Pregunta
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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?

Ask by Pritchard Matthews. in the United States
Jan 24,2025

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a. The account is worth \$14,503.70. b. The initial deposit was \$10,775.00. c. The account has been accumulating interest for 15 years.

Solución

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. ### Given Information: - The equation provided is \( 10750\left(1+\frac{0.02}{6}\right)^{90} \). - Here, \( A(t) = 10750\left(1+\frac{0.02}{6}\right)^{90} \). - The interest rate \( r = 0.02 \) (2%). - The number of compounding periods per year \( n = 6 \) (compounded semi-annually). - The exponent \( 90 \) represents the total number of compounding periods, which is \( nt \). ### Step a: Calculate the value of the account \( A(t) \) We will calculate \( A(t) \) using the given formula. 1. Calculate \( \frac{r}{n} \): \[ \frac{0.02}{6} = 0.0033333 \] 2. Calculate \( 1 + \frac{r}{n} \): \[ 1 + 0.0033333 = 1.0033333 \] 3. Calculate \( (1 + \frac{r}{n})^{90} \): \[ (1.0033333)^{90} \] 4. Finally, calculate \( A(t) = 10750 \times (1.0033333)^{90} \). Let's perform these calculations. ### Step b: Find the initial deposit \( P \) From the formula, we can rearrange it to find \( P \): \[ P = \frac{A(t)}{\left(1+\frac{r}{n}\right)^{nt}} \] ### Step c: Determine the number of years \( t \) Since \( nt = 90 \) and \( n = 6 \), we can find \( t \) as follows: \[ t = \frac{90}{n} = \frac{90}{6} \] Now, let's perform the calculations for parts a, b, and c. ### Performing Calculations 1. Calculate \( (1.0033333)^{90} \). 2. Calculate \( A(t) \). 3. Calculate \( P \). 4. Calculate \( t \). Let's execute these calculations. Calculate the value by following steps: - step0: Calculate: \(\frac{90}{6}\) - step1: Reduce the numbers: \(\frac{15}{1}\) - step2: Calculate: \(15\) Calculate or simplify the expression \( (1.0033333)^90 \). Calculate the value by following steps: - step0: Calculate: \(1.0033333^{90}\) - step1: Convert the expressions: \(\left(\frac{10033333}{10000000}\right)^{90}\) - step2: Use the properties of exponents: \(\frac{10033333^{90}}{10000000^{90}}\) Calculate or simplify the expression \( 10750*(1.0033333)^90 \). Calculate the value by following steps: - step0: Calculate: \(10750\times 1.0033333^{90}\) - step1: Convert the expressions: \(10750\left(\frac{10033333}{10000000}\right)^{90}\) - step2: Simplify: \(10750\times \frac{10033333^{90}}{10000000^{90}}\) - step3: Rewrite the expression: \(250\times 43\times \frac{10033333^{90}}{10000000^{90}}\) - step4: Rewrite the expression: \(250\times 43\times \frac{10033333^{90}}{62500^{90}\times 160^{90}}\) - step5: Rewrite the expression: \(250\times 43\times \frac{10033333^{90}}{250^{180}\times 160^{90}}\) - step6: Reduce the numbers: \(43\times \frac{10033333^{90}}{250^{179}\times 160^{90}}\) - step7: Multiply the fractions: \(\frac{43\times 10033333^{90}}{250^{179}\times 160^{90}}\) ### Step a: Calculate the value of the account \( A(t) \) From our calculations: - \( (1.0033333)^{90} \approx 1.349182 \) Now, substituting this value into the equation for \( A(t) \): \[ A(t) = 10750 \times 1.349182 \approx 14503.701212 \] Rounding to the nearest cent, the value of the account is: \[ \boxed{14503.70} \] ### Step b: Find the initial deposit \( P \) Using the rearranged formula: \[ P = \frac{A(t)}{\left(1+\frac{r}{n}\right)^{nt}} = \frac{14503.70}{(1.0033333)^{90}} = \frac{14503.70}{1.349182} \] Let's calculate \( P \): \[ P \approx \frac{14503.70}{1.349182} \approx 10775.00 \] Thus, the initial deposit made to the account is: \[ \boxed{10775.00} \] ### Step c: Determine the number of years \( t \) From our earlier calculation: \[ t = \frac{90}{6} = 15 \] Thus, the number of years the account had been accumulating interest is: \[ \boxed{15} \] ### Summary of Results: - a. Value of the account: \( \boxed{14503.70} \) - b. Initial deposit: \( \boxed{10775.00} \) - c. Years of accumulation: \( \boxed{15} \)

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The Deep Dive

Starting with part \( a \), we need to compute the value of the investment account after a certain period using the compound interest formula. Plugging in the values: \[ A(90) = 10750\left(1+\frac{0.02}{6}\right)^{90} \] Calculating this step by step: First, calculate the term inside the parentheses: \[ 1 + \frac{0.02}{6} = 1 + 0.0033333 \approx 1.0033333 \] Now raise that to the power of 90: \[ (1.0033333)^{90} \approx 1.348853 \] Finally, multiply by the initial investment of $10,750: \[ A(90) \approx 10750 \cdot 1.348853 \approx 14551.92 \] Thus, the value of the account after 90 periods is approximately **\$14,551.92**. Now for part \( b \), to find the initial deposit made to the account, we can rearrange the compound interest formula to find \( P \): \[ P = \frac{A(t)}{\left(1+\frac{r}{n}\right)^{n t}} = \frac{10750\left(1+\frac{0.02}{6}\right)^{90}}{\left(1+\frac{0.02}{6}\right)^{90}} = 10750 \] Therefore, the initial deposit made to the account was **\$10,750**. Lastly, for part \( c \), to find the number of years the account has been accumulating interest, observe that \( n = 6 \) compounding periods led to a total of 90 periods: \[ t = \frac{90}{6} = 15 \] Thus, the account has been accumulating interest for **15 years**.

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