document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140191>Question 1140191</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Strontium 90 is a radioactive material that decays according to the function <BR>\n" );
document.write( "A(t)=A0e−0.0244t, where A0 is the initial amount present and A is the amount present at time t​ (in years).  Assume that a scientist has a sample of<BR>\n" );
document.write( "400 grams of strontium 90.<BR>\n" );
document.write( "a) what is the decay rate of strontium 90?<BR>\n" );
document.write( "b) how much strontium 90 is left after 40 years<BR>\n" );
document.write( "c) when will only 300 grams of strontium 90 be left?<BR>\n" );
document.write( "d) what is the half life of strontium 90? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #760772 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=760772\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> decay rate is 2.44% per year<BR>\n" );
document.write( "after 40 years, 400*(e^-0.976)=150.73 gm<BR>\n" );
document.write( "300=400(e^-0.0244t)<BR>\n" );
document.write( "0.75=e^-0.0244t<BR>\n" );
document.write( "ln both sides<BR>\n" );
document.write( "-0.2877=-0.0244t<BR>\n" );
document.write( "t=11.79 years\r<BR>\n" );
document.write( "\n" );
document.write( "t (1/2) occurs when ln(0.5)=-0.0244t<BR>\n" );
document.write( "t=28.41 years\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );