document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=36503>Question 36503</A>: <I><SPAN STYLE=\"word-break: break-all;\">  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #302779 by </B><A HREF=/tutors/aboutme.mpl?userid=JWG><B>JWG(21)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JWG><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=302779\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use compound interest formula A = P(1+r/n)^nt \r<BR>\n" );
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document.write( "P = principal amount (the initial amount you borrow or deposit)<BR>\n" );
document.write( "In this question, $500\r<BR>\n" );
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document.write( "r = annual rate of interest (as a decimal)<BR>\n" );
document.write( "In this question, 3% or .03\r<BR>\n" );
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document.write( "t = number of years the amount is deposited or borrowed for.<BR>\n" );
document.write( "In this question, one (1) year\r<BR>\n" );
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document.write( "A = amount of money accumulated after n years, including interest.<BR>\n" );
document.write( "You are solving for A \r<BR>\n" );
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document.write( "n = number of times the interest is compounded per year<BR>\n" );
document.write( "In this question, the rate compounds every six months. One year is 12 months. n=12/6=2. \r<BR>\n" );
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document.write( "P(1+r/n)^nt => 500(1+.03/2)^2+1 => 500(1+.03/2)^3 => 500(203/200)^3 => \r<BR>\n" );
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document.write( "A = $522.84 </SPAN>\n" );
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