Sure, let's tackle these questions step by step.
### 1. What balance will be in an account that has an initial deposit of [tex]$1200 with an APR of 3.5%? The money is compounded annually for 30 years.
To answer this question, we 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 amount of money).
- \( r \) is the annual interest rate (decimal).
- \( n \) is the number of times interest is compounded per year.
- \( t \) is the time the money is invested for in years.
Given the values:
- \( P = 1200 \, \text{dollars} \)
- \( r = 3.5\% = 0.035 \)
- \( n = 1 \) (compounded annually)
- \( t = 30 \, \text{years} \)
Plugging these values into the formula:
\[ A = 1200 \left(1 + \frac{0.035}{1}\right)^{1 \times 30} \]
\[ A = 1200 (1 + 0.035)^{30} \]
\[ A = 1200 (1.035)^{30} \]
After performing the calculations:
\[ A \approx 3368.15 \, \text{dollars} \]
So the balance after 30 years will be approximately $[/tex]3368.15.
### 2. How much interest has been earned for the entire time period?
To find the interest earned, we subtract the initial deposit from the final amount.
Interest earned [tex]\( I \)[/tex] is given by:
[tex]\[ I = A - P \][/tex]
Given:
- [tex]\( A \approx 3368.15 \)[/tex]
- [tex]\( P = 1200 \)[/tex]
Plugging in the values:
[tex]\[ I = 3368.15 - 1200 \][/tex]
[tex]\[ I \approx 2168.15 \][/tex]
So, the interest earned over the 30 years is approximately [tex]$2168.15.
### Summary:
1. The balance in the account after 30 years will be approximately $[/tex]3368.15.
2. The interest earned over the 30-year period will be approximately $2168.15.
