Using the z-distribution, it is found that the 95% confidence interval for the difference is (-1.3, -0.7).
<h3>What are the mean and the standard error for each sample?</h3>
Considering the data given:


<h3>What is the mean and the standard error for the distribution of differences?</h3>
The mean is the subtraction of the means, hence:

The standard error is the square root of the sum of the variances of each sample, hence:

<h3>What is the confidence interval?</h3>
It is given by:

We have a 95% confidence interval, hence the critical value is of z = 1.96.
Then, the bounds of the interval are given as follows:
More can be learned about the z-distribution at brainly.com/question/25890103
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Answer:
This equation has 2 solutions.
1. 
2. 
Answer:
2n+7
Step-by-step explanation:
find the difference
and see the difference from the 2 Times table to the original sequence
and write it as an equation for this sequence
SOLUTION
The mean is 4min
standard deviation is 1min
the z score is

where

then we have

The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0
Hence, the percentage of the calls that lasted less than 3 min is 16%
Answer:
I need to know the numbers then Im able to help :3
Step-by-step explanation: