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:
11 Is One Of Them
Step-by-step explanation:
198 Divided By 18 Equals 11.
It would be: 5/17 * 3/8 = 5*3 / 17*8 = 15/136
In short, Your Answer would be 15/136
Hope this helps!
Answer:
see explanation
Step-by-step explanation:
The 2 equations are
x + 2y = 8 → (1)
2x + 4y = 20 → (2)
4 hours 15 minutes is the answer to the question
