document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1120329>Question 1120329</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If you invest $5,900 in an account paying 12.5% compounded continuously, how much money will be in the account at the end of 3.5 years? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #735999 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=735999\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the formula for continuous compounding is f= p * e^(rn).\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "in your problem, the formula becomes:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "f = 5900 * e^(.125 * 3.5)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "the interest rate per year is .125.<BR>\n" );
document.write( "the number of years is 3.5.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "apply the formula to get f = 9138.098762\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );