Question

The amount of bacteria in a petri dish increases at a rate proportional to the amount present. At time t=0, the amount of bacteria in the dish is 10 grams. At time t=2, the amount of bacteria in the dish is 30 grams. What is the amount of bacteria in the dish at time t=6 ?

Answer 1

Answer:

Step-by-step explanation:

Formula to be used is N=N0e^(kt)

N0 - amount of bacteria at time 0, which is 10g

t=2, 30g

30 = 10*e^(k*2)

3 = e^(2k)

2k = ln3

k = 0.5ln3

At t=6:

N = 10*e^(0.5ln3*6)

N = 10e^(3ln3)

N = 10e^ln27

N = 10*27 =270

So the amount of bacteria at t=6 is 270

Answer 2

The amount of bacteria in the dish at time t =6 is 270.

Given that,

The amount of bacteria in a petri dish increases at a rate proportional to the amount present.

At time t =0, the amount of bacteria in the dish is 10 grams.

At time t =2, the amount of bacteria in the dish is 30 grams.

We have to determine,

What is the amount of bacteria in the dish at time t =6?

According to the question,

The amount of bacteria in a petri dish increases at a rate proportional to the amount present.

The growth of bacteria follows an exponential distribution expressed as:

$\rm P_t = P_0r^{rt}$

Where r is the rate of growth and t is time.

At time t =0, the amount of bacteria in the dish is 10 grams.

Then,

$\rm P_t = P_0e^{rt}\\\\30 = 10\times e^{r\times2}\\\\\dfrac{30}{20} =e^{2r}\\\\e^{2r}=3\\\\Taking \ log \ on \ both \ sides \\\\2r = log3\\\\r = \dfrac{log3}{2}\\\\r = \dfrac{1.09}{2}\\\\r = 0.54$

Therefore,

The amount of bacteria in the dish at time t =6 is,

$\rm P_t = P_0 e^{rt}\\\\P_t = 10 \times e^{0.54\times6}\\\\P_t = 10 \timese^{3.29}\\\\P_t = 10 \times 27\\\\P_t = 270$

Hence, The amount of bacteria in the dish at time t =6 is 270.

To know more about Exponential Function click the link given below.

https://brainly.com/question/22810500