Answer:
The 90% confidence interval is (493.1903, 550.2097), and the critical value to construct the confidence interval is 2.0150
Step-by-step explanation:
Let X be the random variable that represents a measurement of helium gas detected in the waste disposal facility. We have observed n = 6 values, 
= 521.7 and S = 31.6368. We will use
as the pivotal quantity. T has a 
distribution with 5 degrees of freedom. Then, as we want a 90% confidence interval for the mean level of helium gas present in the facility, we should find the 5th quantile of the t distribution with 5 degrees of freedom, i.e., 
, this value is -2.0150. Therefore the 90% confidence interval is given by
, i.e.,
(493.1903, 550.2097)
To find the 5th quantile of the t distribution with 5 degrees of freedom, you can use a table from a book or the next instruction in the R statistical programming language
qt(0.05, df = 5)
Answer:
I think it's c
Step-by-step explanation:
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Answer:
x = -1
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Step-by-step explanation:
Solve for x in the equation: 3(x-4)+6=5(x-1)+1
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Answer:
See below in bold.
Step-by-step explanation:
The y -intercept is when x = 0 so
it is 4 - 0 = 4
That is (0, 4)
For the x intercept g(x) = 0 so
4 - x = 0
x = 4.
x -intercept is at (4,0).
