Answer:
Yes
Step-by-step explanation:
Here is the graph!
Using the z-distribution, it is found that the 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
We are given the <em>standard deviation</em> for the population, which is why the <em>z-distribution </em>is used to solve this question.
The information given is:
- Sample mean of
.
- Population standard deviation of
.
- Sample size of
.
The confidence interval is:
The critical value, using a z-distribution calculator, for a <u>95% confidence interval</u> is z = 1.645, hence:


The 90% confidence interval for mean calories in a 30-gram serving of all chocolate chip cookies is (143, 149).
A similar problem is given at brainly.com/question/16807970
3 7/12 - 1 2/3 = <span>1.916666... hours
So dinner was 2 hours rounded up.</span>
Use the list method to write "The odd integers between 2 and 15". {2, 3, 5, 7, 9, 11, 13, 15} {3, 5, 7, 9, 11, 13} {3, 5, 7, 9,
Sladkaya [172]
The answer would be {3, 5, 7, 9, 11, 13, 15}.