1. **State the problem:** We have data for years since 1995 ($x$) and new crime cases ($y$). We want to find the linear regression equation $y = mx + b$ that fits the data, rounding coefficients to the nearest hundredth. 2. **List the data points:** $(0,1053), (1,1059), (2,1122), (3,1156), (4,1145), (5,1194)$. 3. **Calculate the slope $m$ using the formula:** $$m = \frac{n\sum xy - \sum x \sum y}{n\sum x^2 - (\sum x)^2}$$ where $n=6$ (number of points). 4. **Calculate sums:** $$\sum x = 0+1+2+3+4+5 = 15$$ $$\sum y = 1053+1059+1122+1156+1145+1194 = 6729$$ $$\sum xy = 0\cdot1053 + 1\cdot1059 + 2\cdot1122 + 3\cdot1156 + 4\cdot1145 + 5\cdot1194 = 0 + 1059 + 2244 + 3468 + 4580 + 5970 = 17321$$ $$\sum x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 0 + 1 + 4 + 9 + 16 + 25 = 55$$ 5. **Plug values into slope formula:** $$m = \frac{6 \times 17321 - 15 \times 6729}{6 \times 55 - 15^2} = \frac{103926 - 100935}{330 - 225} = \frac{2991}{105}$$ 6. **Simplify slope:** $$m = \frac{\cancel{2991}}{\cancel{105}} = 28.4857 \approx 28.49$$ 7. **Calculate intercept $b$ using:** $$b = \frac{\sum y - m \sum x}{n} = \frac{6729 - 28.49 \times 15}{6} = \frac{6729 - 427.35}{6} = \frac{6301.65}{6} = 1050.28$$ 8. **Linear regression equation:** $$y = 28.49x + 1050.28$$ 9. **Estimate year when $y=1282$:** $$1282 = 28.49x + 1050.28$$ $$1282 - 1050.28 = 28.49x$$ $$231.72 = 28.49x$$ $$x = \frac{231.72}{28.49} = 8.13$$ 10. **Convert $x$ to calendar year:** $$1995 + 8.13 \approx 2003$$ **Final answer:** The linear regression equation is $y = 28.49x + 1050.28$. The number of new cases reaches 1282 approximately in the year 2003.