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Ksenya-84 [330]
3 years ago
6

The angle measures and/or side lengths of a triangle are described. Determine whether the conditions given

Mathematics
1 answer:
allsm [11]3 years ago
3 0

Answer:

no such tryangle

Step-by-step explanation:

the sum of any 2 sides must be grater then the 3ed

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For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyz
Fed [463]

Answer:

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

Step-by-step explanation:

We use the binomial approximation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 92, p = 0.81

So

\mu = E(X) = np = 92*0.81 = 74.52

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{92*0.81*0.19} = 3.76

Find the probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

This is 1 subtracted by the pvalue of Z when X = 0.79*92 = 72.68. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{72.68 - 74.52}{3.76}

Z = -0.49

Z = -0.49 has a pvalue of 0.3121

1 - 0.3121 = 0.6879

68.79% probability that more than 79% of the sample diamonds fail to qualify as gemstone grade

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What 75% of two hundred
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To find 75% of 200, we can multiply two fractions.

75/100 * 200/1

Now, just multiply the numerators and the denominators separately.

75 * 200 = 15000

100 * 1 = 100

Now, divide.

15000/100 = 150

<h3>75% of 200 is 150.</h3>
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