Using the z-distribution, it is found that the 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
We are given the <u>standard deviation for the population</u>, hence, the z-distribution is used. The parameters for the interval is:
- Sample mean of
- Population standard deviation of
- Sample size of
.
The margin of error is:
In which z is the critical value.
We have to find the critical value, which is z with a p-value of 
, in which 
is the confidence level.
In this problem, 
, thus, z with a p-value of 
, which means that it is z = 1.96.
Then:
The confidence interval is the <u>sample mean plus/minutes the margin of error</u>, hence:
The 95% confidence interval for the mean lifetime of this type of drill, in holes drilled, is (10.4, 13.94).
A similar problem is given at brainly.com/question/22596713
Answer:
B
Step-by-step explanation:
|-82| < |-83|
If the absolute values are replaced by an integer, 82 < 83
Answer:
D. 24.75
Explanation:
literally just subtract 65.25 by 40.5 ??
Answer:
4 liters of 60% solution; 2 liters of 30% solution
Step-by-step explanation:
I like to use a simple, but effective, tool for most mixture problems. It is a kind of "X" diagram as in the attachment.
The ratios of solution concentrations are 3:6:5, so I've used those numbers in the diagram. The constituent solutions are on the left; the desired mixture is in the middle, and the numbers on the other legs of the X are the differences along the diagonals: 6 - 5 = 1; 5 - 3 = 2. This tells you the ratio of 60% solution to 30% solution is 2 : 1.
These ratio units (2, 1) add to 3. We want 6 liters of mixture, so we need to multiply these ratio units by 2 liters to get the amounts of constituents needed. The result is 4 liters of 60% solution and 2 liters of 30% solution.
_____
If you're writing equations, it often works well to let the variable represent the quantity of the greatest contributor—the 60% solution. Let the volume of that (in liters) be represented by v. Then the total volume of iodine in the mixture is ...
... 0.60·v + 0.30·(6 -v) = 0.50·6
... 0.30v = 0.20·6 . . . . subtract 0.30·6, collect terms
... v = 6·(0.20/0.30) = 4 . . . . divide by the coefficient of v
4 liters of 60% solution are needed. The other 2 liters are 30% solution.
The correct answer is x=-2
