Question

A customer deposits $500 in an account that pays 4% annual interest. what is the balance after 3 years if the interest is compounded annually? compound interest formula: mc017-1.jpg t = years since initial deposit n = number of times compounded per year r = annual interest rate (as a decimal)

Answer 1

Going to add on to your variables for the sake of the formula.

Let A = the amount after T years. 
P = Principal amount

$A = P(1 + \frac{r}{n})^{nt}$

$A = 500(1 + \frac{0.04}{1})^{1(3)}$

$A = 500(1 + \frac{0.04}{1})^{3}$

$A = 562.432$ or ≈ $562.43

The customer would have $562.43 at the end of 3 years. 

Answer 2

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,

$A = P(1 +\frac{r}{n})^{nt}$

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,

$A = 500(1 +\frac{0.04}{1})^{1(3)}$  

$A = 500(1+0.04)^{3}$  

 $A = 500(1.04)^{3}$  

 $A =562.432$  

  $A =562.432$  

Therefore, the balance after 3 years if the interest is compounded annually is $562.432.