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At the beginning of a study, a certain culture of bacteria has a population of 340. The population grows according to a continuous exponential growth model. After 11 days, there are 544 bacteria.
(a) Lett be the time (in days) since the beginning of the study, and let y be the number of bacteria at time i. x 5 ? Write a formula relating y tot. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
y = 10:
(b) How many bacteria are there 23 days after the beginning of the study? Do not round any intermediate computations, and round your answer to the nearest whole number.
bacteria

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Answer:

y = 340e^0.0427276t ; 908

Step-by-step explanation:

Given that :

t = time in days since study began

Number of bacteria after 11 days = 544

y = number of bacteria at time t

Exponential growth model:

y = ae^rt

Where a = initial amount ; r = Growth rate

y = 544 ; t = 11 ; a = 340

544 = 340*e^11t

544/340 = e^11t

1.6 = e^11t

Take the In of both sides

In(1.6) = 11t

0.4700036 = 11t

t = 0.4700036 / 11

t = 0.0427276

y = 340e^0.0427276t

Number of bacteria after 23 days :

Using the formula above :

t = 23

y = 340e^0.0427276(23)

y = 340 * 2.6717529

y = 908.39600

y = 908

The required formula is [tex]y = 340e^{0.0427}[/tex]

908 bacteria are there 23 days after the beginning of the study .

Given that,

A certain culture of bacteria has a population of 340.

The population grows according to a continuous exponential growth model. After 11 days, there are 544 bacteria.

We have to find,

Write a formula relating y tot. Use exact expressions to fill in the missing parts of the formula.

How many bacteria are there 23 days after the beginning of the study.

According to the question,

  • Let t be the time (in days) since the beginning of the study,

And let y be the number of bacteria at time,

t = time in days since study began

Number of bacteria after 11 days = 544

y = number of bacteria at time t

Then,

Exponential growth model:

[tex]y = a.e^{rt}[/tex]

Where, a = initial amount ; r = Growth rate

y = 544 ; t = 11 ; a = 340

Then,

[tex]544 = 340.e^{11t}\\\\\dfrac{544}{340} = e^{11t}\\\\1.6 = e^{11t}\\\\Taking \ log \ on \ both \ sides,\\\\ln(1.6) = 11t\\\\0.47 = 11t \\\\t = \dfrac{0.47}{11}\\\\t = 0.0427[/tex]

The required formula is [tex]y = 340e^{0.0427}[/tex]

 

  • There 23 days after the beginning of the study Number of bacteria after 23 days :

[tex]y = 340e^{0.0427}[/tex]

Where, t = 23

[tex]y = 340e^{0.0427276(23)}\\\\y = 340 \times 2.6717529\\\\y = 908.39600\\\\y = 908[/tex]

Therefore, 908 bacteria are there 23 days after the beginning of the study .

To know more about Probability click the link given below.

https://brainly.com/question/21726505

 

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