Answer :
To solve this problem, we need to use the formula for compound interest. The formula for compound interest is given by:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form).
- [tex]\( t \)[/tex] is the number of years the money is invested or borrowed for.
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
Since the interest is compounded yearly, [tex]\( n = 1 \)[/tex].
Here are the given values:
- [tex]\( P = 10000 \)[/tex] (the principal amount)
- [tex]\( r = 0.06 \)[/tex] (the annual interest rate in decimal form)
- [tex]\( t = 4 \)[/tex] (the number of years)
Plug these values into the formula:
[tex]\[ A = 10000 \left(1 + \frac{0.06}{1}\right)^{1 \times 4} \][/tex]
Simplify inside the parentheses:
[tex]\[ A = 10000 \left(1 + 0.06\right)^4 \][/tex]
[tex]\[ A = 10000 \left(1.06\right)^4 \][/tex]
Now, calculate [tex]\((1.06)^4\)[/tex]:
[tex]\[ (1.06)^4 \approx 1.262477 \][/tex]
Then, multiply by the principal:
[tex]\[ A = 10000 \times 1.262477 \approx 12624.77 \][/tex]
So, the account balance after 4 years is approximately [tex]\( \$12,624.77 \)[/tex].
Matching this calculated value to the given choices, we find:
- a.) [tex]$12,689.86 - b.) $[/tex]12,624.77
- c.) [tex]$12,667.70 - d.) $[/tex]12,704.89
The correct answer is:
b.) $12,624.77
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after [tex]\( t \)[/tex] years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial amount of money).
- [tex]\( r \)[/tex] is the annual interest rate (in decimal form).
- [tex]\( t \)[/tex] is the number of years the money is invested or borrowed for.
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
Since the interest is compounded yearly, [tex]\( n = 1 \)[/tex].
Here are the given values:
- [tex]\( P = 10000 \)[/tex] (the principal amount)
- [tex]\( r = 0.06 \)[/tex] (the annual interest rate in decimal form)
- [tex]\( t = 4 \)[/tex] (the number of years)
Plug these values into the formula:
[tex]\[ A = 10000 \left(1 + \frac{0.06}{1}\right)^{1 \times 4} \][/tex]
Simplify inside the parentheses:
[tex]\[ A = 10000 \left(1 + 0.06\right)^4 \][/tex]
[tex]\[ A = 10000 \left(1.06\right)^4 \][/tex]
Now, calculate [tex]\((1.06)^4\)[/tex]:
[tex]\[ (1.06)^4 \approx 1.262477 \][/tex]
Then, multiply by the principal:
[tex]\[ A = 10000 \times 1.262477 \approx 12624.77 \][/tex]
So, the account balance after 4 years is approximately [tex]\( \$12,624.77 \)[/tex].
Matching this calculated value to the given choices, we find:
- a.) [tex]$12,689.86 - b.) $[/tex]12,624.77
- c.) [tex]$12,667.70 - d.) $[/tex]12,704.89
The correct answer is:
b.) $12,624.77
