Compound interest formula: A = P(1 + r/n)^{nt} , where

A = amount of money accumulated after an x number of years

P = principal amount (initial amount deposited/borrowed)

r = annual rate of interest (in decimal form)

t = number of years amount is deposited/borrowed

n = number of times the interest is compounded per year

The first part (a) states that the bank is compounding the interest on the deposit semi-annually. Since semi means half, this means that the interest is being compounded two times per year. Therefore, given the following: P = $3000 , r = 7%/100% = 0.07 , t = 6 , n = 2 , we can calculate A

A = 3000(1 + 0.07/2)^{2·6} = 3000(1 + 0.035)^{12}

= 3000(1.035)^{12} = 4533.2059

**A ≈ $4,533.21**

The second part (b) says that the interest is being compounded monthly, which means that the interest is being compounded 12 times per year since there are 12 months in a year. So, we use the same equation and given info as above but this time n = 12.

A = 3000(1 + 0.07/12)^{12·6} = 3000(1 + 0.0058333333)^{72}

= 3000(1.0058333333)^{72} = 4560.3165

**
A ≈ $4,560.32**

The last part of the problem (c) states that the interest is being compounded continuously, which uses a slightly different formula. The continuous compound interest formula is as follows:

A = Pe^{rt }, where e is a constant the equals approximately 2.7183

Therefore,

A = 3000e^{0.07·6} = 3000e^{0.42} = 3000(2.7183)^{0.42} = 4565.897

**A ≈ $4,565.90**