1. **Problem Statement:** Given data for river width (m) and water speed (km/h), we need to draw a scatter plot, find a line of fit, write its equation, and estimate water speed at 50 m width. 2. **Understanding the Data:** The data points are: River Width $x$: 15, 18, 20, 28, 30, 32, 38, 40, 42, 45 Water Speed $y$: 12.6, 10.7, 11.2, 9.7, 8.1, 8.7, 6.9, 5.4, 3.9, 4.1 3. **Line of Fit Formula:** We use the least squares regression line formula: $$y = mx + b$$ where $$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ and $$b = \frac{\sum y - m \sum x}{n}$$ 4. **Calculate sums:** $$n=10$$ $$\sum x = 15+18+20+28+30+32+38+40+42+45 = 308$$ $$\sum y = 12.6+10.7+11.2+9.7+8.1+8.7+6.9+5.4+3.9+4.1 = 81.3$$ $$\sum x^2 = 15^2+18^2+20^2+28^2+30^2+32^2+38^2+40^2+42^2+45^2 = 10388$$ $$\sum xy = 15\times12.6 + 18\times10.7 + 20\times11.2 + 28\times9.7 + 30\times8.1 + 32\times8.7 + 38\times6.9 + 40\times5.4 + 42\times3.9 + 45\times4.1 = 2340.8$$ 5. **Calculate slope $m$:** $$m = \frac{10 \times 2340.8 - 308 \times 81.3}{10 \times 10388 - 308^2} = \frac{23408 - 25040.4}{103880 - 94864} = \frac{-1632.4}{90016} \approx -0.0181$$ 6. **Calculate intercept $b$:** $$b = \frac{81.3 - (-0.0181) \times 308}{10} = \frac{81.3 + 5.57}{10} = \frac{86.87}{10} = 8.687$$ 7. **Equation of line of fit:** $$y = -0.0181x + 8.687$$ 8. **Estimate water speed at 50 m width:** $$y = -0.0181 \times 50 + 8.687 = -0.905 + 8.687 = 7.782$$ 9. **Interpretation:** As river width increases, water speed decreases slightly, showing a negative relationship. 10. **Relationship for other scenarios:** - Daily attendance vs. hours of rain: Negative relationship (more rain, fewer attendees). - Number of diagonals vs. number of sides: Positive relationship (more sides, more diagonals).