Answer:
Part A) Annual [tex]\$66,480.95[/tex]
Part B) Semiannual [tex]\$66,862.38[/tex]
Part C) Monthly [tex]\$67,195.44[/tex]
Part D) Daily [tex]\$67,261.54[/tex]
Step-by-step explanation:
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
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
[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=1[/tex]
substitute in the formula above
[tex]A=\$47,400(1+\frac{0.07}{1})^{1*5}[/tex]
[tex]A=\$47,400(1.07)^{5}[/tex]
[tex]A=\$66,480.95[/tex]
Part B) Semiannual
in this problem we have
[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=2[/tex]
substitute in the formula above
[tex]A=\$47,400(1+\frac{0.07}{2})^{2*5}[/tex]
[tex]A=\$47,400(1.035)^{10}[/tex]
[tex]A=\$66,862.38[/tex]
Part C) Monthly
in this problem we have
[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=12[/tex]
substitute in the formula above
[tex]A=\$47,400(1+\frac{0.07}{12})^{12*5}[/tex]
[tex]A=\$47,400(1.0058)^{60}[/tex]
[tex]A=\$67,195.44[/tex]
Part D) Daily
in this problem we have
[tex]t=5\ years\\ P=\$47,400\\ r=0.07\\n=365[/tex]
substitute in the formula above
[tex]A=\$47,400(1+\frac{0.07}{365})^{365*5}[/tex]
[tex]A=\$47,400(1.0002)^{1,825}[/tex]
[tex]A=\$67,261.54[/tex]
