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3 years ago

A circle has a circumstance of 22cm what’s the radius using pi

Mathematics

1 answer:

atroni [7]3 years ago

7 0

Answer:

3.50

Step-by-step explanation:

C = 2πr

C = πd

In this case we are using C = 2πr

C = 2πr

22 = 2 x 3.14 x r

22 = 6.28 x r

22/6.28 = 6.28r/6.28

3.503184713375796178343949044585987261 = r

But estimated is 3.50

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An employer with 500 employees gives out gives out 120 surveys but receives only 90 surveys completed by the employees. What are

leva [86]

Answer:

Population 120 given the survey, sample 90 who answered the survey.

Step-by-step explanation:

The entire amount of employees who even received the surveys at all was 120. Because the total number was 120, 120 is the population.

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3 years ago

Evaluate the following -5+4-3-6+9-3

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-4

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3 years ago

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

Vinvika [58]

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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3 years ago