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jeyben [28]
3 years ago
5

Please answer this question only if you know it!

Mathematics
1 answer:
stiv31 [10]3 years ago
8 0

Answer: its B

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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2 years ago
SNOG PLEASE HELP ME! (a+5)(b-3)
Bingel [31]

Answer:

ab-3a+5b-15

Step-by-step explanation:

Once again, FOIL is the way to go!

First, Outside, Inside, Last

ab-3a+5b-15

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3 years ago
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the small and the large intestine are part of the digestive system the small intestine longer than the large intestine and every
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2 years ago
A ski lift is designed with a total load limit of 20,000 pounds. It claims a capacity of 100 persons. An expert in ski lifts thi
Yanka [14]

Answer:

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

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 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.

For the sum of n values of a distribution, the mean is \mu \times n and the standard deviation is \sigma\sqrt{n}

An expert in ski lifts thinks that the weights of individuals using the lift have expected weight of 200 pounds and standard deviation of 30 pounds. 100 individuals.

This means that \mu = 200*100 = 20000, \sigma = 30\sqrt{100} = 300

If the expert is right, what is the probability that a random sample of 100 independent persons will cause an overload

Total load of more than 20,000 pounds, which is 1 subtracted by the pvalue of Z when X = 20000. So

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

Z = \frac{20000 - 20000}{300}

Z = 0

Z = 0 has a pvalue of 0.5

1 - 0.5 = 0.5

0.5 = 50% probability that a random sample of 100 independent persons will cause an overload

5 0
2 years ago
Class A has 15 pupils and class B has 29 pupils.
Vaselesa [24]

The mean score (rounded to 2 DP) in the Math's test for class B is 86.86.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Class A has 15 pupils and class B has 29 pupils. The mean score for class A is 55. Hence:

Mean score of A = (sum of all scores) / 15

55 =  (sum of all scores) / 15

sum of all scores in A = 825

The mean score for both classes is 76. Hence:

[(sum of all scores in A) + (sum of all scores in B)]/(total number of pupils) = mean score of both classes

[825 + (sum of all scores in B)]/(15 + 29) = 76

sum of all scores in B = 2519

Mean score of B = (2519) / 29 = 86.86

The mean score (rounded to 2 DP) in the Math's test for class B is 86.86.

Find out more on equation at: brainly.com/question/13763238

#SPJ1

6 0
1 year ago
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