1. The problem asks us to find the linear regression equation of best fit for the starting salary data given from years 2000 to 2013, using $x$ as years since 2000. 2. We have data points $(x, y)$ for salary: $$ (0, 28000), (1, 29750), (5, 30500), (9, 32000), (11, 36750), (13, 41000) $$ 3. Calculate the necessary sums for linear regression formula $y = mx + b$: - $n = 6$ (number of data points) - $\sum x = 0 + 1 + 5 + 9 + 11 + 13 = 39$ - $\sum y = 28000 + 29750 + 30500 + 32000 + 36750 + 41000 = 197000$ - $\sum xy = (0)(28000) + (1)(29750) + (5)(30500) + (9)(32000) + (11)(36750) + (13)(41000) = 0 + 29750 + 152500 + 288000 + 404250 + 533000 = 1404500$ - $\sum x^2 = 0^2 + 1^2 + 5^2 + 9^2 + 11^2 + 13^2 = 0 + 1 + 25 + 81 + 121 + 169 = 397$ 4. Calculate slope $m$ using formula: $$ m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2} = \frac{6(1404500) - 39(197000)}{6(397) - 39^2} = \frac{8427000 - 7683000}{2382 - 1521} = \frac{744000}{861} \approx 864.542 $$ 5. Calculate intercept $b$: $$ b = \frac{\sum y - m \sum x}{n} = \frac{197000 - 864.542(39)}{6} = \frac{197000 - 33717.1}{6} = \frac{163282.9}{6} \approx 27213.82 $$ 6. The regression line is thus approximately: $$ y = 864.542x + 27213.82 $$ 7. Among the answer choices, the closest equation given is: $$ y = 839.721x - 27541.812 $$ which differs in sign and magnitude, meaning the provided options may involve a typo or reversed terms. The correct slope and intercept match option A in slope magnitude but differ in intercept sign. Final answer based on calculations: $$ y \approx 864.542x + 27213.82 $$ Note: None of the provided choices exactly match correct regression values from data.