Researchers recorded that a certain bacteria population declined from 190,000 to 100 in 48 hours. At this rate of decay, how many bacteria will there be in 15 hours? Round to the nearest whole number.

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raphealnwobi raphealnwobi · Answer 1 · 2020-05-01T05:07:52+02:00

Answer:

17949 bacteria

Step-by-step explanation:

The exponential growth/decay formula is given as:

[tex]N=N_0e^{kt}[/tex]

Where N is the final population, N₀ is the initial population, t is the time and k is the growth or decay rate.

At first, we need to calculate k, given that:

N = 100, N₀ = 190000, t = 48 hours.

[tex]N=N_0e^{kt}\\Substituting:\\100 = 190000e^{48k}\\e^{48k}=\frac{100}{190000} \\e^{48k}=0.0005263\\48k=ln(0.0005263)\\48k=-7.55\\k=-0.1573[/tex]

The decay of bacteria in 15 hours would be:

[tex]N=N_0e^{kt}\\Substituting:\\N = 190000e^{15*-0.1573}\\N=17949[/tex]

ibiscollingwood ibiscollingwood · Answer 2 · 2021-11-28T04:24:15+01:00

In 15 hours, there are 17949 bacteria .

Exponential decay describes the process of reducing an amount by a consistent percentage rate over a period of time.

Exponential decay formula is,

[tex]A=A_{0}e^{kt}[/tex]

Where A is the final population, [tex]A_{0}[/tex] is the initial population and t is time taken to decline the population.

According to question, [tex]A=100,A_{0}=190000,t=48 hours[/tex]

Substituting above values in exponential decay formula.

[tex]100=190000e^{48k} \\\\e^{48k}=\frac{100}{190000}=0.0005263[/tex]

Taking natural log on both side.

We get, [tex]48k=ln(0.0005263)\\\\48k=-7.55\\\\k=-\frac{7.55}{48} =-0.1573[/tex]

In 15 hours, the decay of bacterial will be,

[tex]A=190000e^{-0.1573*15} \\\\A=17949[/tex]

Therefore, In 15 hours, there are 17949 bacteria .

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