document.write( "Question 1147218: Determine the amount of money that will be accumulated in an account that pays compound​ interest, given the initial principal of ​$27,800 invested at 2.97​% annual interest for 9 years compounded
\n" ); document.write( "​(a) daily ​(n=365);
\n" ); document.write( "​(b) continuously.
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Algebra.Com's Answer #768543 by Theo(13342)\"\" \"About 
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discrete compounding formula is 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( "continuous compounding formula is f = p * e ^ (r * t)
\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( "using the discrete compounding formula, your problem becomes:\r
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\n" ); document.write( "\n" ); document.write( "f = 27500 * (1 + .0297/365) ^ (9 * 365) = 35,926.49804.\r
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\n" ); document.write( "\n" ); document.write( "using the continuous compounding formula, your problem becomes;\r
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\n" ); document.write( "\n" ); document.write( "f = 27500 * e ^ (.0297 * 9) = 35,926.88872.\r
\n" ); document.write( "\n" ); document.write( "we could also have done:\r
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\n" ); document.write( "\n" ); document.write( "f = 27500 * e ^ (.0297/365 * 9 * 365) = 35,926.88872.\r
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\n" ); document.write( "\n" ); document.write( "since the compounding is continuous, it doesn't matter what time periods we use as long as we're consistant, i.e. r = annual interest rate / number of compounding periods per year paired with n = number of years * number of compounding periods per year.\r
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\n" ); document.write( "\n" ); document.write( "also note that daily compounding gets pretty close to continuous compounding.
\n" ); document.write( "the more compounding periods per year you get, the closer you get to continuous compounding results.\r
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