A small town has 2100 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? (Do not round k in your calculation. Round the final answer to one decimal place.) hours after the beginning

Answer:2.7 PMStep-by-step explanation:Since, the rate of spread is proportional to the product of fraction y of people who have heard the rumour and the fraction who have not heard,[tex]\frac{dy}{dt}=ky(1-y)[/tex][tex]\frac{dy}{y(1-y)}=kdt[/tex][tex](\frac{1}{y}+\frac{1}{1-y})dy = kdt[/tex]Integrating both sides,[tex]\int (\frac{1}{y}+\frac{1}{1-y})dy = \int kdt[/tex][tex]\ln y - \ln (1-y) = kt + C[/tex][tex]\ln (\frac{y}{1-y}) = kt + C[/tex][tex]\frac{y}{1-y}=e^{kt + C}[/tex][tex]y = e^{kt+C} - y e^{kt+C}[/tex][tex]y(1+e^{kt+C}) = e^{kt+C}[/tex][tex]y = \frac{e^{kt+C}}{1+e^{kt+C}}[/tex]If t = 0, ( at 8 AM ), y = [tex]\frac{80}{2100}[/tex][tex]\frac{80}{2100}= \frac{e^{0+C}}{1+e^{0+C}}[/tex][tex]\frac{4}{105}=\frac{e^C}{1+e^C}[/tex][tex]4 + 4e^C = 105e^C[/tex] [tex]4 = 101e^C[/tex][tex]\implies e^C=\frac{4}{101}[/tex] Now, at noon, i.e t = 4, y = [tex]\frac{1}{2}[/tex][tex]\frac{1}{2}=\frac{e^{4k}.\frac{4}{101}}{1+e^{4k}.\frac{4}{101}}[/tex][tex]\frac{1}{2}=\frac{4e^{4k}}{101+4e^{4k}}[/tex][tex]101 + 4e^{4k}=8 e^{4k}[/tex][tex]101 = 4e^{4k}[/tex][tex]\frac{101}{4}=e^{4k}[/tex][tex](\frac{101}{4})^\frac{1}{4} = e^k[/tex]If [tex]y = \frac{90}{100}=\frac{9}{10}[/tex][tex]\frac{9}{10}= \frac{(\frac{101}{4})^\frac{t}{4}\times \frac{4}{101}}{1+(\frac{101}{4})^\frac{t}{4}\times \frac{4}{101}}[/tex]Using graphing calculator,t ≈ 6.722,Hence, after 6.722 hours since 8 AM, i.e. on 2.7 PM ( approx ) the 90% of the population have heard the rumour.