To find the equation for the line of best fit, we can use the method of linear regression. Given the data:
\[ \begin{align*} \text{Month} (x) & : {1, 2, 3, 4, 5, 6} \ \text{Time (minutes)} (y) & : {46, 42, 40, 41, 38, 36} \end{align*} \]
We need to calculate the slope (m) and the y-intercept (b) of the line, which can be done with the formulas for linear regression.
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Calculate the averages:
- \( \bar{x} = \frac{1+2+3+4+5+6}{6} = 3.5 \)
- \( \bar{y} = \frac{46+42+40+41+38+36}{6} = 40.5 \)
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Calculate \( m \) (slope): \[ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \]
Compute the necessary sums:
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\( \sum (x_i - \bar{x})(y_i - \bar{y}) = (1-3.5)(46-40.5) + (2-3.5)(42-40.5) + (3-3.5)(40-40.5) + (4-3.5)(41-40.5) + (5-3.5)(38-40.5) + (6-3.5)(36-40.5) \)
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This gives:
- \( = (-2.5)(5.5) + (-1.5)(1.5) + (-0.5)(-0.5) + (0.5)(0.5) + (1.5)(-2.5) + (2.5)(-4.5) \)
- \( = -13.75 - 2.25 + 0.25 + 0.25 - 3.75 - 11.25 = -30.5 \)
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\( \sum (x_i - \bar{x})^2 = (1-3.5)^2 + (2-3.5)^2 + (3-3.5)^2 + (4-3.5)^2 + (5-3.5)^2 + (6-3.5)^2 \)
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This gives:
- \( = 6.25 + 2.25 + 0.25 + 0.25 + 2.25 + 6.25 = 17.5 \)
Then, calculating \( m \): \[ m = \frac{-30.5}{17.5} \approx -1.743 \approx -1.74 \]
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Calculate \( b \) (y-intercept): \[ b = \bar{y} - m \bar{x} = 40.5 - (-1.74)(3.5) \approx 40.5 + 6.09 \approx 46.59 \approx 46.6 \]
Now we have: \[ y \approx -1.74x + 46.6 \]
Thus, the equation for the line of best fit is: \[ \boxed{y = -1.74x + 46.6} \]