1. **State the problem:** We have data for 10 students showing their 10km run times (in minutes) and weekly hours spent playing sport. We need to plot the last 5 points, draw a line of best fit, estimate weekly hours for a 35-minute 10km run, and determine the correlation type. 2. **Data points (last 5):** (48,12), (51,7), (56,4), (59,3), (47,9). 3. **Plotting these points:** The x-axis is 10km time, y-axis is hours of sport. 4. **Line of best fit formula:** For linear regression, the line is $y = mx + b$ where $m$ is slope and $b$ is intercept. 5. **Calculate slope $m$ and intercept $b$ using the last 5 points:** Calculate means: $$\bar{x} = \frac{48 + 51 + 56 + 59 + 47}{5} = \frac{261}{5} = 52.2$$ $$\bar{y} = \frac{12 + 7 + 4 + 3 + 9}{5} = \frac{35}{5} = 7$$ Calculate slope $m$: $$m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ Calculate numerator: $$(48 - 52.2)(12 - 7) + (51 - 52.2)(7 - 7) + (56 - 52.2)(4 - 7) + (59 - 52.2)(3 - 7) + (47 - 52.2)(9 - 7)$$ $$= (-4.2)(5) + (-1.2)(0) + (3.8)(-3) + (6.8)(-4) + (-5.2)(2)$$ $$= -21 + 0 - 11.4 - 27.2 - 10.4 = -70$$ Calculate denominator: $$(48 - 52.2)^2 + (51 - 52.2)^2 + (56 - 52.2)^2 + (59 - 52.2)^2 + (47 - 52.2)^2$$ $$= (-4.2)^2 + (-1.2)^2 + (3.8)^2 + (6.8)^2 + (-5.2)^2$$ $$= 17.64 + 1.44 + 14.44 + 46.24 + 27.04 = 106.8$$ So slope: $$m = \frac{-70}{106.8} \approx -0.655$$ Calculate intercept $b$: $$b = \bar{y} - m \bar{x} = 7 - (-0.655)(52.2) = 7 + 34.2 = 41.2$$ 6. **Equation of line of best fit:** $$y = -0.655x + 41.2$$ 7. **Estimate weekly hours for 35-minute 10km run:** $$y = -0.655(35) + 41.2 = -22.925 + 41.2 = 18.275$$ Rounded to nearest whole number: 18 hours. 8. **Correlation type:** Since slope is negative, the graph shows a **negative correlation**. **Final answers:** - Estimated weekly hours for 35-minute run: 18 - Correlation: Negative (n)