Answer:
On 11th day ( approx )
Step-by-step explanation:
Since, if a population change with a constant rate,
Then final population,
![A=P(1-\frac{r}{100})^t](https://tex.z-dn.net/?f=A%3DP%281-%5Cfrac%7Br%7D%7B100%7D%29%5Et)
Where,
P = initial population,
r = rate of change per period,
t = number of periods,
Given,
Initial population = 2500,
Out of which, infected population = 30,
So, healthy people, initially = 2500 - 30 = 2470
Every day after, 18% of those still healthy fall ill.
So, after x days the number of healthy people,
![P=2470(1-\frac{18}{100})^x=2470(1-0.18)^x=2470(0.82)^x----(1)](https://tex.z-dn.net/?f=P%3D2470%281-%5Cfrac%7B18%7D%7B100%7D%29%5Ex%3D2470%281-0.18%29%5Ex%3D2470%280.82%29%5Ex----%281%29)
Now, if 87% of 2470 are ill,
Then reaming healthy population = (100 - 87)% of 2470
= 13% of 2470
![=\frac{13\times 2470}{100}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B13%5Ctimes%202470%7D%7B100%7D)
![=\frac{32110}{100}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B32110%7D%7B100%7D)
= 321.10
If P = 321.10,
From equation (1),
![321.10 = 2470(0.82)^x](https://tex.z-dn.net/?f=321.10%20%3D%202470%280.82%29%5Ex)
![\frac{321.10}{2470}=0.82^x](https://tex.z-dn.net/?f=%5Cfrac%7B321.10%7D%7B2470%7D%3D0.82%5Ex)
![\log(\frac{321.1}{2470}) =x \log(0.82)](https://tex.z-dn.net/?f=%5Clog%28%5Cfrac%7B321.1%7D%7B2470%7D%29%20%3Dx%20%5Clog%280.82%29)
![\implies x = \frac{\log(\frac{321.1}{2470})}{\log(0.82)}=10.2807315584](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%20%5Cfrac%7B%5Clog%28%5Cfrac%7B321.1%7D%7B2470%7D%29%7D%7B%5Clog%280.82%29%7D%3D10.2807315584)
Hence, by 11 day at least 87% of the population be infected.