Question:

Please show me how to solve this!?

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Charlene has $26000 to invest. She is considering depositing her money into an account at a rate of 10% compounding quarterly. If she leaves her money in the account how much interest will she earn over 25 years? Use the formula A = P(1 + i)n.

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  1. 26000 is the present value (P).Number of years (n) is 25. The trick, I believe is with the interest rate. I think you should convert this to annually to solve.  Your (n) is in terms of years, so you want your (i) compounding per year. But in your case it's given as compounded quarterly. So:

    r = nominal interest rate per year

    i = effective interest rate in the time interval

    l = length of the time interval in years

    m= reciprocal of the length of compounding period in years

          (think how many of the compounding period given in one year.

           Eg. compounded weekly... then m=52 )

    i = ( 1 + r/m)^(l*m) - 1

    To simplify things l*m = c which is the number of compounding periods in the time interval. So if you have something compounded monthly with time interval of one year then c=12. If you have something compounded weekly with time interval of one year then c=52.

    In your example:

    i= (1+ 0.1/4)^4 -1  = 0.1038 <-- this is what you plug in the equation.

    so A=(26000)(1+0.1038)(25) = $ 717,470

    For some reason, the number I got doesn't seem right to me :/


  2. "P" should be the amount invested.

    "A" should be the amount after accumulating interest.

    "n" should be the number of times compounded in a year ( 4 in your case)

    I don't know what the "i" is, but usually the following formula is used.

    A=P(1+ r/n)^nt    

    Where "r" is the rate of interest (0.10) and "t" is the time in years (25).

    Interest = A - P

  3. p=26000

    i=0.1

    n=25

    however, tu kompound 4 times a yeer, yu gotta adjust the formula

    N=25*4=100

    r=i/4=0.025

    ------------------------

    anser=307,156.63$

  4. .... you HAVE the formula. Just plug in your numbers.

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