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Math Help - Intrest

  1. #1
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    Intrest

    If $600 is deposited in an account paying 11% compounded continously, then what amount will be in the account after 8 years?
    I just needed to double check if A=Pe^(rt)
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  2. #2
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    Quote Originally Posted by calc_help123 View Post
    If $600 is deposited in an account paying 11% compounded continously, then what amount will be in the account after 8 years?
    I just needed to double check if A=Pe^(rt)
    Hi calc_help123,

    Yes. A=Pe^{rt}

    So, A=600e^{(.11)(8)}\approx 1446.54
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  3. #3
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    Quote Originally Posted by calc_help123 View Post
    If $600 is deposited in an account paying 11% compounded continously, then what amount will be in the account after 8 years?
    I just needed to double check if A=Pe^(rt)
    It is better to use the second formula below because e is only correct for large n as the binomial expansion shows: (1+\frac{1}{n})^n \rightarrow e \: , \: n \rightarrow \infty

    A(t) = A(0)(1+{x})^{t} is a better model


    1. A(t) = Amount at time t
    2. A(0) = Amount at time 0
    3. x = amount of interest in one period
    4. t = time (number of periods)


    A(t) = 600(1+\frac{11}{100})^8 = \$1382.72
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