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Ugo [173]
3 years ago
9

What is the least common multiple of the numbers 5,25, and 15

Mathematics
2 answers:
liubo4ka [24]3 years ago
5 0

Answer:

5

Step-by-step explanation:

Zielflug [23.3K]3 years ago
3 0

Hold on I am doing it right now

Your answer is 75 hope this helps you

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The overhead reach distances of adult females are normally distributed with a mean of 197.5 cm197.5 cm and a standard deviation
fiasKO [112]

Answer:

a) 5.37% probability that an individual distance is greater than 210.9 cm

b) 75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c) Because the underlying distribution is normal. We only have to verify the sample size if the underlying population is not normal.

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 question, we have that:

\mu = 197.5, \sigma = 8.3

a. Find the probability that an individual distance is greater than 210.9 cm

This is 1 subtracted by the pvalue of Z when X = 210.9. So

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

Z = \frac{210.9 - 197.5}{8.3}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463.

1 - 0.9463 = 0.0537

5.37% probability that an individual distance is greater than 210.9 cm.

b. Find the probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

Now n = 15, s = \frac{8.3}{\sqrt{15}} = 2.14

This probability is 1 subtracted by the pvalue of Z when X = 196. Then

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

By the Central Limit Theorem

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

Z = \frac{196 - 197.5}{2.14}

Z = -0.7

Z = -0.7 has a pvalue of 0.2420.

1 - 0.2420 = 0.7580

75.80% probability that the mean for 15 randomly selected distances is greater than 196.00 cm.

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The underlying distribution(overhead reach distances of adult females) is normal, which means that the sample size requirement(being at least 30) does not apply.

5 0
3 years ago
A(n) ____________ chart compares data from three columns and/or rows in a three-dimensional manner
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A(n) is a special type of hyphen that prevents two words separated by a hyphen from splitting at the end of a line.
7 0
3 years ago
if your teacher tells you to do questions 28 through 41 in your math book for homework, how many questions is that
katrin2010 [14]

Its 14.
30-40 is 11, 28-29 is 2, and 41 is 1.

5 0
3 years ago
If a point P(–1, –1) is reflected across the line y = –2, what are the coordinates of its reflection image?
Ksivusya [100]
Since the line of refrection is a horizontal line, the x-coordinate of the image will be the same as the x-coordinate of the point P(-1, -1). P(-1, -1) is one unit above the line y = -2, hence the image will be one unit below the line y = -2. Therefore the coordinate of the image is (-1, -3)
8 0
3 years ago
Write equation of the line containing (-3,4) and (-1,-2)
astra-53 [7]

Answer:

y=-3x+1

Step-by-step explanation:

8 0
2 years ago
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