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)](https://tex.z-dn.net/?f=y-y_1%3D%5Cdfrac%7By_2-y_1%7D%7Bx_2-x_1%7D%28x-x_1%29)
![y-620=\dfrac{749-620}{70-30}(x-30)](https://tex.z-dn.net/?f=y-620%3D%5Cdfrac%7B749-620%7D%7B70-30%7D%28x-30%29)
![y-620=\dfrac{129}{40}(x-30)](https://tex.z-dn.net/?f=y-620%3D%5Cdfrac%7B129%7D%7B40%7D%28x-30%29)
![y-620=\dfrac{129}{40}(x)-\dfrac{129}{40}(30)](https://tex.z-dn.net/?f=y-620%3D%5Cdfrac%7B129%7D%7B40%7D%28x%29-%5Cdfrac%7B129%7D%7B40%7D%2830%29)
![y-620=\dfrac{129}{40}(x)-\dfrac{387}{4}](https://tex.z-dn.net/?f=y-620%3D%5Cdfrac%7B129%7D%7B40%7D%28x%29-%5Cdfrac%7B387%7D%7B4%7D)
Adding 620 on both sides, we get
![y=\dfrac{129}{40}x-\dfrac{387}{4}+620](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B129%7D%7B40%7Dx-%5Cdfrac%7B387%7D%7B4%7D%2B620)
![y=\dfrac{129}{40}x+\dfrac{2480-387}{4}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B129%7D%7B40%7Dx%2B%5Cdfrac%7B2480-387%7D%7B4%7D)
![y=\dfrac{129}{40}x+\dfrac{2093}{4}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B129%7D%7B40%7Dx%2B%5Cdfrac%7B2093%7D%7B4%7D)
We need to find the y-value for
.
![y=\dfrac{129}{40}(57)+\dfrac{2093}{4}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B129%7D%7B40%7D%2857%29%2B%5Cdfrac%7B2093%7D%7B4%7D)
![y=183.825+523.25](https://tex.z-dn.net/?f=y%3D183.825%2B523.25)
![y=707.075](https://tex.z-dn.net/?f=y%3D707.075)
![y\approx 707.1](https://tex.z-dn.net/?f=y%5Capprox%20707.1)
Therefore, the required linear equation for the given situation is
and the average score for persons taking a 57-hour review course is 707.1.