document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=468936>Question 468936</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In 1990, a person places $1,000 in an investment that earned 10% interest, compounded annually. Calculate the value of the investment for the years 1991, 1992, and 1993. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #321714 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=321714\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> In 1990, a person places $1,000 in an investment that earned 10% interest, compounded annually. Calculate the value of the investment for the years 1991, 1992, and 1993.<BR>\n" );
document.write( "---<BR>\n" );
document.write( "Answer for 1992:<BR>\n" );
document.write( "A(t) = P(1 + (r/n))^(nt)<BR>\n" );
document.write( "A(2) = 1000(1+(0.10/1))^(1*2)<BR>\n" );
document.write( "---<BR>\n" );
document.write( "A(2) = 1000(1.10)^2 = $1210.00<BR>\n" );
document.write( "==================================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "================== </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );