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%.
Answer:
3, 5.38, 66, V21
Step-by-step explanation:
I assumed that V was the roman numeral V, which is equivalent to the number 5. I then multiplied 5 by 21 which is 105. If this is not the case and they simply wanted us to replace the v with a 5, then the answer would still be correct because 521 is greater than all the numbers in this set. Hope this helps.
Answer:
a=3.915 and b=1.106
Step-by-step explanation:
This is the answer on edju
Answer:
? = 13
Step-by-step explanation:
Using Pythagoras' identity in the right triangle
?² = 5² + 12² = 25 + 144 = 169 ( take the square root of both sides )
? = = 13
Firstly, seeing that the sequence is based on odd numbers I would say that this is the formula (which will allow you to know whatever term):
nth term = 2n-1
Hope this helps!