1. **State the problem:** We want to find the present deposit amount $P$ that will grow to $150,000$ in 11 years with quarterly compounding at an APR of 5.62%. 2. **Formula used:** The future value $A$ with compound interest is given by: $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ where: - $A$ is the amount after time $t$, - $P$ is the principal (initial deposit), - $r$ is the annual interest rate (decimal), - $n$ is the number of compounding periods per year, - $t$ is the number of years. 3. **Given values:** - $A = 150000$ - $r = 0.0562$ - $n = 4$ (quarterly compounding) - $t = 11$ 4. **Rearrange formula to solve for $P$:** $$P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}}$$ 5. **Substitute values:** $$P = \frac{150000}{\left(1 + \frac{0.0562}{4}\right)^{4 \times 11}} = \frac{150000}{\left(1 + 0.01405\right)^{44}} = \frac{150000}{(1.01405)^{44}}$$ 6. **Calculate the denominator:** $$(1.01405)^{44} \approx 1.747422$$ 7. **Calculate $P$:** $$P = \frac{150000}{1.747422} \approx 85837.68$$ **Final answer:** $85837.68$ should be deposited today to have $150,000$ in 11 years with quarterly compounding at 5.62% APR.