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15 September, 04:16

A person invest $10,000 into a bank the bank pays 4.75% interest compounded semi annually. To the nearest 10th of a year, how long was the person leave the money in the bank until it reaches $19,200 dollars

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  1. 15 September, 07:21
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    T is 13.9 years to the nearest 10th of a year

    Step-by-step explanation:

    In this question, we are to calculate the number of years at which someone who invests a particular amount will have a particular amount based on compound interest.

    To calculate the number of years, what we do is to use the compound interest formula.

    Mathematically,

    A = P (1 + r/n) ^nt

    Where A is the final amount after compounding all interests which is $19,200 according to the question

    P is the initial amount invested which is $10,000 according to the question

    r is the rate which is 4.75% according to the question = 4.75/100 = 0.0475

    n is the number of times per year in which interest is compounded. This is 2 as interest is compounded semi-annually

    t=?

    we plug these values;

    19200 = 10,000 (1+0.0475/2) ^2t

    divide through by 10,000

    1.92 = (1+0.02375) ^2t

    1.92 = (1.02375) ^2t

    We find the log of both sides

    log 1.92 = log [ (1.02375) ^2t)

    log 1.92 = 2tlog 1.02375

    2t = log 1.92/log 1.02375

    2t = 27.79

    t = 27.79/2

    t = 13.89 years

    The question asks to give answer to the nearest tenth of a year and thus t = 13.9 years
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