document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=82474>Question 82474</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A bacteria culture started with a count of 720 at 8:00 A.M., and after (t) hours is expected to grow to f(t)= 720{3/2}t.  Estimate the number of bacteria in the culture at 11:00 A.M. the same day. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #59094 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=59094\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A bacteria culture started with a count of 720 at 8:00 A.M., and after (t) hours is expected to grow to f(t)= 720{3/2}^t.  Estimate the number of bacteria in the culture at 11:00 A.M. the same day.<BR>\n" );
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document.write( "f(t) = 720(3/2)^t<BR>\n" );
document.write( "From 8 to 11 is three hours.<BR>\n" );
document.write( "f(3) = 720(3/2)^3<BR>\n" );
document.write( "f(3) = 2430 bacteria<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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