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A population of 500 E.coli bacteria doubles every 15 minutes.  Find what the...

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lbawa | Student, Undergraduate | eNoter

Posted May 2, 2011 at 11:39 PM via web

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A population of 500 E.coli bacteria doubles every 15 minutes.  Find what the population would be in 87 minutes.  Use an exponential model.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 3, 2011 at 1:08 AM (Answer #1)

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A population of E.coli bacteria doubles every 15 minutes. The initial population of the bacteria is 500.

An exponential model does not imply the use of 'e'. The general formula for exponential growth is P(t) = P0*(r)^(t/T), where P(t) is the population at time t, P0 is the initial population, r is the positive growth factor and T is the time required for the population to increase by a factor of r.

Here P0 = 500, r = 2 and T = 15.

We need to find P(87).

P(87) = 500*2^(87/15)

=> 500*2^5.8

=> 27857

The number of bacteria after 87 minutes would be 27857.

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