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A population of bacteria is growing according to the exponential model P = Cekt, where P is the number of colonies and t is measured in hours. If there are 800 colonies present after 4 hours, and the population's growth rate k = 0.65, how many colonies were present initially? [Round answer to the nearest hundredth.]

Respuesta :

Plugging our values into the equation, we have

[tex]800=Ce^{0.65(4)} \\ \\800=Ce^{2.6} \\\text{Dividing both sides by }e^{2.6} \\ \\\frac{800}{e^{2.6}}=C \\ \\59.42=C[/tex]The correct answer is 59.42.

Solving the exponential equation, it is found that initially, there were 59.42 colonies present.

The population of bacteria after t hours is given by:

[tex]P(t) = Ce^{kt}[/tex]

In which:

  • C is the initial population.
  • k is the growth rate, as a decimal.

In this problem:

  • Growth rate of [tex]k = 0.65[/tex].
  • After 4 hours, there were 800 colonies, thus when [tex]t = 4, P(t) = 800[/tex]. This is used to find C.

[tex]P(t) = Ce^{kt}[/tex]

[tex]800 = Ce^{0.65(4)}[/tex]

[tex]C = \frac{800}{e^{0.65(4)}}[/tex]

[tex]C = 59.42[/tex]

Initially, there were 59.42 colonies present.

A similar problem is given at https://brainly.com/question/14773454

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