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%.
You need to do the distributive property first so the answer is C
5x+4=114
5x= 110
X=22
114+ 3(22)-24=156
180-156=24
2y=24
Y=12
Answer:
Step-by-step explanation:
The area of a sector in terms of radians as opposed to degrees is

Filling in the formula accordingly gives us

The radians cancel out, the pi's cancel out, the 2 in the denominator cancels out (divides into 2-squared once), leaving us with
A = 1.75(2) so
A = 3.5 units squared
Place the part (amount going to the left field), over the total amount of homerun:
(a) 2/57 is your probability
(b) Divide 2/57
2/57 = 0.035, or 3.5%
YES, it is unusual for this player to hit a home run to left field, for he only has a 3.5% chance.
hope this helps