The test statistic for the hypothesis would be 1.413.
Given that the participants in its new diet program lose, on average, more than 13 pounds and the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds.
The objective is to text the advertisement's claim that participants in new diet program lose weight, on average, more than 13 pounds.
Hypothesis:
Null hypothesis:H₀:μ=13
Alternative hypothesis:Hₐ:μ>13
Here, μ be the mean weight loss of all participants.
n=30,
Degree of freedom n-1=30-1=29
=13.8 and s=3.1
To test the null hypothesis H₀, the value of test static would be calculated as follows:

Hence, the value of the test static for the hypothesis with the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds is 1.413.
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Step-by-step explanation:
9.1
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Answer:
80
Step-by-step explanation:
If they drove a total of 5,300 miles in 2 days, and on the second day they drove 10 more miles then the 1st day.
5,300 ÷ 2 = 2,650
they drove 10 miles less on the first day
2,650 - 10 = 2,640
They drove 2,640 miles on the first day.
Hope this helps. :)
Rewrite the equations of the given boundary lines:
<em>y</em> = -<em>x</em> + 1 ==> <em>x</em> + <em>y</em> = 1
<em>y</em> = -<em>x</em> + 4 ==> <em>x</em> + <em>y</em> = 4
<em>y</em> = 2<em>x</em> + 2 ==> -2<em>x</em> + <em>y</em> = 2
<em>y</em> = 2<em>x</em> + 5 ==> -2<em>x</em> + <em>y</em> = 5
This tells us the parallelogram in the <em>x</em>-<em>y</em> plane corresponds to the rectangle in the <em>u</em>-<em>v</em> plane with 1 ≤ <em>u</em> ≤ 4 and 2 ≤ <em>v</em> ≤ 5.
Compute the Jacobian determinant for this change of coordinates:

Rewrite the integrand:

The integral is then
