sorry, didnt mean to reply to this
The question is incomplete. Here is the complete question:
Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit 10 such targets in a row.
Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?
Answer:
40.13%
Step-by-step explanation:
Let 'A' be the event of not missing a target in 10 attempts.
Therefore, the complement of event 'A' is 
Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.
Now, 
We know that the sum of probability of an event and its complement is 1.
So, 
Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.
I saw the image that should have been posted with this word problem.
The rolls of wrapping paper is cylindrical in shape.
Roll A : length = 30 inches ; diameter = 3 inches
Roll B : length = 24 inches ; diameter = unknown
Lateral surface area = 2 π r h
Lateral Surface area of Roll A = Lateral Surface area of Roll B
2 * 3.14 * 1.5in * 30in = 2 * 3.14 * r * 24 in
282.60 in² = 150.72 in * r
282.60 in² / 150.72 in = r
1.875 in = r
diameter is 2 times the radius = 1.875 in * 2 = 3.75 inches
The diameter of Roll B is 3.75 inches.
The median (half-above and half-below) could not have changed, because "corrected value was still lower than any other salary" does not change the position of that observation (still least) in the sorted list of salaries.