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anonymous

  • 5 years ago

how much money must be deposited in an account paying 7.25% annual interest, compounded quarterly, to have a balance of $1000 after 10 years?

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  1. dumbcow
    • 5 years ago
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    General formula: B = D(1+i/n)^nt i = int rate n = num of times per year interest credited t=num years

  2. anonymous
    • 5 years ago
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    what do you mean b=d?

  3. dumbcow
    • 5 years ago
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    b = ending balance d = initial deposit

  4. anonymous
    • 5 years ago
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    im confused

  5. dumbcow
    • 5 years ago
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    ok lets say you deposit $100 and earn 50% 50% of 100 is 50 so your balance at end of year is 150 so 150=100*(1+0.5)

  6. anonymous
    • 5 years ago
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    Compound Formula for calculating compound interest: Where, \[A=P(1+r/t)^{nt}\] A = final amount P = principal amount (initial investment) r = annual nominal interest rate (as a decimal) n = number of times the interest is compounded per year t = number of years

  7. anonymous
    • 5 years ago
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    so just plug in the 1000, the interest rate .0725, number of years and compounds per year

  8. anonymous
    • 5 years ago
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    i do not know how i would set it up like to plug it in or anything

  9. dumbcow
    • 5 years ago
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    1000=D*(1+.0725/4)^40 solve for D

  10. anonymous
    • 5 years ago
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    \[1000/(1+.0725/4)^{40}=P\] That is what you end up with.

  11. anonymous
    • 5 years ago
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    solve for t or p?

  12. anonymous
    • 5 years ago
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    There is no t to solve for

  13. anonymous
    • 5 years ago
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    You already know t, its 10 years

  14. anonymous
    • 5 years ago
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    so solve for p?

  15. anonymous
    • 5 years ago
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    It's already solved, you just have to plug it into a calculator

  16. anonymous
    • 5 years ago
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    or do it by hand if you want to.

  17. anonymous
    • 5 years ago
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    1000/(40725/40000)^40 = 487.48

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