1. **Problem 1: Calculate the current cost of a car given inflation.** The problem states that the cost of automobiles increases by 2.5% each year due to inflation. We want to find the current cost of a car that cost 21,000 five years ago. 2. **Formula used:** The formula for compound interest (or inflation) is: $$ P = P_0 (1 + r)^t $$ where: - $P$ is the current price, - $P_0$ is the initial price, - $r$ is the annual inflation rate (as a decimal), - $t$ is the number of years. 3. **Apply the values:** - $P_0 = 21000$ - $r = 0.025$ - $t = 5$ Calculate: $$ P = 21000 \times (1 + 0.025)^5 = 21000 \times (1.025)^5 $$ 4. **Calculate $(1.025)^5$:** $$ (1.025)^5 = 1.131408 $$ 5. **Calculate the current price:** $$ P = 21000 \times 1.131408 = 23759.57 $$ 6. **Answer:** The car should cost approximately **23759.57** today. --- 1. **Problem 2: Calculate the effective annual interest rate for 6.5% interest compounded daily.** Given an annual nominal interest rate of 6.5% compounded daily (365 days), find the effective annual rate (EAR). 2. **Formula used:** $$ EAR = \left(1 + \frac{r}{n}\right)^n - 1 $$ where: - $r = 0.065$ (nominal annual rate), - $n = 365$ (compounding periods per year). 3. **Apply the values:** $$ EAR = \left(1 + \frac{0.065}{365}\right)^{365} - 1 $$ 4. **Calculate inside the parentheses:** $$ 1 + \frac{0.065}{365} = 1 + 0.00017808 = 1.00017808 $$ 5. **Calculate the power:** $$ (1.00017808)^{365} = e^{365 \times \ln(1.00017808)} $$ Calculate $\ln(1.00017808) \approx 0.00017807$, so: $$ e^{365 \times 0.00017807} = e^{0.065} = 1.06718 $$ 6. **Calculate EAR:** $$ EAR = 1.06718 - 1 = 0.06718 = 6.72\% $$ 7. **Answer:** The effective annual interest rate is approximately **6.72%**.