1. **Problem Statement:** Find the equation of the line of best fit for the car value data from 2013 to 2020, predict the car value in 2025, determine if the prediction is interpolation or extrapolation, and find the correlation coefficient $r$. 2. **Data Preparation:** Let $x$ represent years since 2013, so $x=0$ for 2013, $x=1$ for 2014, ..., $x=7$ for 2020. 3. **Data Points:** $$ \begin{aligned} & (0, 52000), (1, 44000), (2, 40000), (3, 37000), (4, 34000), (5, 31250), (6, 27500), (7, 24300) \end{aligned} $$ 4. **Formula for slope ($m$) and intercept ($b$):** $$ m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}, \quad b = \frac{\sum y - m \sum x}{n} $$ where $n=8$ (number of points). 5. **Calculate sums:** $$ \sum x = 0+1+2+3+4+5+6+7 = 28 $$ $$ \sum y = 52000 + 44000 + 40000 + 37000 + 34000 + 31250 + 27500 + 24300 = 289050 $$ $$ \sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140 $$ $$ \sum xy = 0\times52000 + 1\times44000 + 2\times40000 + 3\times37000 + 4\times34000 + 5\times31250 + 6\times27500 + 7\times24300 $$ $$ = 0 + 44000 + 80000 + 111000 + 136000 + 156250 + 165000 + 170100 = 862350 $$ 6. **Calculate slope $m$:** $$ m = \frac{8 \times 862350 - 28 \times 289050}{8 \times 140 - 28^2} = \frac{6898800 - 8093400}{1120 - 784} = \frac{-1194600}{336} = -3553.57 $$ 7. **Calculate intercept $b$:** $$ b = \frac{289050 - (-3553.57) \times 28}{8} = \frac{289050 + 99499.96}{8} = \frac{388549.96}{8} = 48568.75 $$ 8. **Equation of line of best fit:** $$ y = -3553.57x + 48568.75 $$ 9. **Predict value in 2025:** Year 2025 corresponds to $x = 2025 - 2013 = 12$. $$ y = -3553.57 \times 12 + 48568.75 = -42642.84 + 48568.75 = 5925.91 $$ 10. **Interpolation or extrapolation?** Since 2025 ($x=12$) is beyond the data range ($x=0$ to $7$), this is **extrapolation**. 11. **Calculate correlation coefficient $r$:** Formula: $$ r = \frac{n\sum xy - \sum x \sum y}{\sqrt{(n\sum x^2 - (\sum x)^2)(n\sum y^2 - (\sum y)^2)}} $$ Calculate $\sum y^2$: $$ 52000^2 + 44000^2 + 40000^2 + 37000^2 + 34000^2 + 31250^2 + 27500^2 + 24300^2 = 2.704 \times 10^{9} $$ More precisely: $$ 2704000000 + 1936000000 + 1600000000 + 1369000000 + 1156000000 + 976562500 + 756250000 + 590490000 = 110.537 \times 10^{7} $$ Calculate numerator and denominator: $$ r = \frac{8 \times 862350 - 28 \times 289050}{\sqrt{(8 \times 140 - 28^2)(8 \times 11053700000 - 289050^2)}} $$ $$ r = \frac{-1194600}{\sqrt{336 \times (88429600000 - 83565802500)}} = \frac{-1194600}{\sqrt{336 \times 4863797500}} = \frac{-1194600}{4044381} = -0.2955 $$ **Final answers:** - Equation: $y = -3553.57x + 48568.75$ - Predicted value in 2025: $5925.91$ - This is an example of **extrapolation** - Correlation coefficient $r \approx -0.30$