Question

the number of bacteria in a culture increases rapidly. the table below gives the number ( n(t) ) of bacteria at a few times ( t ) (in hours) after the moment when ( n = 1000 ).

time ( t ) (hours)	number of bacteria ( n(t) )
3.4	1510
6.8	2292
10.2	3856
13.6	5080

(a) find the average rate of change for the number of bacteria from 0 hours to 6.8 hours.

bacteria per hour

(b) find the average rate of change for the number of bacteria from 10.2 hours to 13.6 hours.

bacteria per hour

Explanation:

Part (a)

Step1: Recall the formula for average rate of change

The average rate of change of a function \( N(t) \) from \( t = a \) to \( t = b \) is given by \(\frac{N(b)-N(a)}{b - a}\).

Step2: Identify the values from the table

For \( t = 0 \) hours, \( N(0)=1000 \) and for \( t = 6.8 \) hours, \( N(6.8)=2292 \). Here, \( a = 0 \) and \( b = 6.8 \).

Step3: Substitute the values into the formula

Substitute \( N(6.8) = 2292 \), \( N(0)=1000 \), \( b = 6.8 \) and \( a = 0 \) into the formula \(\frac{N(b)-N(a)}{b - a}\).
We get \(\frac{2292 - 1000}{6.8-0}=\frac{1292}{6.8}\).

Step4: Calculate the value

\(\frac{1292}{6.8}=190\) (after performing the division: \(1292\div6.8 = 190\)).

Step1: Recall the formula for average rate of change

The average rate of change of a function \( N(t) \) from \( t = a \) to \( t = b \) is given by \(\frac{N(b)-N(a)}{b - a}\).

Step2: Identify the values from the table

For \( t = 10.2 \) hours, \( N(10.2)=3856 \) and for \( t = 13.6 \) hours, \( N(13.6)=5080 \). Here, \( a = 10.2 \) and \( b = 13.6 \).

Step3: Substitute the values into the formula

Substitute \( N(13.6) = 5080 \), \( N(10.2)=3856 \), \( b = 13.6 \) and \( a = 10.2 \) into the formula \(\frac{N(b)-N(a)}{b - a}\).
We get \(\frac{5080 - 3856}{13.6 - 10.2}=\frac{1224}{3.4}\).

Step4: Calculate the value

\(\frac{1224}{3.4} = 360\) (after performing the division: \(1224\div3.4=360\)).

Answer:

\(190\)

Part (b)