Respuesta :
Going to add on to your variables for the sake of the formula.
Let A = the amount after T years.
P = Principal amount
[tex]A = P(1 + \frac{r}{n})^{nt} [/tex]
[tex]A = 500(1 + \frac{0.04}{1})^{1(3)} [/tex]
[tex]A = 500(1 + \frac{0.04}{1})^{3} [/tex]
[tex]A = 562.432 [/tex] or ≈ $562.43
The customer would have $562.43 at the end of 3 years.
Let A = the amount after T years.
P = Principal amount
[tex]A = P(1 + \frac{r}{n})^{nt} [/tex]
[tex]A = 500(1 + \frac{0.04}{1})^{1(3)} [/tex]
[tex]A = 500(1 + \frac{0.04}{1})^{3} [/tex]
[tex]A = 562.432 [/tex] or ≈ $562.43
The customer would have $562.43 at the end of 3 years.
Answer:
The balance after 3 years if the interest is compounded annually is $562.432.
Step-by-step explanation:
Given : A customer deposits $500 in an account that pays 4% annual interest.
To find : What is the balance after 3 years if the interest is compounded annually?
Solution :
The compound interest formula,
[tex]A = P(1 +\frac{r}{n})^{nt}[/tex]
Where, A is the amount
P is the principal P=$500
r is the interest rate r=4%=0.04
t is the time t=3 years
n is number of times compounded per year n=1
Substitute the value in the formula,
[tex]A = 500(1 +\frac{0.04}{1})^{1(3)}[/tex]
[tex]A = 500(1+0.04)^{3}[/tex]
[tex]A = 500(1.04)^{3}[/tex]
[tex]A =562.432[/tex]
[tex]A =562.432[/tex]
Therefore, the balance after 3 years if the interest is compounded annually is $562.432.