1. **State the problem:** We are given two sets of data: Times at Bat ($x$) and Hits ($y$). We need to find the correlation coefficient $r$ of the best-fit line for these data points. 2. **Recall the formula for the correlation coefficient:** $$r = \frac{n\sum xy - \sum x \sum y}{\sqrt{\left(n\sum x^2 - (\sum x)^2\right)\left(n\sum y^2 - (\sum y)^2\right)}}$$ where $n$ is the number of data points. 3. **List the data points:** $$x = [4, 5, 8, 12, 22]$$ $$y = [1, 0, 2, 4, 6]$$ 4. **Calculate the necessary sums:** $$n = 5$$ $$\sum x = 4 + 5 + 8 + 12 + 22 = 51$$ $$\sum y = 1 + 0 + 2 + 4 + 6 = 13$$ $$\sum xy = (4)(1) + (5)(0) + (8)(2) + (12)(4) + (22)(6) = 4 + 0 + 16 + 48 + 132 = 200$$ $$\sum x^2 = 4^2 + 5^2 + 8^2 + 12^2 + 22^2 = 16 + 25 + 64 + 144 + 484 = 733$$ $$\sum y^2 = 1^2 + 0^2 + 2^2 + 4^2 + 6^2 = 1 + 0 + 4 + 16 + 36 = 57$$ 5. **Plug values into the formula:** $$r = \frac{5(200) - 51(13)}{\sqrt{(5)(733) - 51^2} \times \sqrt{(5)(57) - 13^2}}$$ 6. **Calculate numerator:** $$5 \times 200 = 1000$$ $$51 \times 13 = 663$$ $$\text{Numerator} = 1000 - 663 = 337$$ 7. **Calculate denominator parts:** $$5 \times 733 = 3665$$ $$51^2 = 2601$$ $$5 \times 57 = 285$$ $$13^2 = 169$$ 8. **Calculate denominator:** $$\sqrt{3665 - 2601} \times \sqrt{285 - 169} = \sqrt{1064} \times \sqrt{116}$$ 9. **Simplify denominator:** $$\sqrt{1064} \approx 32.61$$ $$\sqrt{116} \approx 10.77$$ $$\text{Denominator} = 32.61 \times 10.77 \approx 351.0$$ 10. **Calculate $r$:** $$r = \frac{337}{351.0} \approx 0.960$$ 11. **Interpretation:** The correlation coefficient is approximately $0.959$, indicating a strong positive linear relationship between Times at Bat and Hits. **Final answer:** $$r \approx 0.959$$