Given:
A = $45,000
i = 3.9% or 0.039 (annual interest rate)
n = 4 years
We can plug these values into the formula and solve for d:
$45,000 = d(0.00325(1+0.00325)^48 - 1) / 0.00325
$45,000 = d(0.00325(1.00325)^48 - 1) / 0.00325
$45,000 = d(0.00325(1.166095958) - 1) / 0.00325
$45,000 = d(0.003796166) / 0.00325
$45,000 = d(1.1672748)
d ≈ $38,568.37
Therefore, Jin needs to contribute approximately $38,568 every month in order to end up with $45,000 after 4 years.
Jin is saving money and plans on making monthly contributions into an account earning an annual interest rate of 3.9% compounded monthly. If Jin would like to end up with dollar sign, 45, comma, 000$45,000 after 4 years, how much does he need to contribute to the account every month, to the nearest dollar? Use the following formula to determine your answer. A, equals, d, left bracket, start fraction, left bracket, 1, plus, i, right bracket, to the power n , minus, 1, divided by, i, end fraction, right bracket A=d( i (1+i) n −1 ) A, equalsA= the future value of the account after n periods d, equalsd= the amount invested at the end of each period i, equalsi= the interest rate per period n, equalsn= the number of periods
1 answer