As your testing H0:p=0.79, then you have a two-tailed test.
The p-value at two-tailed test is given by:
Then, you need to check into the standard normal cumulative table to find at which -z you do have a probability of 0.395. Thus:
As you can see in the picture at z=-0.26 you have a P(Z<=-z)=0.3974
And at z=-0.27 you have a P(Z<=-z)=0.3936.
The average of these values is:
Then, you have a probability of 0.395 at a z of:
Then -z=-0.27, thus z=0.27
Answer:
1092
Step-by-step explanation:
We have been given that the number of bacteria in the colony t minutes after the initial count modeled by the function . 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.
Upon substituting our given values, we will get:
Therefore, the average rate of change in the number of bacteria is 1092 bacteria per minute.
Answer:
Look at the image below
