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%.
The class 4S because you divide each number by its mean mark and standard division and 4S is the highest.
Answer:
Expected value would be $ 0.896
Step-by-step explanation:
Given,
The price of the lottery ticket = $44800000,
Also, the probability of winning the grand prize = .000000020,
Thus, the expected value of the lottery ticket = value of the lottery ticket × probability of getting the lottery ticket
= 44800000 × .000000020
= $0.896
Note : value of lottery ticket = prize amount - cost of each ticket,
Here the cost price of a ticket is not given,
That's why we did not consider it.
Answer:
826
Step-by-step explanation:
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Answer:
Step-by-step explanation:
First you need to convert it to a linear expression.
3x - y = 4
+3x +3x
-1y = 3x + 4
--- ----- -----
-1 -1 -1
y = -3x -4
6x - 2y = 7
-6x -6x
-2y = -6x + 7
---- ----- -----
-2 -2 -2
y = 3x - 3.5
-3x - 4 = 3x - 3.5
+3x +3x
-4 = 6x - 3.5
+3.5 +3.5
-0.5 = 6x
----- -----
6 6
-0.5
------ = x
6
