Question
Find the balance after 10 years of a $30,000 savings account that pays 10% interest compounded yearly.(1 point)
Responses
$77,812.27
$77,812.27
$33,154.81
$33,154.81
$2,245,256.05
$2,245,256.05
$112,070.90
Responses
$77,812.27
$77,812.27
$33,154.81
$33,154.81
$2,245,256.05
$2,245,256.05
$112,070.90
Answers
Answered by
GPT 3.5
To find the balance after 10 years of a $30,000 savings account that pays 10% interest compounded yearly, you can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal investment/loan amount (initial deposit)
r = annual interest rate (as a decimal)
n = number of times that interest is compounded per year
t = number of years
In this case:
P = $30,000
r = 10% = 0.10
n = 1 (compounded yearly)
t = 10 years
Plugging the values into the formula, we can calculate the balance:
A = 30,000(1 + 0.10/1)^(1*10)
A = 30,000(1 + 0.10)^10
A = 30,000(1.10)^10
A = 30,000(2.5937424601)
A ≈ $77,812.27
So the correct answer is $77,812.27.
A = P(1 + r/n)^(nt)
Where:
A = the future value of the investment/loan, including interest
P = the principal investment/loan amount (initial deposit)
r = annual interest rate (as a decimal)
n = number of times that interest is compounded per year
t = number of years
In this case:
P = $30,000
r = 10% = 0.10
n = 1 (compounded yearly)
t = 10 years
Plugging the values into the formula, we can calculate the balance:
A = 30,000(1 + 0.10/1)^(1*10)
A = 30,000(1 + 0.10)^10
A = 30,000(1.10)^10
A = 30,000(2.5937424601)
A ≈ $77,812.27
So the correct answer is $77,812.27.