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=(y_2-y_1)/(x_2-x_1)(x-x_1)`

`y-620=(749-620)/(70-30)(x-30)`

`y-620=(129)/(40)(x-30)`

`y-620=(129)/(40)(x)-(129)/(40)(30)`

`y-620=(129)/(40)(x)-(387)/(4)`

Adding 620 on both sides, we get

`y=(129)/(40)x-(387)/(4)+620`

`y=(129)/(40)x+(2480-387)/(4)`

`y=(129)/(40)x+(2093)/(4)`

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

`y=(129)/(40)(57)+(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=(129)/(40)x+(2093)/(4)` and the average score for persons taking a 57-hour review course is 707.1.