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Sagot :
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:
[tex]y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]
[tex]y-620=\dfrac{749-620}{70-30}(x-30)[/tex]
[tex]y-620=\dfrac{129}{40}(x-30)[/tex]
[tex]y-620=\dfrac{129}{40}(x)-\dfrac{129}{40}(30)[/tex]
[tex]y-620=\dfrac{129}{40}(x)-\dfrac{387}{4}[/tex]
Adding 620 on both sides, we get
[tex]y=\dfrac{129}{40}x-\dfrac{387}{4}+620[/tex]
[tex]y=\dfrac{129}{40}x+\dfrac{2480-387}{4}[/tex]
[tex]y=\dfrac{129}{40}x+\dfrac{2093}{4}[/tex]
We need to find the y-value for [tex]x=57[/tex].
[tex]y=\dfrac{129}{40}(57)+\dfrac{2093}{4}[/tex]
[tex]y=183.825+523.25[/tex]
[tex]y=707.075[/tex]
[tex]y\approx 707.1[/tex]
Therefore, the required linear equation for the given situation is [tex]y=\dfrac{129}{40}x+\dfrac{2093}{4}[/tex] and the average score for persons taking a 57-hour review course is 707.1.
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