Answer:
I think it is 3.
Step-by-step explanation:
Sometimes because it might not always be true
Answer: 578.7 m 47.33 kg. 789.5 cm
Step-by-step explanation: every whole number is a sig fig (1-9). Count the first four whole numbers and if the last whole number can be rounded up using the rounding rule (5 or more bumps the previous # up), then round
Answer:
The number of students we expect to have an interval that does not contain the true mean value is,
.
Step-by-step explanation:
A [100(1 - α)%] confidence interval for true parameter implies that if 100 confidence intervals are created then [100(1 - α)] of these 100 confidence intervals will consist the true population parameter value.
Here α is the significance level. It is defined as the probability rejecting the claim that the true parameter value is not included in the 100(1 - α)% confidence interval.
It is provided that 255 students create the same confidence interval, correctly.
Then the number of students we expect to have an interval that does not contain the true mean value is, ![255\times [\alpha\%]](https://tex.z-dn.net/?f=255%5Ctimes%20%5B%5Calpha%5C%25%5D)
For instance, if the students are creating a 95% confidence interval for mean then the number of students we expect to have an interval that does not contain the true mean will be:
The significance level is:

Number of students we expect to have an interval that does not contain the true mean will be: ![255\times [\alpha\%]=255\times 0.05=12.75\approx13](https://tex.z-dn.net/?f=255%5Ctimes%20%5B%5Calpha%5C%25%5D%3D255%5Ctimes%200.05%3D12.75%5Capprox13)
Thus, 13 of the 255 confidence intervals will not consist the true mean value.
Y=6/4x-4.5 would be an answer