document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=981587>Question 981587</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how much interest would i earn if I invested 3000 for 5 years into a saving account with a intrest rate of 0.35<BR>\n" );
document.write( "then compare at 3 different compounding periods monthly daily continuously<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #602566 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=602566\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A=Ao{1+r/n)^nt, where n is the number of compounding periods per year.<BR>\n" );
document.write( "Look at {1+r/n}^nt<BR>\n" );
document.write( "For monthly, (1+(.35/12)^60, since 60 months in 5 years.<BR>\n" );
document.write( "That is 5.6123, and the total is 3000* that, or $16,836.97\r<BR>\n" );
document.write( "\n" );
document.write( "For daily, (1+(.35/365)}^1825, or 5.7498, and total is $17,249.34\r<BR>\n" );
document.write( "\n" );
document.write( "For continuously<BR>\n" );
document.write( "A=Aoe^.35*5=3000*e^1.75=$17263.81 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );