Answer:
Explanation:
given,
initial count of bacteria(P)=4,80,000 ,
increasing rate of bacteria(R)=2.5%,
time(T)= 2 hours
amount(A)=p(1+R/100)^T
A=480000(1+2.5/100)^2
A=480000*(1681/1600)
A=504300
therefore, the bacteria at the end of 2 hours is 504300(approx)