document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1087644>Question 1087644</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the principal needed now to get the given amount, that is, find the present value.\r<BR>\n" );
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document.write( "To get $90 after 1 1/2 years at 6% compounded continuously\r<BR>\n" );
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document.write( "The present value of $90 is $?<BR>\n" );
document.write( "(round to the nearest cent as needed) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #701945 by </B><A HREF=/tutors/aboutme.mpl?userid=mathmate><B>mathmate(429)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=mathmate><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=mathmate><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=701945\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Question:<BR>\n" );
document.write( "Find the principal needed now to get the given amount, that is, find the present value.<BR>\n" );
document.write( "To get $90 after 1 1/2 years at 6% compounded continuously<BR>\n" );
document.write( " <BR>\n" );
document.write( "Solution:<BR>\n" );
document.write( "Amount after t years compounded continuously at rate r from present value P is given by:<BR>\n" );
document.write( "F=Pe^(rt)<BR>\n" );
document.write( "substituting<BR>\n" );
document.write( "$90=Pe^(0.06*1.5)<BR>\n" );
document.write( "=><BR>\n" );
document.write( "P=$90/(e^(0.05*1.5)=$90/1.09417=$82.25 [present value] </SPAN>\n" );
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