document.write( "Question 1192292: Jane found an account at Big Bank that pays out 2.05% compounded monthly. If she needs $1300 in 4 years, how much should she place in the account? Assume that she never adds additional money to the account. \n" ); document.write( "
Algebra.Com's Answer #824215 by Theo(13342)\"\" \"About 
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f = p * (1 + r) ^ n
\n" ); document.write( "f is the future value
\n" ); document.write( "p is the present value
\n" ); document.write( "r is the interest rate per time period.
\n" ); document.write( "n is the number of time periods.\r
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\n" ); document.write( "\n" ); document.write( "2.05% compounded monthly becomes .0205/12.
\n" ); document.write( "the formula uses the rate, not the percent.\r
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\n" ); document.write( "\n" ); document.write( "4 years * 12 months per year = (4 * 12) months.\r
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\n" ); document.write( "\n" ); document.write( "formula becomes:\r
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\n" ); document.write( "\n" ); document.write( "1300 = p * (1 + .0205/12) ^ (4 * 12)\r
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\n" ); document.write( "\n" ); document.write( "solve for p to get:\r
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\n" ); document.write( "\n" ); document.write( "p = 1300 / ((1 + .0205/12) ^ (4 * 12)) = 1197.737339.\r
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\n" ); document.write( "\n" ); document.write( "round to the nearest penny to get 1197.74.\r
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\n" ); document.write( "\n" ); document.write( "that's how much she needs to invest today at 2.05% per year compounded monthly so that she can have 1300 at the end of the 4 year investment period.\r
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