Compound Interest 8Cfdfb

1. **State the problem:** We want to find the value of an account after 10 years with two different interest rates and compounding periods. 2. **Formula for compound interest:** $$A = P \left(1 + \frac{r}{n}\right)^{nt}$$ 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. **Calculate amount after first 5 years:** - $P = 3000$ - $r = 0.08$ - $n = 2$ (semiannual compounding) - $t = 5$ $$A_1 = 3000 \left(1 + \frac{0.08}{2}\right)^{2 \times 5} = 3000 \left(1 + 0.04\right)^{10} = 3000 \times 1.04^{10}$$ Calculate $1.04^{10}$: $$1.04^{10} \approx 1.48024$$ So, $$A_1 = 3000 \times 1.48024 = 4440.72$$ 4. **Calculate amount after next 5 years with new rate:** - New principal $P = A_1 = 4440.72$ - New rate $r = 0.0852$ - New compounding periods $n = 4$ (quarterly) - Time $t = 5$ $$A_2 = 4440.72 \left(1 + \frac{0.0852}{4}\right)^{4 \times 5} = 4440.72 \left(1 + 0.0213\right)^{20} = 4440.72 \times 1.0213^{20}$$ Calculate $1.0213^{20}$: $$1.0213^{20} \approx 1.49182$$ So, $$A_2 = 4440.72 \times 1.49182 = 6623.45$$ 5. **Final answer:** The value of the account after 10 years is approximately $6623$ (rounded to the nearest dollar).