Question

$750 is invested at 11% compounded quarterly. how much is the account after 10 years?

Answer 1

Answer:

F= $2219.91

Step-by-step explanation:

$F=P(1+i)^{n}$

Here

F= value of money after 10 years

P= present value

i= interest

n= number of compound on money

P=$750   i = 11%/4=0.0275 n=10x4= 40

$F=750(1+0.0275)^{40}$

$F=750(1.0275)^{40}$

$F=750(2.96)$

F= $2219.91