1. **Problem Statement:** We have data points representing hours spent studying ($x$) and money earned ($y$) by 10 students: (18, 23), (20, 21), (23, 20), (25, 19), (25, 21), (27, 18), (32, 16), (38, 17), (40, 16), (41, 23). We need to analyze the relationship and predict income for 10 hours of study. 2. **Analyzing the Relationship:** Plotting these points suggests a trend. To quantify it, we find the line of best fit using linear regression: $$y = mx + c$$ where $m$ is the slope and $c$ the intercept. 3. **Calculate Means:** $$\bar{x} = \frac{18+20+23+25+25+27+32+38+40+41}{10} = \frac{289}{10} = 28.9$$ $$\bar{y} = \frac{23+21+20+19+21+18+16+17+16+23}{10} = \frac{194}{10} = 19.4$$ 4. **Calculate Slope $m$:** $$m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ Calculate numerator: $$\sum (x_i - 28.9)(y_i - 19.4) = (18-28.9)(23-19.4) + (20-28.9)(21-19.4) + \ldots + (41-28.9)(23-19.4)$$ $$= (-10.9)(3.6) + (-8.9)(1.6) + (-5.9)(0.6) + (-3.9)(-0.4) + (-3.9)(1.6) + (-1.9)(-1.4) + (3.1)(-3.4) + (9.1)(-2.4) + (11.1)(-3.4) + (12.1)(3.6)$$ $$= -39.24 -14.24 -3.54 +1.56 -6.24 +2.66 -10.54 -21.84 -37.74 +43.56 = -85.8$$ Calculate denominator: $$\sum (x_i - 28.9)^2 = (-10.9)^2 + (-8.9)^2 + (-5.9)^2 + (-3.9)^2 + (-3.9)^2 + (-1.9)^2 + (3.1)^2 + (9.1)^2 + (11.1)^2 + (12.1)^2$$ $$= 118.81 + 79.21 + 34.81 + 15.21 + 15.21 + 3.61 + 9.61 + 82.81 + 123.21 + 146.41 = 628.9$$ 5. **Slope:** $$m = \frac{-85.8}{628.9} \approx -0.1364$$ 6. **Calculate Intercept $c$:** $$c = \bar{y} - m\bar{x} = 19.4 - (-0.1364)(28.9) = 19.4 + 3.94 = 23.34$$ 7. **Regression Line:** $$y = -0.1364x + 23.34$$ 8. **Interpretation:** The negative slope indicates that as study hours increase, money earned tends to decrease slightly. 9. **Prediction for $x=10$ hours:** $$y = -0.1364(10) + 23.34 = -1.364 + 23.34 = 21.976$$ **Final answer:** The predicted income for a student studying 10 hours is approximately 22 units of money.