Question

Colby and Jaquan are growing bacteria in an experiment in a laboratory. Colby starts with 50 bacteria in his culture and the number of bacteria doubles every 2 hours. Jaquan start with 80 of a different type of bacteria that doubles every 3 hours.
Let x equal number of days.
( * equals x in this case )

Colby’s experiment follows the model:
Ⓐ y = 50 • 2*
Ⓑ y = 50 • 2⁸*
Ⓒ y = 50 • 2¹²*

Jaquan’s experiment follows the model:
Ⓐ y = 80 • 2*
Ⓑ y = 80 • 2⁸*
Ⓒ y = 80 • 2¹²*

Answer 1

Colby's experiment follows model B.

Jaquan's experiment follows model B as well.

Because they are doubling, you will raise 2 to the power that represents how many times in a 24 hour period It will double.

24÷2 = 12.

24÷3 = 8.

Answer 2

Answer:

A) $y = 50 . 2^{12x}$

B) $y=80.2^{8x}$

Step-by-step explanation:

Since, the population of bacteria after x days who are growing with a constant factor is,

$P(x)=ab^{nx}$

Where a is the initial population of bacteria

b is the growth factor per period,

n is the number of periods in a day,

Given,

A) In Colby’s experiment,

The initial number of bacteria, a = 50,

Also, they are doubling in every 2 hours.

⇒ Number of period, n = 24/2 = 12,   ( 1 day = 24 hours )

Growth factor, b = 2

Thus, the population of bacteria after x days is,

$P(x)=50.2^{12x}$

B) In Jaquan’s experiment ,

The initial number of bacteria, a = 80,

Also, they are doubling in every 3 hours,

⇒ Number of period, n = 24/3 =  8,

Growth factor, b = 2

Thus, the population of bacteria after x days is,

$P(x)=80.2^{8x}$