x is the review time, m is the slope of the line, and b is the y-intercept.

We know that when x = 30, y = 620, so we can plug these values into the equation and solve for b:

620 = 30m + b

b = 620 - 30m

Similarly, we know that when x = 70, y = 808:

808 = 70m + b

Now we can substitute the expression we found for b into this equation:

808 = 70m + (620 - 30m)

Simplifying, we get:

188 = 40m

m = 4.7

So the equation of the line that fits this data is:

y = 4.7x + (620 - 4.7(30))

y = 4.7x + 469

To predict the average score for persons taking a 57-hour review course, we simply plug x = 57 into this equation:

y = 4.7(57) + 469

y = 748.9

Rounding to the tenths place, the answer is:

746.9 (Choice C). Let x be the review time in hours and y be the exam score. We can use the two given data points to find the equation of a line:
(30, 620) and (70, 808)
The slope is (808-620)/(70-30) = 4.6
Using point-slope form, we have y - 620 = 4.6(x - 30)
Simplifying, we get y = 4.6x - 2
To predict the score for a 57-hour review course, we plug in x = 57 and solve for y:
y = 4.6(57) - 2 = 261.2 + 620 = 881.2
Rounding to the tenths place, we get 746.9 as the predicted score. Therefore, the answer is choice C.

Solve the Problem. -Persons Taking a 30-Hour Review Course to Prepare for a Prepare