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mr Goodwill [35]
3 years ago
7

Find the value of X. Write your answer in simplest form.What does X equal? ​

Mathematics
1 answer:
drek231 [11]3 years ago
7 0

<u>Given</u>:

Given that the figure of similar triangles.

The altitude of the triangle is x.

The length of the left part is 30.

The length of the right part is 15.

We need to determine the value of x.

<u>Value of x:</u>

The value of x can be determined using the geometric mean theorem.

Thus, we have;

\frac{\text { left }}{\text { altitude }}=\frac{\text { altitude }}{\text { right }}

Substituting the values, we have;

\frac{30}{x}=\frac{x}{15}

Cross multiplying, we get;

30 \times 15= x^2

     450=x^2

Taking square root on both sides, we have;

15 \sqrt{2}=x

Thus, the value of x is 15√2

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The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
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0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
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