5.9-1:please help with part C: A population of E. coli bacteria grows exponentially with time. You believe that the mean time between divisions is t, = 44 min, and that cell death occurs on averageafter tm = 500 hr. The population starts with 3000 cells. Complete parts (a) through (c).The number of cells present after 2 hours is 19811 .(Do not round until the final answer. Then round to the nearest integer as needed.)(b) In fact, you measure that there are 42,000 cells present in the population after 2 hours. Can the discrepancy between the model and data be explained by yourestimate for t, being wrong? Can it be explained by the estimate for tm being wrong?O A. t, and tm must be wrong.O B. tm must be wrong.C. t, must be wrong.(c) Assuming that tm is correct, calculate a revised estimate for t, to make the mathematical model fit your experimental data.= 0.356 hour(s)to(Do not round until the final answer. Then round to three decimal places as needed.)

Name: 5.9-1:please help with part C: A population of E. coli bacteria grows
Uploaded: 2024-03-20T06:46:55.000Z
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Description: 5.9-1:please help with part C: A population of E. coli bacteria grows exponentially with time. You believe that the mean time between divisions is t, = 44 min, and that cell death occurs on averageafter tm = 500 hr. The population starts with 3000 ce

Answer for question c: Given that the population of e.coli bacteria grows exponentially with time. Hence the function that gives the population is, Pt=3000ekt…1 where k>0 is the growth constant. To find tb such that P2=42,000 Substitute t=2 in 1, to find k: P2=3000e2k42000=3000e2k420003000=e2k14=e2kln14=lne2k2.63905733=2k2.639057332=k1.319528665=k To find tb: The formula to calculate the doubling timetb is: tb=ln 2k where k>0 is the growth constant. When P2=42000, from step 2, k=1.319528665 Therefore, tb=ln21.319528665=0.693147181.319528665=0.52529907 Hence the revised estimate for tbwould be tb=0.525 hours.