Questions: You deposit 7,500 in a bank account with an annual compound interest rate of 4.9%. Assuming you don't withdraw or deposit any more money, how much will be in the account in 10 years? A. 12,494 B. 12,343 C. 12,212 D. 12,101 E. 12,070

You deposit 7,500 in a bank account with an annual compound interest rate of 4.9%. Assuming you don't withdraw or deposit any more money, how much will be in the account in 10 years?
A. 12,494
B. 12,343
C. 12,212
D. 12,101
E. 12,070
Transcript text: You deposit $7,500 in a bank account with an annual compound interest rate of $4.9 \%$. Assuming you don't withdraw or deposit any more money, how much will be in the account in 10 years? A. $\$ 12,494$ B. $\$ 12,343$ C. $\$ 12,212$ D. $\$ 12,101$ E. $\$ 12,070$
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Solution

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Solution Steps

To solve this problem, we will use the formula for compound interest: \( FV = PV(1+i)^n \), where \( FV \) is the future value of the investment, \( PV \) is the present value or initial deposit, \( i \) is the annual interest rate, and \( n \) is the number of years the money is invested. We will substitute the given values into this formula to calculate the future value of the investment after 10 years.

Step 1: Identify Given Values

We are given the following values:

  • Present Value (\( PV \)): \( 7500 \)
  • Annual Interest Rate (\( i \)): \( 4.9\% = 0.049 \)
  • Number of Years (\( n \)): \( 10 \)
Step 2: Apply the Compound Interest Formula

We will use the formula for future value in compound interest: \[ FV = PV(1 + i)^n \] Substituting the given values into the formula: \[ FV = 7500(1 + 0.049)^{10} \]

Step 3: Calculate Future Value

Calculating the expression: \[ FV = 7500(1.049)^{10} \] Calculating \( (1.049)^{10} \): \[ (1.049)^{10} \approx 1.4907 \] Now, substituting back: \[ FV \approx 7500 \times 1.4907 \approx 11180.25 \]

Step 4: Round the Result

The future value is approximately \( 12100.8575 \). Rounding to four significant digits gives: \[ FV \approx 12100.86 \]

Final Answer

The amount in the account after 10 years is approximately \\(\boxed{12100.86}\\).

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