Question

a population of e. coli bacteria grows exponentially with time. you believe that the mean time between divisions is ( t_{b}=32 mathrm{~min} ), and that cell death occurs on average after ( t_{m}=200 mathrm{hr} ). the population starts with 7000 cells. complete parts (a) through (c). (a) use the equation ( n(t)=n_{0} e^{(b - m) t} ) to predict how many cells are present after 5 hours. the number of cells present after 5 hours is ( square ). (do not round until the final answer. then round to the nearest integer as needed.)

Explanation:

Step1: Calculate \(b\) and \(m\)

• For \(b\): The mean - time between divisions \(t_b = 32\) min. Since \(b=\frac{\ln2}{t_b}\) (in per - minute units), \(b=\frac{\ln2}{32}\approx\frac{0.6931}{32}\approx0.02166\) per minute.
• For \(m\): The mean - time until death \(t_m = 200\) hr \(=200\times60 = 12000\) min. Since \(m=\frac{1}{t_m}\) (in per - minute units), \(m=\frac{1}{12000}\approx8.333\times10^{-5}\) per minute.
• The time \(t = 5\) hr \(=5\times60=300\) min, and \(N_0 = 7000\).

Step2: Substitute into the formula \(N(t)=N_0e^{(b - m)t}\)

• First, calculate \(b - m\): \(b - m=0.02166-8.333\times 10^{-5}\approx0.021577\).
• Then, calculate \((b - m)t\): \((b - m)t=0.021577\times300 = 6.4731\).
• Now, \(N(t)=7000\times e^{6.4731}\).
• Using the property \(e^x\approx436.7\) when \(x = 6.4731\) (since \(e^6\approx403.43\) and \(e^{0.4731}\approx1.605\), and \(e^{6 + 0.4731}=e^6\times e^{0.4731}\)), \(N(t)=7000\times436.7\).

Answer:

\(3056900\)