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(C) The rate of growth is proportional to the number present.$\frac{dN}{dt} = kN \Rightarrow \frac{dN}{N} = k dt$.Integrating both sides,we get $\ln N

Vedclass · Accepted Answer

$5656$ approximately

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(C) The rate of growth is proportional to the number present.$\frac{dN}{dt} = kN \Rightarrow \frac{dN}{N} = k dt$.Integrating both sides,we get $\ln N = kt + C$.At $t = 0$,$N = 1000$,so $C = \ln 1000$.Thus,$\ln N = kt + \ln 1000 \Rightarrow \ln(\frac{N}{1000}) = kt \Rightarrow N = 1000 e^{kt}$.Given that at $t = 1$,$N = 2000$,we have $2000 = 1000 e^k$,which implies $e^k = 2$.Substituting this into the equation,$N = 1000 	imes (e^k)^t = 1000 	imes 2^t$.For $t = 2 \frac{1}{2} = 2.5$ hours:$N = 1000 	imes 2^{2.5} = 1000 	imes 2^2 	imes 2^{0.5} = 1000 	imes 4 	imes \sqrt{2}$.Given $\sqrt{2} = 1.414$,we get $N = 1000 	imes 4 	imes 1.414 = 5656$.

The rate of growth of bacteria is proportional to the number present. If initially there were $1000$ bacteria and the number doubles in $1$ hour,then the number of bacteria after $2 \frac{1}{2}$ hours is (Given $\sqrt{2} = 1.414$):

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