Answer :
Sure, let's walk through each step to find the account balance using the compound-interest formula. The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the account balance after time [tex]\( t \)[/tex]
- [tex]\( P \)[/tex] is the principal amount (initial investment)
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal)
- [tex]\( n \)[/tex] is the number of compounding periods per year
- [tex]\( t \)[/tex] is the time the money is invested for in years
Given:
- [tex]\( P = \$51,445 \)[/tex]
- [tex]\( r = 4 \frac{1}{2}\% = 4.5\% = 0.045 \)[/tex] (as a decimal)
- Compounding is Quarterly, so [tex]\( n = 4 \)[/tex]
- [tex]\( t = 7 \frac{1}{2} \)[/tex] years = 7.5 years
Now, plug these values into the formula:
[tex]\[ A = 51445 \left(1 + \frac{0.045}{4}\right)^{4 \cdot 7.5} \][/tex]
Breaking it down:
1. Calculate [tex]\( \frac{r}{n} = \frac{0.045}{4} = 0.01125 \)[/tex]
2. Add 1 to this value: [tex]\( 1 + 0.01125 = 1.01125 \)[/tex]
3. Raise this result to the power of [tex]\( nt \)[/tex]:
[tex]\[ (1.01125)^{4 \times 7.5} = (1.01125)^{30} \][/tex]
4. Finally, multiply this result by [tex]\( P \)[/tex]:
[tex]\[ A = 51445 \times (1.01125)^{30} = 51445 \times 1.39856 = 71961.33475973115 \][/tex]
So, the account balance [tex]\( A \)[/tex] after 7.5 years is approximately \$71,961.33.
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the account balance after time [tex]\( t \)[/tex]
- [tex]\( P \)[/tex] is the principal amount (initial investment)
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal)
- [tex]\( n \)[/tex] is the number of compounding periods per year
- [tex]\( t \)[/tex] is the time the money is invested for in years
Given:
- [tex]\( P = \$51,445 \)[/tex]
- [tex]\( r = 4 \frac{1}{2}\% = 4.5\% = 0.045 \)[/tex] (as a decimal)
- Compounding is Quarterly, so [tex]\( n = 4 \)[/tex]
- [tex]\( t = 7 \frac{1}{2} \)[/tex] years = 7.5 years
Now, plug these values into the formula:
[tex]\[ A = 51445 \left(1 + \frac{0.045}{4}\right)^{4 \cdot 7.5} \][/tex]
Breaking it down:
1. Calculate [tex]\( \frac{r}{n} = \frac{0.045}{4} = 0.01125 \)[/tex]
2. Add 1 to this value: [tex]\( 1 + 0.01125 = 1.01125 \)[/tex]
3. Raise this result to the power of [tex]\( nt \)[/tex]:
[tex]\[ (1.01125)^{4 \times 7.5} = (1.01125)^{30} \][/tex]
4. Finally, multiply this result by [tex]\( P \)[/tex]:
[tex]\[ A = 51445 \times (1.01125)^{30} = 51445 \times 1.39856 = 71961.33475973115 \][/tex]
So, the account balance [tex]\( A \)[/tex] after 7.5 years is approximately \$71,961.33.