Part 1 :
P = $200
r = 3.5% or 0.035
t = 3
n = 4
Compound interest formula =
![P(1+(r)/(n))^(nt)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5sjqdtzxeafln2cjbq8vqd94sq9c9qvcyk.png)
=
![200(1+(0.035)/(4))^(3*4)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/bo1evz5azacoubk44mekhn3g39s38dguxx.png)
=
=
![200*1.1102](https://img.qammunity.org/2020/formulas/mathematics/middle-school/crcbjxekd4mip6y9u11pa4x8rzs9j9jq1j.png)
= 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 =
![Pe^(rt)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/fpugrvjzmtxfbj8l6z5zlawz5yr8ywu1ty.png)
Putting e = 2.71828 we get.
A =
![200(2.71828)^(0.035*3)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ndhzlr7tjmp3iyoypnapxq11iwwgql2f7x.png)
=
![200(2.71828)^(0.105)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4gnrkvg7bxytk8g1tnynznm5pwltyhh6bh.png)
= 200*1.11071 = 222.14
Hence, amount in account after 3 years will be= 200+222.14 = $422.14