Suppose that an outbreak of cholera follows severe flooding in an isolated town of 2500 people. Initially (Day 0), 30 people are infected. Every day after, 18% of those still healthy fall ill. By what day will at least 87% of the population be infected?

Answer:On 11th day ( approx )Step-by-step explanation:Since, if a population change with a constant rate,Then final population,[tex]A=P(1-\frac{r}{100})^t[/tex]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 = 2470Every day after, 18% of those still healthy fall ill.So, after x days the number of healthy people,[tex]P=2470(1-\frac{18}{100})^x=2470(1-0.18)^x=2470(0.82)^x----(1)[/tex]Now, if 87% of 2470 are ill,Then reaming healthy population = (100 - 87)% of 2470= 13% of 2470[tex]=\frac{13\times 2470}{100}[/tex][tex]=\frac{32110}{100}[/tex]= 321.10If P = 321.10,From equation (1),[tex]321.10 = 2470(0.82)^x[/tex][tex]\frac{321.10}{2470}=0.82^x[/tex][tex]\log(\frac{321.1}{2470}) =x \log(0.82)[/tex][tex]\implies x = \frac{\log(\frac{321.1}{2470})}{\log(0.82)}=10.2807315584[/tex]Hence, by 11 day at least 87% of the population be infected.