Answer:
Step-by-step explanation:
We will need 2 different equations for this problem. For compounding a certain number of times a year, the formula is
[tex]A(t)=P(1+\frac{r}{n})^{nt}[/tex] so let's do that first. Filling in we get
[tex]A(t)=10000(1+\frac{.03}{4})^{(4)(5)[/tex] and simplifying a bit gives us
[tex]A(t)=10000(1+.0075)^{20}[/tex] and a bit more,
[tex]A(t)=10000(1.0075)^{20}[/tex] and then
A(t) = 10000(1.161184142) so
A(t) = 11611.84
Then for the compounding continuously, the formula is
[tex]A(t)=Pe^{rt[/tex]
[tex]A(t)=10000e^{(.03)(5)}[/tex] and simplify a bit to\
[tex]A(t)=10000e^{.15}[/tex] Now raise e to the power of .15 on your calculator to get
A(t) = 10000(1.161834243) so
A(t) = 11618.34
Not too much of a difference between the 2 values at the end.
