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16 December, 08:39

A person inventes 4500 dollars in a bank. The bank pays 4.75% interest compounded semi-annually. To the nearest tenth of a year, how long mist the person leave the money in the bank until it reaches 5900 dollars

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  1. 16 December, 09:12
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    It should take 5.8 years to reach his goal.

    Step-by-step explanation:

    In order to solve this question we must use the compounded interest's formula shown below:

    M = C * (1 + r/n) ^ (t*n)

    Where M is the final amount, C is the initial amount, r is the interest rate, n is the compound rate and t is the time elapsed in years. Applying the data from the problem we have:

    5900 = 4500 * (1 + 0.0475/2) ^ (2*t)

    5900 = 4500 * (1 + 0.02375) ^ (2*t)

    5900 = 4500 * (1.02375) ^ (2*t)

    4500 * (1.02375) ^ (2*t) = 5900

    (1.02375) ^ (2*t) = 5900/4500

    (1.02375) ^ (2*t) = 59/45

    log[1.02375^ (2*t) ] = log (59/45)

    2*t*log (1.02375) = log (59/45)

    t = log (59/45) / [2*log (1.02375) ] = 5.77008

    It should take 5.8 years to reach his goal.
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