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at the beginning of a story, a certain culture of bacteria has a population of 80. the population grows according to a continuous exponential growth model. after 14 days, there are 216 bacteria.

At The Beginning Of A Story A Certain Culture Of Bacteria Has A Population Of 80 The Population Grows According To A Continuous Exponential Growth Model After 1 class=

Sagot :

Given:

The population of the bacteria at the beginning = 80

the population grows according to a continuous exponential growth model.

after 14 days, there are 216 bacteria. ​

y = the number of bacteria after time t

So, the general relation between y and t will be:

[tex]y=a\cdot e^{bt}[/tex]

We need to find the values of a and b

At t = 0 y = 80

So,

[tex]\begin{gathered} 80=a\cdot e^0 \\ a=80 \end{gathered}[/tex]

When t = 14 , y = 216

So,

[tex]\begin{gathered} 216=80e^{14b} \\ \frac{216}{80}=e^{14b} \\ \text{2}.7=e^{14b} \\ \ln 2.7=14b \\ \text{0}.99325=14b \\ b=\frac{0.99325}{14}=0.071 \end{gathered}[/tex]

so, the function will be:

[tex]y=80\cdot e^{0.071t}[/tex]

Part b: we need to find the number of bacteria after 23 days

So, substitute with t = 23

so,

[tex]y=80\cdot e^{0.071\cdot23}=409[/tex]

so, after 23 days the number of bacteria = 409

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