USE THE ATTACHED IMAGE BELOW TO HELP DO MY HOMEWORK PLEASE

Part A

Sue

A = P*(1+r/n)^(n*t) is the compound interest formula

A = 2300*(1+0.024/1)^(1*3)

A = 2469.6061952

A = 2469.61

A - P = 2469.61-2300 = 169.61

Bill

A = P*(1+r/n)^(n*t)

A = 1800*(1+0.034/1)^(1*3)

A = 1989.9131472

A = 1989.91

A-P = 1989.91-1800 = 189.91

Bill has earned more in interest.

Part B

By year 2, Bill has 1924.48 pounds in his account based on the work shown below

A = P*(1+r/n)^(n*t)

A = 1800*(1+0.034/1)^(1*2)

A = 1924.4808

A = 1924.48

This amount is the new deposit, so to speak, when we change the interest rate. Now r = 0.034 changes to r = 0.04. We only go for one year so t = 1

A = P*(1+r/n)^(n*t)

A = 1924.48*(1+0.04/1)^(1*1)

A = 2001.4592

A = 2001.46

Bill has 2001.46 pounds in his account after 3 years if the interest rate for the 1234 account changes to 4% in the third year.

Now subtract off the original amount Bill deposited to get

2001.46-1800 = 201.46

For this scenario, Bill earns 201.46 pounds in interest.

Therefore, Bill has earned the most interest for both cases of the interest rate staying at 3.4% or changing to 4% for that third year.