Pregunta
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?. QUESTION Ravi invested \( \$ 4800 \) in an account that pays an annual interest rate of \( 3.3 \% \), compounded daily. Assume there are 365 days in each year. Answer each par If necessary, refer to the list of financial formulas. (a) Find the amount in the account after one year, assuming no withdrawals are made. Do not round any intermediate computations, and round your answer to the nearest cent. \[ \square \] (b) Find the effective annual interest rate, expressed as a percentage. Do not round any intermediate computations, and round your answer to the nearest hundredth of a percent. \( \square \% \)

Ask by Jimenez Harmon. in the United States
Feb 02,2025

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After one year, Ravi has approximately \$4961.04 in the account. The effective annual interest rate is approximately 3.35%.

Solución

To find the amount in the account after one year, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount in the account after \( t \) years, - \( P \) is the principal amount (initial investment), - \( r \) is the annual interest rate, - \( n \) is the number of times the interest is compounded per year, - \( t \) is the time in years. Given: - \( P = \$4800 \), - \( r = 3.3\% = 0.033 \), - \( n = 365 \) (compounded daily), - \( t = 1 \) year. Substitute the values into the formula: \[ A = 4800 \left(1 + \frac{0.033}{365}\right)^{365 \cdot 1} \] Now, we can calculate the amount in the account after one year. Calculate the value by following steps: - step0: Calculate: \(4800\left(1+\frac{0.033}{365}\right)^{365}\) - step1: Divide the terms: \(4800\left(1+\frac{33}{365000}\right)^{365}\) - step2: Add the numbers: \(4800\left(\frac{365033}{365000}\right)^{365}\) - step3: Simplify: \(4800\times \frac{365033^{365}}{365000^{365}}\) - step4: Rewrite the expression: \(200\times 24\times \frac{365033^{365}}{365000^{365}}\) - step5: Rewrite the expression: \(200\times 24\times \frac{365033^{365}}{200^{365}\times 1825^{365}}\) - step6: Reduce the numbers: \(24\times \frac{365033^{365}}{200^{364}\times 1825^{365}}\) - step7: Rewrite the expression: \(8\times 3\times \frac{365033^{365}}{200^{364}\times 1825^{365}}\) - step8: Rewrite the expression: \(8\times 3\times \frac{365033^{365}}{8^{364}\times 25^{364}\times 1825^{365}}\) - step9: Reduce the numbers: \(3\times \frac{365033^{365}}{8^{363}\times 25^{364}\times 1825^{365}}\) - step10: Multiply the fractions: \(\frac{3\times 365033^{365}}{8^{363}\times 25^{364}\times 1825^{365}}\) The amount in the account after one year is approximately $4961.035188. Now, to find the effective annual interest rate, we can use the formula: \[ \text{Effective Annual Interest Rate} = \left(1 + \frac{r}{n}\right)^n - 1 \] Substitute the values into the formula: \[ \text{Effective Annual Interest Rate} = \left(1 + \frac{0.033}{365}\right)^{365} - 1 \] Now, we can calculate the effective annual interest rate. Calculate the value by following steps: - step0: Calculate: \(\left(1+\frac{0.033}{365}\right)^{365}-1\) - step1: Divide the terms: \(\left(1+\frac{33}{365000}\right)^{365}-1\) - step2: Add the numbers: \(\left(\frac{365033}{365000}\right)^{365}-1\) - step3: Rewrite the expression: \(\frac{365033^{365}}{365000^{365}}-1\) - step4: Reduce fractions to a common denominator: \(\frac{365033^{365}}{365000^{365}}-\frac{365000^{365}}{365000^{365}}\) - step5: Transform the expression: \(\frac{365033^{365}-365000^{365}}{365000^{365}}\) The effective annual interest rate is approximately 3.35%. Therefore, the answers are: (a) The amount in the account after one year is approximately $4961.035188. (b) The effective annual interest rate is approximately 3.35%.

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The Deep Dive

To determine the amount in the account after one year and the effective annual interest rate, we can use the formula for compound interest: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( A \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial sum of money). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per unit \( t \). - \( t \) is the time the money is invested or borrowed for, in years. ### (a) Given: - \( P = 4800 \) - \( r = 3.3\% = 0.033 \) - \( n = 365 \) - \( t = 1 \) Plugging in the values: \[ A = 4800 \left(1 + \frac{0.033}{365}\right)^{365 \cdot 1} \] Calculating the expression inside the parenthesis: \[ 1 + \frac{0.033}{365} \approx 1 + 0.00009041096 \approx 1.00009041096 \] Then, raising it to the power of 365: \[ A \approx 4800 \times (1.00009041096)^{365} \approx 4800 \times 1.033682 \approx 4944.123 \] So, rounding to the nearest cent: \[ A \approx \$ 4944.12 \] ### (b) To find the effective annual interest rate (EAR), we can use the formula: \[ EAR = \left(1 + \frac{r}{n}\right)^{n} - 1 \] Plugging in the values: \[ EAR = \left(1 + \frac{0.033}{365}\right)^{365} - 1 \] We already calculated: \[ \left(1 + \frac{0.033}{365}\right)^{365} \approx 1.033682 \] Therefore: \[ EAR \approx 1.033682 - 1 \approx 0.033682 \] To convert this to a percentage: \[ EAR \approx 0.033682 \times 100 \approx 3.3682\% \] Rounding to the nearest hundredth of a percent: \[ EAR \approx 3.37\% \] In summary: (a) \$ 4944.12 (b) 3.37%

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