We need to follow the formula to find the size of the bacterial population is,
![y=a(b)^t](https://img.qammunity.org/2023/formulas/mathematics/college/ib8jujslo10whah3idluno9bt2ut8tnn6t.png)
Here, t is the total amount of bacteria, a is the initial bacteria at time = 0, abd b is the growth factor ( which is 2 according to the question )
At t = 110 mins , y = 60,000 we have
![60,000=a(2)^{(110)/(15)}](https://img.qammunity.org/2023/formulas/mathematics/college/jiawwhnbhtxzxz27veunebr3lpahzd5v5i.png)
![\begin{gathered} a=(60000)/(2^(7.33)) \\ a=372.92 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d7e93h38w42p1dpigcyj1xie9qlpjhxeil.png)
Now, the size of the population after 3 hours
![2^{(3*60)/(15)}p=2^(12)p](https://img.qammunity.org/2023/formulas/mathematics/college/q34ex5deleupw9c2qnfsimfadotnivkmd1.png)
![\begin{gathered} 2^(12)*372.92=4096*372.92 \\ =1,527,480.32 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/54s817m8ovbkxebbsfc2qg0u7fhqvmhwtd.png)
The answer is 1,527,480.32