1. **State the problem:** We are given data for cost (in millions) and profit (in thousands) for a motorcycle company over six months. We need to find the least squares regression line that models profit as a function of cost. 2. **Recall the formula for the least squares line:** The regression line is given by $$y = a + bx$$ where $$b = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ and $$a = \bar{y} - b\bar{x}$$ Here, $x$ is the cost, $y$ is the profit, and $n$ is the number of data points. 3. **List the data:** $$x: 9.5, 10.2, 10.9, 11.6, 12.4, 13.1$$ $$y: 230, 270, 310, 350, 390, 430$$ 4. **Calculate sums:** $$\sum x = 9.5 + 10.2 + 10.9 + 11.6 + 12.4 + 13.1 = 67.7$$ $$\sum y = 230 + 270 + 310 + 350 + 390 + 430 = 1980$$ $$\sum x^2 = 9.5^2 + 10.2^2 + 10.9^2 + 11.6^2 + 12.4^2 + 13.1^2 = 90.25 + 104.04 + 118.81 + 134.56 + 153.76 + 171.61 = 772.99$$ $$\sum xy = (9.5)(230) + (10.2)(270) + (10.9)(310) + (11.6)(350) + (12.4)(390) + (13.1)(430) = 2185 + 2754 + 3379 + 4060 + 4836 + 5633 = 22847$$ 5. **Calculate slope $b$:** $$b = \frac{6(22847) - (67.7)(1980)}{6(772.99) - (67.7)^2} = \frac{137082 - 134046}{4637.94 - 4582.09} = \frac{3036}{55.85}$$ Intermediate step with cancellation: $$b = \frac{\cancel{6}(22847) - (67.7)(1980)}{\cancel{6}(772.99) - (67.7)^2}$$ 6. **Simplify slope:** $$b \approx 54.36$$ 7. **Calculate intercept $a$:** $$\bar{x} = \frac{67.7}{6} \approx 11.28, \quad \bar{y} = \frac{1980}{6} = 330$$ $$a = 330 - 54.36 \times 11.28 = 330 - 613.5 = -283.5$$ 8. **Write the least squares line:** $$y = -283.5 + 54.36x$$ **Final answer:** The least squares regression line is $$\boxed{y = -283.5 + 54.36x}$$ where $y$ is profit in thousands and $x$ is cost in millions.