1. **State the problem:** We have a data set with independent variable $x$ and response variable $y$. We want to find the linear regression line $y = mx + b$ and then predict $y$ when $x = 57.4$. 2. **Formula for linear regression:** The slope $m$ and intercept $b$ are given by: $$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ $$b = \frac{\sum y - m \sum x}{n}$$ where $n$ is the number of data points. 3. **Calculate sums:** - $n = 9$ - $\sum x = 89.1 + 74.7 + 51.1 + 87.7 + 88.6 + 80.2 + 63.8 + 71.9 + 80.6 = 687.7$ - $\sum y = 12.4 + 41.9 + 35.5 + (-1.8) + 12.6 + 28.2 + 39.8 + 38.1 + 25.7 = 232.4$ - $\sum x^2 = 89.1^2 + 74.7^2 + 51.1^2 + 87.7^2 + 88.6^2 + 80.2^2 + 63.8^2 + 71.9^2 + 80.6^2 = 54,010.15$ - $\sum xy = (89.1)(12.4) + (74.7)(41.9) + (51.1)(35.5) + (87.7)(-1.8) + (88.6)(12.6) + (80.2)(28.2) + (63.8)(39.8) + (71.9)(38.1) + (80.6)(25.7) = 17,153.15$ 4. **Calculate slope $m$:** $$m = \frac{9 \times 17153.15 - 687.7 \times 232.4}{9 \times 54010.15 - (687.7)^2} = \frac{154378.35 - 159774.48}{486091.35 - 472933.29} = \frac{\cancel{154378.35} - 159774.48}{\cancel{486091.35} - 472933.29} = \frac{-5396.13}{13158.06} \approx -0.410$$ 5. **Calculate intercept $b$:** $$b = \frac{232.4 - (-0.410) \times 687.7}{9} = \frac{232.4 + 281.96}{9} = \frac{514.36}{9} \approx 57.15$$ 6. **Regression line equation:** $$y = -0.410x + 57.15$$ 7. **Predict $y$ when $x = 57.4$:** $$y = -0.410 \times 57.4 + 57.15 = -23.53 + 57.15 = 33.62$$ Rounded to 1 decimal place, $y = 33.6$. **Final answer:** The predicted value of $y$ when $x=57.4$ is **33.6**.