document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=407895>Question 407895</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose the sales of a particular brand of appliance satisfy the linear model y=100x+2300, where y represents the number of sales in a year x with x = 0 corresponding to 1982, with x = 1 corresponding to 1983 etc.  find the number of sales in 1999. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #287420 by </B><A HREF=/tutors/aboutme.mpl?userid=graphmatics><B>graphmatics(170)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=graphmatics><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=graphmatics><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=287420\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If we have to let x=0 correspond to 1982 then the x for 1999 is found as <BR>\n" );
document.write( "x = 1999-1983<BR>\n" );
document.write( "x = 16<BR>\n" );
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document.write( "so<BR>\n" );
document.write( "y = 100*16 + 2300<BR>\n" );
document.write( "y = 1600 + 2300<BR>\n" );
document.write( "y = 3900\r<BR>\n" );
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