document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=773262>Question 773262</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A protected wooded area was stocked with 1500 deer of a certain species in 1995. The deer population in 2010 was found to be 3350 deer.\r<BR>\n" );
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document.write( "a. Let t be the number of years after 1995. Find the deer population at t=0.\r<BR>\n" );
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document.write( "b. Find the growth function that gives the deer population t years after 1995.\r<BR>\n" );
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document.write( "c. Predict the deer population in 2020.\r<BR>\n" );
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document.write( "d. In what year will the deer population reach 9,000? Give the exact value for t and then use your calculator to approximate t to get the year.\r<BR>\n" );
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document.write( "e. Are your answers for parts (c) and (d) consistent with the given population data in you found to the answer in part (b)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #471434 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=471434\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a) 1500<BR>\n" );
document.write( "b) A = P*e^kt <BR>\n" );
document.write( "   3350 = 1500*e^k*15<BR>\n" );
document.write( "   2.23 = e^15k<BR>\n" );
document.write( "   ln(2.23) = 15k<BR>\n" );
document.write( "   k = ln(2.23) / 15<BR>\n" );
document.write( "   k = 0.05<BR>\n" );
document.write( "   A = P*e^0.05t<BR>\n" );
document.write( "c) A = 1500*e^0.05*25<BR>\n" );
document.write( "   A = 1500*2.718^(0.05*25)<BR>\n" );
document.write( "   A = 5235<BR>\n" );
document.write( "d) 9000 = 1500 *e^0.05*t<BR>\n" );
document.write( "   6 = e^0.05t<BR>\n" );
document.write( "   ln6 = 0.05t<BR>\n" );
document.write( "   t = ln6 / 0.05<BR>\n" );
document.write( "   t = 35.84 years<BR>\n" );
document.write( "approx year is 1995 + 35.84 = 2030.84 = 2031<BR>\n" );
document.write( "e) A = 1500*e^0.05*15<BR>\n" );
document.write( "   A = 1500*2.718^0.05*15<BR>\n" );
document.write( "   A = 3281<BR>\n" );
document.write( "approx equal, model is consistent </SPAN>\n" );
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