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anonymous

  • 4 years ago

Fernando has a savings account balance of $2,471.23. The interest rate on the account is 2.4% compounded quarterly. If he opened the account nine years ago, what was the value of his initial deposit?

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  1. anonymous
    • 4 years ago
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    In 9 yrs there are 36 quarters. The interest rate is 2.4/4 per quarter just stick those into the compound interest formula - it needs the rate & number of periods to be THE SAME TIME UNITS.

  2. anonymous
    • 4 years ago
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    Fernando is such a boss name. don't you think? :)

  3. Directrix
    • 4 years ago
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    A = P [( 1 + ( r / n) ] ^ (nt) P = principal amount (the initial amount you borrow or deposit) r = annual rate of interest (as a decimal) t = number of years the amount is deposited or borrowed for. A = amount of money accumulated after n years, including interest. n = number of times the interest is compounded per year

  4. Directrix
    • 4 years ago
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    2471.23 = P [ ( 1 + (.024)/ 4 ] ^ [ (4)(9)]

  5. Directrix
    • 4 years ago
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    2471.23 = P [ (1 + .006) ] ^ 36

  6. Directrix
    • 4 years ago
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    2471.23 = P [ 1.006 ] ^ 36

  7. Directrix
    • 4 years ago
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    2471.23 = 1.24 P P = $ 1 992 . 93 approximately

  8. Directrix
    • 4 years ago
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    @ merengat --> Will you post your work? Or, find the error in mine, please. Thanks.

  9. dumbcow
    • 4 years ago
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    correct, without rounding i get 1992.44

  10. anonymous
    • 4 years ago
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    Call a i deposit on a monthly x is the interest rate n is the number of months post we have After January, the amount of a + ax = a (x + 1) fra In early February: a×(x+1)+a=a×(x+1+1)=(ax)[(1+x)2−1] After February: (ax)[(1+x)2−1]+(ax)[(1+x)2−1]x=(ax)[(1+x)2−1]x=(ax)[(1+x)2−1]((x+1)) ...... After n months, the amount of principal and interest are: (ax)(x+1)[(1+x)n−1] Applying the formula, instead of on and

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