1. **Problem Statement:** Ms. Casas has 80,000 and wants to deposit it for 10 years. There are two accounts: - Account 1: 6% interest compounded monthly - Account 2: 6.5% interest compounded semi-annually We want to find which account yields more money and the difference between the two. 2. **Formula for compound interest:** $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ where: - $A$ = amount after time $t$ - $P$ = principal (initial amount) - $r$ = annual interest rate (decimal) - $n$ = number of compounding periods per year - $t$ = number of years 3. **Calculate for Account 1:** - $P = 80000$ - $r = 0.06$ - $n = 12$ (monthly) - $t = 10$ $$ A_1 = 80000 \left(1 + \frac{0.06}{12}\right)^{12 \times 10} = 80000 \left(1 + 0.005\right)^{120} = 80000 \times (1.005)^{120} $$ Calculate $(1.005)^{120}$: $$ (1.005)^{120} = e^{120 \ln(1.005)} \approx e^{120 \times 0.0049875} = e^{0.5985} \approx 1.8194 $$ So, $$ A_1 \approx 80000 \times 1.8194 = 145,552 $$ 4. **Calculate for Account 2:** - $P = 80000$ - $r = 0.065$ - $n = 2$ (semi-annually) - $t = 10$ $$ A_2 = 80000 \left(1 + \frac{0.065}{2}\right)^{2 \times 10} = 80000 \left(1 + 0.0325\right)^{20} = 80000 \times (1.0325)^{20} $$ Calculate $(1.0325)^{20}$: $$ (1.0325)^{20} = e^{20 \ln(1.0325)} \approx e^{20 \times 0.03199} = e^{0.6398} \approx 1.896 $$ So, $$ A_2 \approx 80000 \times 1.896 = 151,680 $$ 5. **Compare the two amounts:** - Account 1: 145,552 - Account 2: 151,680 Account 2 yields more money. 6. **Difference between the two accounts:** $$ \text{Difference} = A_2 - A_1 = 151,680 - 145,552 = 6,128 $$ 7. **Answer to question 39:** B. 2nd account 8. **Answer to question 40:** B. 6,115.29 (closest to calculated difference 6,128) --- **Slug:** compound interest **Subject:** finance **Desmos:** {"latex":"y=80000(1+0.06/12)^{12x}","features":{"intercepts":true,"extrema":true}} **q_count:** 2