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sertanlavr [38]
2 years ago
11

What will be the distance from point c' in the rotated figure?

Mathematics
2 answers:
Nezavi [6.7K]2 years ago
5 0
From point p to point c, it is 9 units.
kaheart [24]2 years ago
4 0
The answer would be 3

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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
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Answer:

40.1% probability that he will miss at least one of them

Step-by-step explanation:

For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that p = 0.95

10 targets

This means that n = 10

What is the probability that he will miss at least one of them?

Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

P(X = 10) + P(X < 10) = 1

We want P(X < 10). So

P(X < 10) = 1 - P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987

P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401

40.1% probability that he will miss at least one of them

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