document.write( "Question 1093899: Please help me understand this
\n" ); document.write( "The formula when interest is compounded n time per year is A=p(1+r/n)nt \r
\n" ); document.write( "\n" ); document.write( "Where A is the accrued amount after t years, P is the starting principal, and r is the interest rate, expressed as a decimal, that is compounded n times a year. If you invest $1000 at an interest rate of 7%, and leave it there for 30 years, determine your ending balance if the interest is compounded.
\n" ); document.write( "1. Once a year 2. Twice a year
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Algebra.Com's Answer #708515 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
\"+A+\" is the amount you end up with after \"+t+\" years
\n" ); document.write( "\"+P+\" is the amount you begin investing with
\n" ); document.write( "\"+r+\" is the interest rate
\n" ); document.write( "\"+n+\" is the number of times you compound per year
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\n" ); document.write( "Note that unless you know what \"+n+\" and \"+t+\" are
\n" ); document.write( "for instance, if \"+nt+=+6+\", you could have \"+1%2A6+=+6+\" or
\n" ); document.write( "\"+2%2A3+=+6+\".
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\n" ); document.write( "(1) \"+n+=+1+\"
\n" ); document.write( "\"+A+=+P%2A%28+1+%2B+r%2Fn+%29%5E%28nt%29+\"
\n" ); document.write( "\"+A+=+1000%2A%28+1+%2B+.07%2F1+%29%5E%28+1%2A30+%29+\"
\n" ); document.write( "\"+A+=+1000%2A1.07%5E30+\"
\n" ); document.write( "\"+A+=+1000%2A7.6123+\"
\n" ); document.write( "\"+A+=+7612.3+\"
\n" ); document.write( "$7,612.30 ending balance
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\n" ); document.write( "(2) \"+n+=+2+\"
\n" ); document.write( "\"+A+=+P%2A%28+1+%2B+r%2Fn+%29%5E%28nt%29+\"
\n" ); document.write( "\"+A+=+1000%2A%28+1+%2B+.07%2F2+%29%5E%282%2A30%29+\"
\n" ); document.write( "\"+A+=+1000%2A1.035%5E60+\"
\n" ); document.write( "\"+A+=+1000%2A7.8781+\"
\n" ); document.write( "\"+A+=+7878.1+\"
\n" ); document.write( "$7,878.10 ending balance
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\n" ); document.write( "get a 2nd opinion, too, if needed
\n" ); document.write( "on the problems
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