document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=998534>Question 998534</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If $1400 is invested at an interest rate of 6.75% per year, compounded quarterly, find the value of the investment after 5 years. (Round your answers to the nearest cent.)  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #616302 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=616302\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Formula<BR>\n" );
document.write( "P=Po{1+r/n}^nt<BR>\n" );
document.write( "=1400{1+.0675/4}^20<BR>\n" );
document.write( "=$1956.50. Round only at end<BR>\n" );
document.write( "Rule of 70 would show doubling in about 10 years, so this makes sense. <BR>\n" );
document.write( "Divide r by compoundings per year then raise to the total number of compoundings  </SPAN>\n" );
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