document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1165018>Question 1165018</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Tom Jones and Bill Smith set out to create their own thermometers. Jones calls the freezing point of<BR>\n" );
document.write( "8' water on his scale 40 degrees, while Smith calls his freezing point 25 degrees. Jones makes the boiling<BR>\n" );
document.write( "L{ 6 TC point of water 280 degrees, and Smith, 125 degrees. What temperature on the Smith scale is equivalent<BR>\n" );
document.write( "'1 8x6 to 97 degrees on the Jones scale (to 1/10 of a degree)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #789522 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=789522\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i'm assuming you are comparing their freezing point and their boiling points with comparable units.<BR>\n" );
document.write( "if not, then this answer is not good.<BR>\n" );
document.write( "the method is the same as used to convert from fahrenheit to centigrade and vice versa.\r<BR>\n" );
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document.write( "you have:\r<BR>\n" );
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document.write( "smith = 25 degrees freezing point and 125 degrees boiling point.<BR>\n" );
document.write( "jones = 40 degrees freezing point and 280 degrees boiling point.\r<BR>\n" );
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document.write( "the number of degrees between freezing point and boiling point for smith is 100 degrees.<BR>\n" );
document.write( "the number of degrees between freezing point and boiling point for jones is 240 degrees.\r<BR>\n" );
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document.write( "each increment on the jones scale is 240 / 100 = 2.4 times each increment on the smith scale.\r<BR>\n" );
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document.write( "to make the freezing point the same, you need to adjust the jones scale by -20.\r<BR>\n" );
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document.write( "consider freezing point for smith is 25 degrees and freezing point for jones is 40.<BR>\n" );
document.write( "since jones increment is 2.4 * smith, then you get:<BR>\n" );
document.write( "jones freezing point is 25 * 2.4 = 60 - 20 = 40.\r<BR>\n" );
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document.write( "jones scale is equal to smith scale by the formula of j = s * 2.4 - 20.\r<BR>\n" );
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document.write( "while the in between may not be exact if the change in degrees is not linear, the freeze and boil points should be right on.\r<BR>\n" );
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document.write( "when smith = 25 degrees, jones = 25 * 2.4 - 20 = 40.<BR>\n" );
document.write( "their freeze points are consistent with what each is supposed to be.\r<BR>\n" );
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document.write( "when smith = 125 degrees, jones = 125 * 2.4 - 20 = 280.<BR>\n" );
document.write( "their boiling points are consistent with what each is supposed to be.\r<BR>\n" );
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document.write( "therefore, to convert from smith scale to jones scale, your formula is j = 2.4 * s - 20\r<BR>\n" );
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document.write( "to convert from jones scale to smith scale, your formula is s = (j + 20) / 2.4\r<BR>\n" );
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document.write( "at the freezing point, you have s = 25 and j = 40.<BR>\n" );
document.write( "when you convert from s to j, you get j = 2.4 * s - 20 = 2.4 * 25 - 20 = 40.<BR>\n" );
document.write( "when you convert from j to s, you get s = (j + 20) / 2.4 = (40 + 20) / 2.4 = 60 / 2.4 = 25\r<BR>\n" );
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document.write( "at the boiling point, you  have s = 125 and j = 280.<BR>\n" );
document.write( "when you convert from s to j, you get j = 2.4 * 125 - 20 = 280.<BR>\n" );
document.write( "when you convert from j to s, you get s = (280 + 20) / 2.4 = 125.\r<BR>\n" );
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document.write( "if your problem is not what i described above, then i have no idea what you would do.\r<BR>\n" );
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