document.write( "Question 475305: An anthropologist finds there is so little remaining Carbon-14 in a prehistoric bone that instruments cannot measure it. This means that there is less than 0.5% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die? Round your answer to the nearest thousand. (22,000, etc) \n" ); document.write( "
Algebra.Com's Answer #325972 by ankor@dixie-net.com(22740)\"\" \"About 
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An anthropologist finds there is so little remaining Carbon-14 in a prehistoric bone that instruments cannot measure it.
\n" ); document.write( " This means that there is less than 0.5% of the amount of Carbon-14 the bones would have contained when the person was alive.
\n" ); document.write( " How long ago did the person die? Round your answer to the nearest thousand. (22,000, etc)
\n" ); document.write( ":
\n" ); document.write( "The half life formula
\n" ); document.write( "A = Ao*2^(-t/h)
\n" ); document.write( "Where
\n" ); document.write( "A = remaining amt after t yrs
\n" ); document.write( "Ao = initial amt
\n" ); document.write( "t = time in yrs
\n" ); document.write( "h = half-life of substance
\n" ); document.write( ":
\n" ); document.write( "The generally accepted half-life for Carbon 14 = 5730 yrs
\n" ); document.write( "We can assume the initial amt = 1, the resulting amt =.005
\n" ); document.write( ":
\n" ); document.write( "2^(-t/5730) = .005
\n" ); document.write( "using nat logs
\n" ); document.write( "\"-t%2F5730\"ln(2) = ln(.005)
\n" ); document.write( "\"-t%2F5730\" = \"ln%28.005%29%2Fln%282%29\"
\n" ); document.write( "\"-t%2F5730\" = -7.6438562
\n" ); document.write( "t = -5730 * -7.6438562
\n" ); document.write( "t = 43,800 yrs
\n" ); document.write( " \n" ); document.write( "
\n" );