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
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document.write( "What was the initial population at time t = 0 ? \r
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document.write( "Find the size of the baterial population after 4 hours \n" );
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Algebra.Com's Answer #535033 by ankor@dixie-net.com(22740) You can put this solution on YOUR website! 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 \n" ); document.write( " |