document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=182605>Question 182605</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Population studies of fiddler crabs on a tropical island reported 1.1 x 10^4 in 1980 and 1.5 x 10^4 in 1982. Predict the time that the maximum population, 2.0 x 10^4, that can be supported by the island's resources will be reached.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #137096 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=137096\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Population studies of fiddler crabs on a tropical island reported 1.1 x 10^4 in 1980 and 1.5 x 10^4 in 1982. Predict the time that the maximum population, 2.0 x 10^4, that can be supported by the island's resources will be reached. <BR>\n" );
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document.write( "Generate an equation based on (1982, 1.5x10^4) and (1980, 1.1x10^4)<BR>\n" );
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document.write( "slope = (1.5x10^4 - 1.1x10^4) / (1982-1980) = 0.4x10^4/2 = 2000<BR>\n" );
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document.write( "So the crab population is growing by 2000 each year.<BR>\n" );
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document.write( "2.0x10^4 - 1.5x10^4 = 0.5(1x10^4) = (1/2)(10,000 = 5000<BR>\n" );
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document.write( "5000/(2000 per year) = 2.5 years<BR>\n" );
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document.write( "The population will be 2x10^4 in 1982 + 2.5 = 1984.5 or 1985<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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