Part 1 :
P = $200
r = 3.5% or 0.035
t = 3
n = 4
Compound interest formula = [tex]P(1+\frac{r}{n})^{nt}[/tex]
= [tex]200(1+\frac{0.035}{4})^{3*4}[/tex]
= [tex]200(1.00875)^{12}[/tex]
= [tex]200\times1.1102[/tex]
= 222.04
So, money in the bank will be = 200+222.04 = $422.04
Part 2 :
P = $200
r = 3.5% or 0.035
t = 3
Continuous compound interest formula is =
A = [tex]Pe^{rt}[/tex]
Putting e = 2.71828 we get.
A = [tex]200(2.71828)^{0.035*3}[/tex]
= [tex]200(2.71828)^{0.105}[/tex]
= 200*1.11071 = 222.14
Hence, amount in account after 3 years will be= 200+222.14 = $422.14
