document.write( "Question 885241: The doubling period of a baterial population is 15 minutes. At time t = 90 minutes, the baterial population was 50000. Round your answers to at least 1 decimal place.\r
\n" ); document.write( "\n" ); document.write( "What was the initial population at time t = 0 ? \r
\n" ); document.write( "\n" ); document.write( "Find the size of the baterial population after 4 hours
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Algebra.Com's Answer #535033 by ankor@dixie-net.com(22740)\"\" \"About 
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The doubling period of a bacterial population is 15 minutes.
\n" ); document.write( " At time t = 90 minutes, the bacterial population was 50000.
\n" ); document.write( " Round your answers to at least 1 decimal place.
\n" ); document.write( ":
\n" ); document.write( "We can use the formula:
\n" ); document.write( "A = Ao*2^(t/d); where:
\n" ); document.write( "A = amt after t time
\n" ); document.write( "Ao = initial amt (t=0)
\n" ); document.write( "t = time period in question
\n" ); document.write( "d = doubling time of substance
\n" ); document.write( "In our problem
\n" ); document.write( "d = 15 min
\n" ); document.write( "t = 90 min
\n" ); document.write( "A = 50000
\n" ); document.write( "What was the initial population at time t = 0
\n" ); document.write( "Ao * 2^(90/15) = 50000
\n" ); document.write( "Ao * 2^6 = 50000
\n" ); document.write( "We know 2^6 = 64
\n" ); document.write( "64(Ao) = 50000
\n" ); document.write( "Ao = 50000/64
\n" ); document.write( "Ao = 781.25 is the initial population
\n" ); document.write( " :
\n" ); document.write( "Find the size of the bacterial population after 4 hours
\n" ); document.write( "Change 4 hr to 240 min
\n" ); document.write( "A = 781.25 * 2^(240/15
\n" ); document.write( "A = 781.25 * 2^16
\n" ); document.write( "A= 781.25 * 65536
\n" ); document.write( "A = 51,199,218.75 after 4 hrs
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