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AlekseyPX
3 years ago
11

PLZ Answer these questions/Help

Mathematics
1 answer:
arlik [135]3 years ago
5 0
1. They are all round.
2.The kids got each other sick. They were contagious.
3. 15: 15, 30, 45
20: 20, 40, 60, 80
25: 25, 50, 75, 100
The multiples of five will either end in 0 or 5.
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Julian has to read 4 articles for school. He has 8 nights to read them. He decides to read the same number of articles each nigh
daser333 [38]

<u><em>Answer:</em></u>

a. He will have to read half an article per night

b. Fraction of the reading assignment read each night = \frac{0.5}{4}=\frac{1}{8}

<u><em>Explanation:</em></u>

<u>Part a:</u>

<u>We are given that:</u>

number of articles to read = 4 articles

number of nights = 8 nights

To get the number of articles that he should read per night, we will simply divide the the number of articles by the number of nights

<u>Therefore:</u>

number of articles to read per night = \frac{4}{8}=\frac{1}{2} articles per night

<u>Part b:</u>

Now, we know that he will read half an article each night from a total of 4 articles

To get the fraction of the reading assignment read each night, we will divide the number of articles read each night by the total number of assignments

<u>Therefore:</u>

Fraction of the reading assignment read each night = \frac{0.5}{4}=\frac{1}{8}

Hope this helps :)

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3 years ago
Which of the following are true?
Alex73 [517]
D : because this make so much since duh . && watch yuu get it right !
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Help! I can't seem to answer this. (question in photo)
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The answer would be 48...

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Please help with math problem​
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Read 2 more answers
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
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