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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You are given the masses of four planets.
Add them together, and then divide your sum by 4.
That will be the "average planet mass" for these 4 planets.
Get on algebra calculator (mathpapa) and you should get the answer
Answer:
<em>Answer is </em><em>given below with explanations</em><em>. </em>
Step-by-step explanation:

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<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
Answer:
28 hours
Step-by-step explanation:
let the time period at which the temperature be equal in both Brownsville and Mesquite Texas be 'x'.
Given,
In Brownsville Texas, the temperature is expected to drop 1.5° each hour and In Mesquite Texas, the temperature is expected to drop 2° each hour.
Now, according to the question,
after 'x' hours,
⇒ 80° - 1.5°(x) = 94° - 2°(x)
⇒ 0.5°(x) = 14°
⇒ x = 28
Hence, Till 28 more hours the temperature in Mesquite Texas be greater than that in Brownsville.