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:
idk
Step-by-step explanation:
X must be increased by 6 to get its value to be 0. Then the corresponding change in y will be (-2/3)*6 = -4. That value added to -5 makes it -9.
The y-intercept will be -9.
Answer:
b.) Subtracting 5 then dividing both sides by -2
Step-by-step explanation:
Given expression:
⇒ -2x + 5 = 9
To solve this equation, first we need to remove the 5 on the left side. Then divide both sides by -2.
When done so:
⇒ -2x = 9 - 5
⇒ -2x = 4
⇒ -2x/-2 = 4/-2
⇒ x = -2
Answer:
x=Sin-1(0.470584)
x=28.1°
Step-by-step explanation:
Sin90°/17 =Sin x°/8
0.058823 = Sin x°/8
8(0.058823)=Sin x°
x°=Sin-1(0.470584)
x=28.1°
