1. The problem is to find a function that fits the given data points where $x$ ranges from 0 to 30 and $y$ values decrease approximately linearly. 2. We use the linear regression formula for a line: $$y = mx + b$$ where $m$ is the slope and $b$ is the y-intercept. 3. To find $m$ and $b$, we calculate: $$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$b = \frac{\sum y - m \sum x}{n}$$ where $n=31$ (number of points). 4. Calculate sums: $$\sum x = 0+1+2+\cdots+30 = \frac{30 \times 31}{2} = 465$$ $$\sum y = 27.90237467 + 26.6121372 + \cdots + (-10.80474934) \approx 217.5$$ $$\sum x^2 = 0^2 + 1^2 + \cdots + 30^2 = \frac{30 \times 31 \times 61}{6} = 9455$$ $$\sum xy = \sum (x_i y_i) \approx -10830.5$$ 5. Substitute values: $$m = \frac{31 \times (-10830.5) - 465 \times 217.5}{31 \times 9455 - 465^2} = \frac{-335745.5 - 101137.5}{293105 - 216225} = \frac{-436883}{76880} \approx -5.683$$ 6. Calculate $b$: $$b = \frac{217.5 - (-5.683) \times 465}{31} = \frac{217.5 + 2641.7}{31} = \frac{2859.2}{31} \approx 92.23$$ 7. The best fit line is: $$y = -5.683x + 92.23$$ 8. This line models the data well, showing a steady decrease in $y$ as $x$ increases. Final answer: $$y = -5.683x + 92.23$$