Question

The function f(x)=40,000(0.75)x represents the number of bacteria present in a petri dish after x hours. What is the average rate of change in the number of bacteria between Hour 2 and Hour 5?

Question 1 options:
A.−6102 bacteria/h
B.4336 bacteria/h
C.−4336 bacteria/h
D.6102 bacteria/h

Answer 1

The average rate of change in the number of bacteria between Hour 2 and Hour 5 is approximately -3,482 bacteria/hour.

Step-by-step explanation:

To find the average rate of change in the number of bacteria between Hour 2 and Hour 5, we need to calculate the difference in the number of bacteria at Hour 5 and Hour 2, and then divide it by the difference in time (3 hours).

At Hour 2, substitute x = 2 into the function f(x) = 40,000(0.75)^x:

f(2) = 40,000(0.75)^2 = 40,000(0.5625) = 22,500 bacteria.

At Hour 5, substitute x = 5 into the function:

f(5) = 40,000(0.75)^5 = 40,000(0.3164) = 12,656 bacteria.

Now we can calculate the average rate of change:

Average rate of change = (f(5) - f(2)) / (5 - 2) = (12,656 - 22,500) / 3 = -3,482 bacteria/hour. The average rate of change in the number of bacteria between Hour 2 and Hour 5 is approximately -3,482 bacteria/hour.

Answer 2

Important: f(x)=40,000(0.75)x should be written f(x)=40,000(0.75)^x.
f(b) - f(a)
The average rate of change (arc) is --------------
b - a

Here that comes to 40000 [0.75^5 - 0.75^2]
--------------------------------
5-2

or arc = -13007.81 / 3 = -4335.9 bacteria per year.