document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=288104>Question 288104</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In 1920, the record for a certyain race was 46.4 seconds. In 1950, it wq 46.1 seconds. Let R(t)= the record in the race and t= the number of years since 1920. <BR>\n" );
document.write( "a.)find a linear fucntion that fits the data<BR>\n" );
document.write( "b.) Use the function in (a) to predict the record in 2003 and in 2006.<BR>\n" );
document.write( "c.) find the year that the record will be 45.53 seconds.\r<BR>\n" );
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document.write( "Find a linear equation function that fits the data.<BR>\n" );
document.write( " r(t)=______<BR>\n" );
document.write( "ROUND TO THE NEAREST HUNDRETH </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #208930 by </B><A HREF=/tutors/aboutme.mpl?userid=amnd><B>amnd(23)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=amnd><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=amnd><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=208930\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a) The basic form of the linear equation would be: R(t) = At + 46.4\r<BR>\n" );
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document.write( "This is formed from the record for the year 1920 (46.4), and the fact that for the year 1920, t (which is the number of years since the same year, 1920) is zero. This can be proven mathematically as:\r<BR>\n" );
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document.write( "R(0) = At + 46.4 = 0 + 46.4 = 46.4\r<BR>\n" );
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document.write( "Now, to find the value of the constant for the variable <i>t</i>. The record in 1950 (with t = 1950 - 1920 = 30) would help. Insert it into the linear equation with the unknown constant A:\r<BR>\n" );
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document.write( "R(30)= A*30 +46.4 <BR>\n" );
document.write( "46.1 = 30A+46.4<BR>\n" );
document.write( "30A  = -0.3<BR>\n" );
document.write( "A    =-0.01\r<BR>\n" );
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document.write( "Knowing A, we can form the linear equation:<BR>\n" );
document.write( "<b>R(t) = -0.01t +46.4</b> or 46.4 - 0.01t<BR>\n" );
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document.write( "b) Simply input the value for t for both years (for 2003, t = 83, and for 2006, t = 86). This is easy enough math so I'm not going to jot down the whole thing, but these are the answers if you want to check yours:<BR>\n" );
document.write( "For 2003: <b>R(83)= 45.57</b><BR>\n" );
document.write( "For 2006: <b>R(86)= 45.54</b><BR>\n" );
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document.write( "c) R(t) = -0.01t + 46.4<BR>\n" );
document.write( "45.53 = -0.01t + 46.4<BR>\n" );
document.write( "0.01t = 46.4 - 45.53<BR>\n" );
document.write( "t=0.87*100=87\r<BR>\n" );
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document.write( "The year would be 1920 + t = 1920 + 87 = <b>2007</b> </SPAN>\n" );
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