1. **State the problem:** We have data points representing the number of successful baskets versus the distance from the basketball net. We need to create a scatter plot, draw a line of best fit, describe the correlation, and predict the number of baskets at 7 m. 2. **Scatter plot:** The points given are approximately (1,6), (1,5), (1,3), (2,4), (2,2), and (2,1) where the x-axis is "number of baskets" and the y-axis is "distance from net". 3. **Line of best fit:** We can estimate a line that best fits these points. The points suggest a negative correlation: as the number of baskets increases, the distance decreases. 4. **Correlation description:** The correlation is negative because as the distance from the net decreases, the number of baskets tends to increase. 5. **Prediction using line of best fit:** To predict the number of baskets at 7 m, we first find the equation of the line. Using two points (1,6) and (2,1): $$m = \frac{1 - 6}{2 - 1} = \frac{-5}{1} = -5$$ Equation form: $$y = mx + b$$ Using point (1,6): $$6 = -5(1) + b \Rightarrow b = 6 + 5 = 11$$ So the line is: $$y = -5x + 11$$ To find the number of baskets $x$ when distance $y=7$: $$7 = -5x + 11$$ $$-5x = 7 - 11 = -4$$ $$x = \frac{-4}{-5} = \frac{4}{5} = 0.8$$ So, at 7 m, the predicted number of baskets is approximately 0.8. **Final answer:** The predicted number of baskets at 7 m is about 0.8.