Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
2 answers:
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%.
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Answer:
Final graph is attached.
Solution is x=2, y=3
Step-by-step explanation:
As we have to solve by graphing to let's graph y=2x-1 first.
we are free to plug any number for x say x=0 and x=2
when x=0 then y=2x-1 changes to
y=2(0)-1=0-1=-1
Hence point is (0,-1)
Similarly
when x=2 then y=2x-1 changes to
y=2(2)-1=4-1=3
Hence point is (2,3).
Now we graph both points (0,-1) and (2,3) then join them by a straight line.
We repeat exactly same process for other line y=-x+5 and get points (0,5) and (2,3)
Now graph both to see the intersection point of both lines
We see that both lines intersect at point (2,3)
<u>Hence final answer is (2,3) or we can also write x=2, y=3.</u>
