## Thinking Mathematically (6th Edition)

We can find the total number of crimes. total crimes = 446 + 526 + 835 + 227 + 338 + 456 total crimes = 2828 We can find the standard divisor. $standard ~divisor = \frac{total ~crimes}{officers}$ $standard ~divisor = \frac{2828}{180}$ $standard ~divisor = 15.71$ If we use the standard divisor and round each standard quota to the nearest whole number, the sum of the apportioned officers will be 179 officers. To obtain a sum of 180 officers, we need to find a modified divisor that is slightly less than the standard divisor. Let's choose a modified divisor of 15.7. Note that it may take some trial-and-error to find a modified divisor that works. We can find the modified quota for each precinct. Precinct A: $modified~quota = \frac{crimes}{modified~divisor}$ $modified~quota = \frac{446}{15.7}$ $modified~quota = 28.41$ Precinct B: $modified~quota = \frac{crimes}{modified~divisor}$ $modified~quota = \frac{526}{15.7}$ $modified~quota = 33.503$ Precinct C: $modified~quota = \frac{crimes}{modified~divisor}$ $modified~quota = \frac{835}{15.7}$ $modified~quota = 53.18$ Precinct D: $modified~quota = \frac{crimes}{modified~divisor}$ $modified~quota = \frac{227}{15.7}$ $modified~quota = 14.46$ Precinct E: $modified~quota = \frac{crimes}{modified~divisor}$ $modified~quota = \frac{338}{15.7}$ $modified~quota = 21.53$ Precinct F: $modified~quota = \frac{crimes}{modified~divisor}$ $modified~quota = \frac{456}{15.7}$ $modified~quota = 29.04$ Using Webster's method, we apportion the officers by rounding the modified quota to the nearest whole number. The officers are apportioned to the precincts as follows: Precinct A is apportioned 28 officers. Precinct B is apportioned 34 officers. Precinct C is apportioned 53 officers. Precinct D is apportioned 14 officers. Precinct E is apportioned 22 officers. Precinct F is apportioned 29 officers. Note that the sum of the apportioned officers is 180, so using a modified divisor of 15.7 is acceptable.