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suppose the population of a town increases by 3% each year. the population of the town today is 25,000. use compound interest table to find its population ten years from now. (must give the answer to the nearest thousand)

  • math -

    the calculation is

    25000(1.03)^10
    which on a calculator is a piece of cake

    If you have to use "compound interest tables ??" you probably have a column for 4%, go down that table to the row n=10 to find 1.34392
    multiply that by 25000
    you should get $33597.91

  • math -

    the above contains a typo, I obviously meant 3% in the third last line

  • math -

    Compound interest table for 3% p.a., compounded each year, obtained by multiplying the factor of the previous year by 1.03. The value is also equal to 1.03n where n is the number of years compounded.
    Year growth factor
    0 1.0000
    1 1.0300
    2 1.0609
    3 1.0927
    4 1.1255
    5 1.1593
    6 1.1941
    7 1.2299
    8 1.2668
    9 1.3048
    10 1.3439

    After 10 years, the population is
    25000*1.3439
    =33598
    =34000 (to the nearest thousand).

  • math -

    Oops, it does not seem to like pasting from Excel!

    Here is the compound interest table:
    year growth factor
    0 1.0000
    1 1.0300
    2 1.0609
    3 1.0927
    4 1.1255
    5 1.1593
    6 1.1941
    7 1.2299
    8 1.2668
    9 1.3048
    10 1.3439
    After the tenth year, the population is
    25000*1.3439
    =33598
    =34000 (to the nearest thousand)

  • math -

    OK, I give up pasting!
    The growth factor for 10 years is 1.3439.
    Multiply this with 25000 gives 33598, and rounding to the nearest thousand gives 34000.

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