Using the z-distribution, it is found that:
- The 95% confidence interval is of -1.38 to 1.38.
- The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.
<h3>What is the z-distribution confidence interval?</h3>
The confidence interval is:
In which:
is the difference between the population means.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of 
, so the critical value is z = 1.96.
The estimate and the standard error are given by:
Hence the bounds of the interval are given by:
1.74 is outside the interval, hence:
- The 95% confidence interval is of -1.38 to 1.38.
- The value of the sample mean difference is of 1.74, which falls outside the 95% confidence interval.
More can be learned about the z-distribution at brainly.com/question/25890103
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Answer:
<u>The probability of winning three games in a row is 1/64 or 1.6%</u>
Step-by-step explanation:
Probability of winning a game = 1/4
Probability of winning two games in a row = 1/4 * 1/4 = 1/16
Probability of winning three games in a row = 1/4 * 1/4 * 1/4 = 1/64
Probability of winning three games in a row = (1/4)³ = 1/64
1/64 = 0.015625
1/64 = 1.6% (rounding to the next tenth)
For question 1:
Since each person is going to eat 1/4 of pizza and there are two of them, there are going to be 8 slices. There are 5 people eating, that's 5 slices gone, leaving 3.
It would be <u>D.) 3/4 pizza</u>
For question 2:
three people order the same meal. Now, if I'm correct, the meal total is AFTER the coupons were used. so that would bring total cost up to $30.75.
Divide that by 3 for all three friends:
30.75/3 = <u>D.) $10.25</u>
Answer:
$2.55
Step-by-step explanation:
2d + $14.00 – $7.00 = $12.10; $2.55
