Sterile fruit flies are used in an experiment where the proportion that survives at least t days is given by e^−0.15t. If the experiment begins with 400 fruit flies, and flies are added at the rate of 6 per hour, how many flies are present 15 days after the start of the experiment? (Round your answer to the nearest integer.)

Answer:901 fliesStep-by-step explanation:Let the total population of flies is P,While [tex]P_s[/tex] and [tex]P_i[/tex] represent the number of flies that survived and that are initially respectively.Here,[tex]\frac{P_s}{P_i}=e^{-0.15t}[/tex][tex]\implies P_s = P_i e^{-0.15t}[/tex]Differentiating w. r. t. t ( time ),[tex]\frac{dP_s}{dt}=-0.15 P_i e^{-0.15t}=-0.15 P_s-----(1)[/tex]Now, if 6 flies are added per hour,Then the number of flies added per day = 24 × 6 = 144 ( ∵ 1 day = 24 hours),So,[tex]\frac{dP}{dt}=144 + \frac{dP_s}{dt}[/tex][tex]\frac{dP}{dt}=144 -0.15 P_s[/tex][tex]\implies \frac{dP}{dt}=144 -0.15 P[/tex]When P = [tex]P_s[/tex] = [tex]P_i[/tex][tex]\implies \frac{dP}{144 -0.15 P}=dt[/tex]∵ if t = 0, P = 400,[tex] \int_{400}^{P}\frac{dP}{144 -0.15 P}=\int_{t=0}^{15}dt[/tex][tex][\frac{\ln(144-0.15P)}{-0.15}}]_{400}^{P} = [t]_{0}^{15}[/tex][tex]\frac{\ln (144-0.15P)- \ln(144-60)}{-0.15}}=15[/tex][tex]\ln (144-0.15P) - \ln(84) = -2.25[/tex][tex]-\ln (144-0.15P) + \ln(84) = 2.25[/tex][tex] \ln (\frac{84}{144-0.15P})= 2.25[/tex][tex]\frac{84}{144-0.15P} = e^{2.25}[/tex][tex]\frac{84}{e^{2.25}} =144-0.15P [/tex][tex]8.853 = 144 - 0.15P[/tex][tex]0.15P = 144 - 8.853[/tex][tex]\implies P = \frac{135.147}{0.15}=\frac{135.147}{0.15}\approx 901[/tex]i.e. 901 flies would present.