The doubling period of a bacterial population is 15 minutes. At time t= 80 minutes, the bacterial population was 90000​

unknown please expainsiate

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Step-by-step explanation:

The doubling period of a bacterial population is 15 minutes.

At time t = 90 minutes, the bacterial population was 50000.

Round your answers to at least 1 decimal place.

:

We can use the formula:

A = Ao*2^(t/d); where:

A = amt after t time

Ao = initial amt (t=0)

t = time period in question

d = doubling time of substance

In our problem

d = 15 min

t = 90 min

A = 50000

What was the initial population at time t = 0

Ao * 2^(90/15) = 50000

Ao * 2^6 = 50000

We know 2^6 = 64

64(Ao) = 50000

Ao = 50000/64

Ao = 781.25 is the initial population

:

Find the size of the bacterial population after 4 hours

Change 4 hr to 240 min

A = 781.25 * 2^(240/15

A = 781.25 * 2^16

A= 781.25 * 65536

A = 51,199,218.75 after 4 hrs

it’s supposed to be -4+45=41

step-by-step explanation:

the correct answer is b. negative exponential growth.

step-by-step explanation:

you can tell it's an exponential function because it decreases rapidly then levels out and never touches the x-axis.

it is negative because the y-axis decreases as the x-axis increases.

therefore it is a negative exponential growth graph.

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