Question

If 47400 dollars is invested at an interest rate of 7 percent per year, find the value of the investment at the end of 5 years for the following compounding methods, to the nearest cent.
(a) Annual: $
(b) Semiannual: $
(c) Monthly: $
(d) Daily: $

Answer 1

Answer:

Part A) Annual $\ 66,480.95$  

Part B) Semiannual $\ 66,862.38$  

Part C) Monthly $\ 67,195.44$  

Part D) Daily $\ 67,261.54$  

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

so

Part A) Annual

in this problem we have  

$t=5\ years\\ P=\ 47,400\\ r=0.07\\n=1$  

substitute in the formula above  

$A=\ 47,400(1+\frac{0.07}{1})^{1*5}$  

$A=\ 47,400(1.07)^{5}$  

$A=\ 66,480.95$  

Part B) Semiannual

in this problem we have  

$t=5\ years\\ P=\ 47,400\\ r=0.07\\n=2$  

substitute in the formula above  

$A=\ 47,400(1+\frac{0.07}{2})^{2*5}$  

$A=\ 47,400(1.035)^{10}$  

$A=\ 66,862.38$  

Part C) Monthly

in this problem we have  

$t=5\ years\\ P=\ 47,400\\ r=0.07\\n=12$  

substitute in the formula above  

$A=\ 47,400(1+\frac{0.07}{12})^{12*5}$  

$A=\ 47,400(1.0058)^{60}$  

$A=\ 67,195.44$  

Part D) Daily

in this problem we have  

$t=5\ years\\ P=\ 47,400\\ r=0.07\\n=365$  

substitute in the formula above  

$A=\ 47,400(1+\frac{0.07}{365})^{365*5}$  

$A=\ 47,400(1.0002)^{1,825}$  

$A=\ 67,261.54$