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A sample of bacteria is decaying according to a half life model. if the sample begins...

A sample of bacteria is decaying according to a half life model. if the sample begins with 900 bacteria and after 19 minutes there are 360 bacteria, after how many minutes will there be 10 bacteria remaining? Round answer to nearest whole number
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Answer:

Data given:

A sample of bacteria is decaying according to a half life model.

The sample begins with 900 bacteria and after 19 minutes there are 360 bacteria.

Time taken for the count of bacteria to reduce to 10 bacteria = ?

As we know, the formula for the exponential decay, according to the half-life model, is given as -

N(t) = N_{0} \ e^{-\lambda t}

where   NO - is the initial quantity

  Nt - is the remaining quantity after time t

  \lambda - decay constant

  t - elapsed time

Using the above information, we can find the time taken for the count of bacteria to reduce to 10 bacteria as -

1. The sample begins with 900 bacteria and after 19 minutes there are 360 bacteria.

N_{0} = 900

At \ \ t = 19 \ minutes, \ \ \ N(19) = 360

N(t) = N_{0} \ e^{-\lambda t}

\therefore \ N(19) = 360 = 900 \ e^{-\lambda(19)}

360 = 900 \ e^{- 19\lambda}

\frac{360}{900}= e^{- 19\lambda}

0.4= e^{- 19\lambda}

\ln 0.4= - 19\lambda(From \ the \ natural \ logarithm \ property: \ \ \ a = e^{b} \ \ \Rightarrow \ \ \ln a = b)

\frac{1}{(-19)} \ln 0.4= \lambda

\lambda = 0.04823

\therefore \ \ \ N(t) = 900 \ e^{-0.04823t}

2. Now, we can find the time taken for the count of bacteria to reduce to 10 bacteria as -

N(t) = 10, \ \ \ t = ?

N(t) = 900 \ e^{-0.04823t}

10 = 900 \ e^{-0.04823t}

\frac{10}{900} = e^{-0.04823t}

\frac{1}{90} = e^{-0.04823t}

\ln \frac{1}{90}= - 0.04823t(From \ the \ natural \ logarithm \ property: \ \ \ a = e^{b} \ \ \Rightarrow \ \ \ln a = b)

\frac{1}{(- 0.04823)} \ln \frac{1}{90}= t

t = 93.299

Thus, the time taken for the count of bacteria to reduce to 10 bacteria is approximately 93 minutes (rounded to nearest whole number) .

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