1. **State the problem:** Lina deposits 600 monthly for 4 years in an account with 6.2% annual interest compounded monthly, then stops depositing and leaves the money to earn 4.8% interest compounded monthly for 5 more years. Calculate the total amount after 9 years. 2. **Identify given values:** - Monthly deposit, \( P = 600 \) - Annual interest rate for first 4 years, \( r_1 = 6.2\% = 0.062 \) - Annual interest rate for next 5 years, \( r_2 = 4.8\% = 0.048 \) - Time with deposits, \( t_1 = 4 \) years - Time after deposits, \( t_2 = 5 \) years - Compounding monthly means \( n = 12 \) times per year 3. **Calculate monthly interest rates:** - \( i_1 = \frac{r_1}{n} = \frac{0.062}{12} = 0.0051667 \) - \( i_2 = \frac{r_2}{n} = \frac{0.048}{12} = 0.004 \) 4. **Calculate total number of months:** - Deposits period \( m_1 = t_1 \times n = 4 \times 12 = 48 \) - Interest-only period \( m_2 = t_2 \times n = 5 \times 12 = 60 \) 5. **Calculate amount after 4 years of deposits using the future value of an annuity formula:** $$ A_1 = P \times \frac{(1 + i_1)^{m_1} - 1}{i_1} $$ Substitute values: $$ A_1 = 600 \times \frac{(1 + 0.0051667)^{48} - 1}{0.0051667} $$ Calculate \( (1 + 0.0051667)^{48} \): $$ (1.0051667)^{48} \approx 1.270243 $$ Then, $$ A_1 = 600 \times \frac{1.270243 - 1}{0.0051667} = 600 \times \frac{0.270243}{0.0051667} = 600 \times 52.272 = 31363.2 $$ 6. **Calculate final amount after additional 5 years of compound interest (no new deposits):** $$ A_2 = A_1 \times (1 + i_2)^{m_2} $$ Substitute values: $$ A_2 = 31363.2 \times (1 + 0.004)^{60} $$ Calculate \( (1.004)^{60} \): $$ (1.004)^{60} \approx 1.270477 $$ Then, $$ A_2 = 31363.2 \times 1.270477 = 39839.7 $$ 7. **Final answer:** The total amount in the account at the end of 9 years is approximately **39839.7**.