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Given:
- Initial Investment = 8000
- Time = 10 years
- Rate = 6% = .06
Assumptions:
- Assuming compounded annually (i.e., every year, the investment is increased by 6%). I will also solve for compounded monthly.
Formula and Variables:
- A = P[1+(r/n)]nt, where:
- A = total money after some time
- P = Initial Amount
- r = Interest Rate as a decimal (annual)
- n = how many times interested is compounded per year (we are using 1)
- t = time in years
From there, just plug in everything into the formula:
A = P[1+(r/n)]nt
A = (8000)[1+(.06/1)](1)(10)
A = (8000)(1.06)10
A ~= 14326.782
because currency only goes to the cent, round to the cent
A = 14326.78 (if interest is compounded annually)
If interest where compounded monthly:
A = P[1+(r/n)]nt
A = (8000)[1+(.06/12)](12)(10)
A = (8000)(1.005)120
A = 14555.17 (about 288.39 more than compounded annually)
I've given you the formula, and two possible answers, all bolded. If your interested is compounded at another interval, just use the formula. Example, if interested is compounded quarterly, use n=4 instead.
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