Question

if $6,000 is invested at an annual interest rate of 1.83%, compounded daily, what will the investment be worth after 10 years?

Answer 1

Answer:

$\ 7,204.85$  

Step-by-step explanation:

we know that    

The compound interest formula is equal to  

$A=P(1+\frac{r}{n})^{nt}$  

where  

A is the Final Investment Value  

P is the Principal amount of money to be invested  

r is the rate of interest  in decimal

t is Number of Time Periods  

n is the number of times interest is compounded per year

in this problem we have  

$t=10\ years\\ P=\ 6,000\\ r=0.0183\\n=365$  

substitute in the formula above  

$A=\ 6,000(1+\frac{0.0183}{365})^{365*10}=\ 7,204.85$