Use the compound interest formula, \( A(t)=P\left(1+\frac{r}{n}\right)^{n t} \cdot \) After a certain amount of years, the value of an investment account is represented by the equation \( 10750\left(1+\frac{0.02}{6}\right)^{90} \cdot \) \( a \). Rounded to the nearest cent, what is the value of the account? \( \$ \) \( b \). What was the initial deposit made to the account? \( \$ \) c. How many years had the account been accumulating interest?

To solve the problem, we will use the compound interest formula: \[ A(t) = P\left(1+\frac{r}{n}\right)^{nt} \] where: - \( A(t) \) is the amount of money accumulated after n years, including interest. - \( P \) is the principal amount (the initial deposit). - \( r \) is the annual interest rate (decimal). - \( n \) is the number of times that interest is compounded per year. - \( t \) is the time the money is invested for in years. From the given equation: \[ A(t) = 10750\left(1+\frac{0.02}{6}\right)^{90} \] ### Step 1: Calculate the value of the account \( A(t) \) We will first calculate \( A(t) \) using the provided values. - \( r = 0.02 \) - \( n = 6 \) - \( t = 90/6 = 15 \) years (since the interest is compounded 6 times a year for 15 years) Now, we will calculate \( A(t) \): \[ A(t) = 10750\left(1+\frac{0.02}{6}\right)^{90} \] Calculating the expression inside the parentheses and then raising it to the power of 90: \[ A(t) = 10750\left(1+\frac{0.02}{6}\right)^{90} \] Let's calculate this step by step. 1. Calculate \( \frac{0.02}{6} \) 2. Add 1 to the result. 3. Raise the result to the power of 90. 4. Multiply by 10750. Let's perform these calculations. Calculate the value by following steps: - step0: Calculate: \(\frac{0.02}{6}\) - step1: Convert the expressions: \(\frac{\frac{1}{50}}{6}\) - step2: Multiply by the reciprocal: \(\frac{1}{50}\times \frac{1}{6}\) - step3: Multiply the fractions: \(\frac{1}{50\times 6}\) - step4: Multiply: \(\frac{1}{300}\) Calculate or simplify the expression \( (1+(0.02/6))^90 \). Calculate the value by following steps: - step0: Calculate: \(\left(1+\left(\frac{0.02}{6}\right)\right)^{90}\) - step1: Divide the terms: \(\left(1+\frac{1}{300}\right)^{90}\) - step2: Add the numbers: \(\left(\frac{301}{300}\right)^{90}\) - step3: Use the properties of exponents: \(\frac{301^{90}}{300^{90}}\) Calculate or simplify the expression \( 10750*(1+(0.02/6))^90 \). Calculate the value by following steps: - step0: Calculate: \(10750\left(1+\left(\frac{0.02}{6}\right)\right)^{90}\) - step1: Divide the terms: \(10750\left(1+\frac{1}{300}\right)^{90}\) - step2: Add the numbers: \(10750\left(\frac{301}{300}\right)^{90}\) - step3: Simplify: \(10750\times \frac{301^{90}}{300^{90}}\) - step4: Rewrite the expression: \(50\times 215\times \frac{301^{90}}{300^{90}}\) - step5: Rewrite the expression: \(50\times 215\times \frac{301^{90}}{50^{90}\times 6^{90}}\) - step6: Reduce the numbers: \(215\times \frac{301^{90}}{50^{89}\times 6^{90}}\) - step7: Rewrite the expression: \(5\times 43\times \frac{301^{90}}{50^{89}\times 6^{90}}\) - step8: Rewrite the expression: \(5\times 43\times \frac{301^{90}}{25^{89}\times 2^{89}\times 6^{90}}\) - step9: Rewrite the expression: \(5\times 43\times \frac{301^{90}}{5^{178}\times 2^{89}\times 6^{90}}\) - step10: Reduce the numbers: \(43\times \frac{301^{90}}{5^{177}\times 2^{89}\times 6^{90}}\) - step11: Multiply the fractions: \(\frac{43\times 301^{90}}{5^{177}\times 2^{89}\times 6^{90}}\) ### Step 1: Calculate the value of the account \( A(t) \) 1. **Calculate \( \frac{0.02}{6} \)**: \[ \frac{0.02}{6} \approx 0.003333 \] 2. **Calculate \( 1 + \frac{0.02}{6} \)**: \[ 1 + 0.003333 \approx 1.003333 \] 3. **Raise to the power of 90**: \[ (1.003333)^{90} \approx 1.349186 \] 4. **Multiply by 10750**: \[ A(t) = 10750 \times 1.349186 \approx 14503.744579 \] Rounding to the nearest cent, the value of the account is: \[ \boxed{14503.74} \] ### Step 2: Determine the initial deposit \( P \) From the compound interest formula, we can rearrange it to find \( P \): \[ P = \frac{A(t)}{\left(1+\frac{r}{n}\right)^{nt}} \] Using the values we have: - \( A(t) = 14503.744579 \) - \( r = 0.02 \) - \( n = 6 \) - \( t = 15 \) Now we can calculate \( P \): \[ P = \frac{14503.744579}{(1+\frac{0.02}{6})^{90}} \] We already calculated \( (1+\frac{0.02}{6})^{90} \approx 1.349186 \). Now we can find \( P \). Calculate the value by following steps: - step0: Calculate: \(\frac{14503.744579}{\left(1+\left(\frac{0.02}{6}\right)\right)^{90}}\) - step1: Divide the terms: \(\frac{14503.744579}{\left(1+\frac{1}{300}\right)^{90}}\) - step2: Add the numbers: \(\frac{14503.744579}{\left(\frac{301}{300}\right)^{90}}\) - step3: Convert the expressions: \(\frac{\frac{14503744579}{1000000}}{\left(\frac{301}{300}\right)^{90}}\) - step4: Use the properties of exponents: \(\frac{\frac{14503744579}{1000000}}{\frac{301^{90}}{300^{90}}}\) - step5: Multiply by the reciprocal: \(\frac{14503744579}{1000000}\times \frac{300^{90}}{301^{90}}\) - step6: Rewrite the expression: \(\frac{14503744579}{1000000}\times \frac{100^{90}\times 3^{90}}{301^{90}}\) - step7: Rewrite the expression: \(\frac{14503744579}{100^{3}}\times \frac{100^{90}\times 3^{90}}{301^{90}}\) - step8: Reduce the numbers: \(14503744579\times \frac{100^{87}\times 3^{90}}{301^{90}}\) - step9: Multiply: \(\frac{14503744579\times 100^{87}\times 3^{90}}{301^{90}}\) ### Step 2: Determine the initial deposit \( P \) Using the rearranged formula, we calculated: \[ P = \frac{14503.744579}{(1+\frac{0.02}{6})^{90}} \approx \frac{14503.744579}{1.349186} \approx 10750 \] Thus, the initial deposit made to the account is: \[ \boxed{10750} \] ### Step 3: Determine how many years the account had been accumulating interest From the problem, we know that the interest is compounded 6 times a year, and the total number of compounding periods is 90. To find the number of years \( t \): \[ t = \frac{90}{n} = \frac{90}{6} = 15 \] Thus, the account had been accumulating interest for: \[ \boxed{15} \] ### Summary of Results: - a. The value of the account: \( \boxed{14503.74} \) - b. The initial deposit made to the account: \( \boxed{10750} \) - c. The number of years the account had been accumulating interest: \( \boxed{15} \)