SOLUTION: If $3000 is invested at an interest rate of 5% each year, find the amount of the investment at the end of 4 years for the following compounding methods. B)semiannually, which is

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Question 591829: If $3000 is invested at an interest rate of 5% each year, find the amount of the investment at the end of 4 years for the following compounding methods.
B)semiannually, which is six months, but it is also half a year, so I'm not sure which variable this is, as well as what number it is.
Note: I know that the formula I need to use is this: A(t)= P[1+(r/n)]^nt
Initially I tried plugging in the values: A(t)= 3000(1+0.05/4)^4, this was just a guess (this is one is for Annually though). I couldn't figure that one out, but since I think B) might be harder I tried that one. I had no success getting the right answer. I'd appreciate your help on this one, Thank you so much.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A=P%281%2Br%2Fn%29%5E%28n%2At%29 Start with the compound interest formula


A=3000%281%2B0.05%2F2%29%5E%282%2A4%29 Plug in P=3000, r=0.05, n=2 (since we're compounding twice a year) and t=4.


A=3000%281%2B0.025%29%5E%282%2A4%29 Evaluate 0.05%2F2 to get 0.025


A=3000%281.025%29%5E%282%2A4%29 Add 1 to 0.025 to get 1.025


A=3000%281.025%29%5E%288%29 Multiply 2 and 4 to get 8.


A=3000%281.21840289750992%29 Evaluate %281.025%29%5E%288%29 to get 1.21840289750992.


A=3655.20869252975 Multiply 3000 and 1.21840289750992 to get 3655.20869252975.


A=3655.21 Round to the nearest hundredth (ie to the nearest penny).


So there is $3655.21 in the account after 4 years (where $3000 is invested at an interest rate of 5% each year and it's compounded semiannually).