Answer:
See below
Step-by-step explanation:
Step 1.
P = $400
r = 0.04
t = 10 years
n = 12 ( as there are 12 months in a year).
Step 2.
A(10) = 400(1 + 0.04/12)^12^10
= 400 * 1.00333333^120
= $596.33 to the nearest hundredth (answer).
Answer and Explanation:
Given : The value of $400 invested at 4% interest compounded monthly for 10 years.
To find : The value of each of the following for this problem ?
Solution :
The interest formula is [tex]A(t)=P(1+\frac{r}{n})^{nt}[/tex]
Step 1 -
P is the amount invested, P=$400
r is the interest rate, r=4%=0.04
t is the time , t=10 years
n is the number of compounding periods per year, n=12
Step 2 - To find A(10),
Substitute all the values in the formula,
[tex]A(t)=P(1+\frac{r}{n})^{nt}[/tex]
[tex]A(10)=400(1+\frac{0.04}{12})^{12\times 10}[/tex]
[tex]A(10)=400(1+0.0033)^{120}[/tex]
[tex]A(10)=400(1.0033)^{120}[/tex]
[tex]A(10)=400(1.490)[/tex]
[tex]A(10)=596.33[/tex]
Therefore, The amount after 10 year is $596.33.
