Compound Interest F1E7C9

1. **State the problem:** We want to find the present value (the amount to deposit today) that will grow to $13,000 in 2 years with an interest rate of 4.5% compounded monthly. 2. **Formula used:** The compound interest formula is $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount accumulated after $t$ years, - $P$ is the principal (initial deposit), - $r$ is the annual interest rate (decimal), - $n$ is the number of compounding periods per year, - $t$ is the time in years. 3. **Given values:** - $A = 13000$ - $r = 0.045$ - $n = 12$ (monthly compounding) - $t = 2$ 4. **Rearrange the formula to solve for $P$:** $$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}$$ 5. **Substitute the values:** $$P = \frac{13000}{\left(1 + \frac{0.045}{12}\right)^{12 \times 2}} = \frac{13000}{\left(1 + 0.00375\right)^{24}} = \frac{13000}{(1.00375)^{24}}$$ 6. **Calculate the denominator:** $$(1.00375)^{24} \approx 1.093443$$ 7. **Calculate $P$:** $$P = \frac{13000}{1.093443} \approx 11883.68$$ **Final answer:** The amount to deposit today is approximately **11883.68**.