Using line of best-fit, it is found that:
- The linear regression equation is [tex]y = 22.6x + 52.33[/tex]
- The profit would reach 234 thousand dollars during the calendar year of 2020.
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The line of best-fit is given by:
[tex]y = bx + a[/tex]
[tex]b = \frac{\sum (x-\overline{x})(y - \overline{y})}{\sum (x - \overline{x})^2}[/tex]
- With the slope, we replace y and x by their means in the line, to find a.
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[tex]\overline{x} = \frac{0 + 1 + 2 + 3 + 4 + 5}{6} = 2.5[/tex]
[tex]\overline{y} = \frac{38 + 92 + 98 + 119 + 147 + 159}{6} = 108.8[/tex]
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[tex]\sum x - \overline{x} = (0 - 2.5) + (1 - 2.5) + ... (5 - 2.5)[/tex]
[tex]\sum y - \overline{y} = (38 - 108.8) + ... (159 - 108.8)[/tex]
Using a calculator:
[tex]\sum (x-\overline{x})(y - \overline{y}) = 395.5[/tex]
[tex]\sum (x - \overline{x})^2 = 17.5[/tex]
Thus, the slope is:
[tex]b = \frac{\sum (x-\overline{x})(y - \overline{y})}{\sum (x - \overline{x})^2} = \frac{395.5}{17.5} = 22.6[/tex]
Then
[tex]y = 22.6x + a[/tex]
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Replacing the means, the intercept a is given by:
[tex]108.8 = 22.6(2.5) + a[/tex]
[tex]a = 108.8 - 22.6(2.5)[/tex]
[tex]a = 52.33[/tex]
Thus, the linear regression equation is:
[tex]y = 22.6x + 52.33[/tex]
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- The profit will reach 234 thousand dollars in x years after 2012, and x is found when y = 234, thus:
[tex]y = 22.6x + 52.33[/tex]
[tex]234 = 22.6x + 52.33[/tex]
[tex]22.6x = 234 - 52.33[/tex]
[tex]x = \frac{234 - 52.33}{22.6}[/tex]
[tex]x = 8.03[/tex]
2012 + 8 = 2020
The profit would reach 234 thousand dollars during the calendar year of 2020.
A similar problem is given at https://brainly.com/question/22992800
