Answer :
Answer: There will be 4977 bacteria present after 10 hours.
Step-by-step explanation:
The exponential function for continuous growth in t years is given by :-
[tex]P=Ae^{kt}[/tex] (i)
, where A = initial population, k= rate of growth.
As per given, A= 2000,
After t= 2 hours, P=2400
Put these values in (i), we get
[tex]2400=2000e^{2k}\\\\\Rightarrow\ 1.2=e^{2k}[/tex]
Taking log on both sides
[tex]\ln 1.2=2k\\\\\Rightarrow\ k=\dfrac{\ln1.2}{2}=\dfrac{0.182321556794}{2}\\\\\Rightarrow\ k\approx0.09116[/tex]
put value of A=2000, k= 0.09116 and t= 10 , we get
[tex]P=2000e^{0.09116\times10}\\\\=2000e^{0.9116}\\\\=2000\times2.4883\\\\=4976.6\approx4977[/tex]
Hence, there will be 4977 bacteria present after 10 hours.