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notsponge [240]
2 years ago
9

If you get the answer just give me the full answer please. No matter the length.

Mathematics
1 answer:
sammy [17]2 years ago
5 0

Answer:

CD = 4

Step-by-step explanation:

I believe you know the geometric mean for altitudes?

CD = √AD * DB = √8 * 2 = √16 = 4

If you are not aware of this theorem, then use the similarity of triangles.

ADC ~ CDB

AD/DC = CD/DB (cross-multiply)

CD^2 = AD*DB,

CD = √AD * DB = √8 * 2 = √16 = 4

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Determine the standard variation of the data below. (1, 2, 3, 4, 5)
Ne4ueva [31]
Calculate for the mean/ average of the given numbers:
 
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The mean height of women in a country (ages 20-29) is 64 4 inches A random sample of 50 women in this age group is selected What
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Answer:

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 64.4 inches, standard deviation of 2.91

This means that \mu = 64.4, \sigma = 2.91

Sample of 50 women

This means that n = 50, s = \frac{2.91}{\sqrt{50}}

What is the probability that the mean height for the sample is greater than 65 inches?

This is 1 subtracted by the p-value of Z when X = 65. So

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

By the Central Limit Theorem

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

Z = \frac{65 - 64.4}{\frac{2.91}{\sqrt{50}}}

Z = 1.46

Z = 1.46 has a p-value of 0.9279

1 - 0.9279 = 0.0721

0.0721 = 7.21% probability that the mean height for the sample is greater than 65 inches.

8 0
2 years ago
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