document.write( "Question 1024155: Predict the population of bacteria for each situation and time period. This is exponential growth. Can you help me solve for A and B?\r
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document.write( "1) 55 bacteria that double every hour after:
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document.write( "A) after 3 hours B) after 5 hours. \r
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document.write( "2) 33 bacteria that double every 30 minutes
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document.write( "A) after 1 hour B) after 6 hours. \n" );
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Algebra.Com's Answer #639610 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! Predict the population of bacteria for each situation and time period. This is exponential growth. Can you help me solve for A and B?\r \n" ); document.write( "\n" ); document.write( "1) 55 bacteria that double every hour after: \n" ); document.write( "A) after 3 hours \n" ); document.write( "> 1 hour = 55*2 = 110 \n" ); document.write( "> 2 hour = 110*2 = 220 \n" ); document.write( "> 1 hour = 220*2 = 440 \n" ); document.write( "--> 55*2^t, t in hours \n" ); document.write( "--- \n" ); document.write( "B) after 5 hours \n" ); document.write( "55*2^5 = 1760 \n" ); document.write( "======================\r \n" ); document.write( "\n" ); document.write( "2) 33 bacteria that double every 30 minutes \n" ); document.write( "A) after 1 hour \n" ); document.write( "= 33*2^t, t in 30 min periods \n" ); document.write( "= 33*2^2 = 132 \n" ); document.write( "---------- \n" ); document.write( "B) after 6 hours \n" ); document.write( "6 hrs = 12 periods \n" ); document.write( "= 33*2^12 \n" ); document.write( "= 135168 \n" ); document.write( " |