We have been given that the number of bacteria in the colony t minutes after the initial count modeled by the function $B(t)=9(3)^t$ . We are asked to find the average rate of change in the number of bacteria over the first 6 minutes of the experiment.

We will use average rate of change formula to solve our given problem.

$\text{Average rate of change}=\frac{f(b)-f(a)}{b-a}$

Upon substituting our given values, we will get:

$\text{Average rate of change}=\frac{b(6)-b(0)}{6-0}$

$\text{Average rate of change}=\frac{9(3)^6-9(3)^0}{6}$

$\text{Average rate of change}=\frac{9(729)-9(1)}{6}$

$\text{Average rate of change}=\frac{6561-9}{6}$

$\text{Average rate of change}=\frac{6552}{6}$

$\text{Average rate of change}=1092$

Therefore, the average rate of change in the number of bacteria is 1092 bacteria per minute.

8 0

2 years ago

What is the value of b?<br> 50 points in it for you plus a thanks, 5 star and brainliest.

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

4 0

3 years ago

A bag contains nine red balls, four green balls, and fifteen yellow balls. What is the probability of

dmitriy555 [2]

Answer:

Step-by-step explanation:

When you subtract the 9-1=8

6 0

3 years ago

danny is installing a fence around his rectangular yard. His yard is 20feet by 45feet. If the fence cost 25 per foot based on th

3241004551 [841]

Answer:

$225

Step-by-step explanation:

The area to be fenced in is 20 ft by 45 ft, or 900 ft^2. But "25 per foot based on the area" is illogical. Please ensure that you have copied this problem down accurately, and be sure to provide units of measurement: did you mean $25 per square foot, or per foot, or what?

The fencing cost is usually dependent on the unit cost and the length of the yard perimeter. The perimeter is P = 2L + 2W, which here is P = 2(45 ft) + 2(20 ft), or 130 ft. If the fencing is 25 cents per foot, $130 ft of fencing would come to $32.50.

6 0

2 years ago

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