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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:
- k=27
- y=27x
Step-by-step explanation:
3 tickets to attend an Off-Broadway show cost $81,
4 tickets cost $108; and
5 tickets cost $135.
Let the price of a ticket=k
Therefore:
- 3k=81
- 4k=108
- 5k=135
Solving any of the equation above gives: x=27
We therefore have:
Let y be the total cost of x tickets
- From the above, the constant of proportionality, k=27
An equation for the relationship is therefore:
y=27x (where x is the number of tickets and y is the total costs (in $)).