Question

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 749. Find a linear equation which fits this data, and use this equation to predict an average score for persons taking a 57-hour review course. Round your answer to the tenths place.

Answer 1

Given:

30-hour review course average a score of 620 on that exam.

70-hour review course average a score of 749.

To find:

The linear equation which fits this data, and use this equation to predict an average score for persons taking a 57-hour review course.

Solution:

Let x be the number of hours of review course and y be the average score on that exam.

30-hour review course average a score of 620 on that exam. So, the linear function passes through the point (30,620).

70-hour review course average a score of 749. So, the linear function passes through the point (70,749).

The linear function passes through the points (30,620) and (70,749). So, the linear equation is:

$y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y-620=\dfrac{749-620}{70-30}(x-30)$

$y-620=\dfrac{129}{40}(x-30)$

$y-620=\dfrac{129}{40}(x)-\dfrac{129}{40}(30)$

$y-620=\dfrac{129}{40}(x)-\dfrac{387}{4}$

Adding 620 on both sides, we get

$y=\dfrac{129}{40}x-\dfrac{387}{4}+620$

$y=\dfrac{129}{40}x+\dfrac{2480-387}{4}$

$y=\dfrac{129}{40}x+\dfrac{2093}{4}$

We need to find the y-value for $x=57$.

$y=\dfrac{129}{40}(57)+\dfrac{2093}{4}$

$y=183.825+523.25$

$y=707.075$

$y\approx 707.1$

Therefore, the required linear equation for the given situation is $y=\dfrac{129}{40}x+\dfrac{2093}{4}$ and the average score for persons taking a 57-hour review course is 707.1.