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Leviafan [203]
2 years ago
5

A sample of bacteria is being eradicated by an experimental procedure. The population is following a pattern of exponential deca

y and approaching a population of 0. If the sample begins with 500 bacteria and after 11 minutes there are 200 bacteria, after how many minutes will there be 50 bacteria remaining? Round your answer to the nearest whole number, and do not include units. answer
Mathematics
1 answer:
77julia77 [94]2 years ago
3 0

Answer:

There will be 50 bacteria remaining after 28 minutes.

Step-by-step explanation:

The exponential decay equation is

N=N_0e^{-rt}

N= Number of bacteria after t minutes.

N_0 = Initial number of bacteria when t=0.

r= Rate of decay per minute

t= time is in minute.

The sample begins with 500 bacteria and after 11 minutes there are 200 bacteria.

N=200

N_0 = 500

t=11 minutes

r=?

N=N_0e^{-rt}

\therefore 200=500e^{-11r}

\Rightarrow e^{-11r}=\frac{200}{500}

Taking ln both sides

\Rightarrow ln| e^{-11r}|=ln|\frac{2}{5}|

\Rightarrow {-11r}=ln|\frac{2}{5}|

\Rightarrow r}=\frac{ln|\frac{2}{5}|}{-11}

To find the time when there will be 50 bacteria remaining, we plug N=50, N_0= 500 and  r}=\frac{ln|\frac{2}{5}|}{-11} in exponential decay equation.

50=500e^{-\frac{ln|\frac25|}{-11}.t}

\Rightarrow \frac{50}{500}=e^{\frac{ln|\frac25|}{11}.t}

Taking ln both sides

\Rightarrow ln|\frac{50}{500}|=ln|e^{\frac{ln|\frac25|}{11}.t}|

\Rightarrow ln|\frac{1}{10}|={\frac{ln|\frac25|}{11}.t}

\Rightarrow t= \frac{ln|\frac{1}{10}|}{\frac{ln|\frac25|}{11}.}

\Rightarrow t= \frac{11\times ln|\frac{1}{10}|}{{ln|\frac25|}}

\Rightarrow t\approx 28 minutes

There will be 50 bacteria remaining after 28 minutes.

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Solution to the problem

For this case we have the following info related to the time to prepare a return

\mu =90 , \sigma =14

And we select a sample size =49>30 and we are interested in determine the standard deviation for the sample mean. From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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And the best answer would be

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