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- anonymous

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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- anonymous

- schrodinger

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- anonymous

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.

- anonymous

Fernando is such a boss name. don't you think? :)

- Directrix

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

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- Directrix

2471.23 = P [ ( 1 + (.024)/ 4 ] ^ [ (4)(9)]

- Directrix

2471.23 = P [ (1 + .006) ] ^ 36

- Directrix

2471.23 = P [ 1.006 ] ^ 36

- Directrix

2471.23 = 1.24 P
P = $ 1 992 . 93 approximately

- Directrix

@ merengat --> Will you post your work? Or, find the error in mine, please. Thanks.

- dumbcow

correct, without rounding i get 1992.44

- anonymous

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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