1. Problem 3a: Write the formula for calculating population after $T$ years given an initial population $P_0$ and annual growth rate $r$. The formula for population after $T$ years with compound growth is: $$P = P_0 (1 + r)^T$$ where $P_0 = 57245$ and $r = 0.07$ (7%). 2. Problem 3b: Find the population 2 years ago. We use the formula backward: $$P_{-2} = \frac{P}{(1 + r)^2}$$ Substitute values: $$P_{-2} = \frac{57245}{(1 + 0.07)^2} = \frac{57245}{1.07^2}$$ Calculate $1.07^2 = 1.1449$: $$P_{-2} = \frac{57245}{1.1449}$$ Intermediate step with cancellation: $$P_{-2} = \frac{57245}{\cancel{1.1449}}$$ Calculate: $$P_{-2} \approx 49980$$ So, the population 2 years ago was approximately 49980. 3. Problem 3c: If the growth rate was 9% last year, find the present population. Let $P_{-1}$ be the population one year ago. Then: $$P = P_{-1} (1 + 0.07)$$ But last year growth was 9%, so: $$P_{-1} = \frac{P}{1.07}$$ And one year before that (2 years ago), population was: $$P_{-2} = \frac{P_{-1}}{1.09} = \frac{P}{1.07 \times 1.09}$$ Calculate population now with 9% growth last year: $$P = P_{-2} \times 1.09 \times 1.07$$ Since $P_{-2} = 49980$ (from 3b), then: $$P = 49980 \times 1.09 \times 1.07$$ Calculate intermediate: $$P = 49980 \times 1.1663$$ $$P \approx 58288$$ So, the present population would be approximately 58288 if last year's growth was 9%.