1. **State the problem:** We are given a linear regression model $$\hat{y} = b_0 + b_1 x$$ where $x$ is the Price in Dollars and $\hat{y}$ is the predicted Number of Bids. We want to find the best fit line for the data points: Price = [30, 32, 43, 45, 50], Number of Bids = [2, 4, 5, 6, 7]. 2. **Formula used:** The slope $b_1$ and intercept $b_0$ of the least squares regression line are given by: $$b_1 = \frac{n \sum xy - \sum x \sum y}{n \sum x^2 - (\sum x)^2}$$ $$b_0 = \bar{y} - b_1 \bar{x}$$ where $n$ is the number of data points, $\sum xy$ is the sum of products of $x$ and $y$, $\sum x$ and $\sum y$ are sums of $x$ and $y$ values, and $\bar{x}$, $\bar{y}$ are means. 3. **Calculate sums and means:** $$n = 5$$ $$\sum x = 30 + 32 + 43 + 45 + 50 = 200$$ $$\sum y = 2 + 4 + 5 + 6 + 7 = 24$$ $$\sum x^2 = 30^2 + 32^2 + 43^2 + 45^2 + 50^2 = 900 + 1024 + 1849 + 2025 + 2500 = 8298$$ $$\sum xy = 30\times2 + 32\times4 + 43\times5 + 45\times6 + 50\times7 = 60 + 128 + 215 + 270 + 350 = 1023$$ $$\bar{x} = \frac{200}{5} = 40$$ $$\bar{y} = \frac{24}{5} = 4.8$$ 4. **Calculate slope $b_1$:** $$b_1 = \frac{5 \times 1023 - 200 \times 24}{5 \times 8298 - 200^2} = \frac{5115 - 4800}{41490 - 40000} = \frac{315}{1490}$$ Intermediate step with cancellation: $$b_1 = \frac{\cancel{315}}{\cancel{1490}}$$ (no common factors to cancel here, so fraction remains $\frac{315}{1490}$) Simplify fraction by dividing numerator and denominator by 5: $$b_1 = \frac{315 \div 5}{1490 \div 5} = \frac{63}{298} \approx 0.2114$$ 5. **Calculate intercept $b_0$:** $$b_0 = 4.8 - 0.2114 \times 40 = 4.8 - 8.456 = -3.656$$ 6. **Final regression equation:** $$\hat{y} = -3.656 + 0.2114 x$$ This means for each increase of 1 dollar in price, the predicted number of bids increases by approximately 0.2114, starting from about -3.656 when price is zero (extrapolated).