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disa [49]
3 years ago
6

Find the value of X in each triangle (pls I need help this is a test grade)

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
8 0

Answer:

x=125°

Step-by-step explanation:

100° + 25° =125°

180°- 125° = 55°

and since adjacent angles on a line equal 180°

180° - 55° = 125°

x= 125°

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74-34=20

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Vilka [71]

Answer:

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 75, \sigma = 4, n = 64, s = \frac{4}{\sqrt{64}} = 0.5

Find the probability that the average score of the 64 golfers exceeded 76.

This is 1 subtracted by the pvalue of Z when X = 64.

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

By the Central Limit Theorem

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

Z = \frac{76 - 75}{0.5}

Z = 2

Z = 2 has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the average score of the 64 golfers exceeded 76.

6 0
4 years ago
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