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Ilia_Sergeevich [38]
3 years ago
7

PLEASE HELP ASAP!

Mathematics
1 answer:
Andrej [43]3 years ago
3 0
B, C, E are all negatives
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E=mc²
E/m = c²
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MATH HELP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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So the first one, 1.2 >11/5. and the second one 1/3> 1/2
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The length of a garden can be expressed as 2X +6 feet and the width is X +7 feet what is the perimeter of the garden what is the
Virty [35]

Answer:

6x + 26 feet

Step-by-step explanation:

Assuming the garden is a rectangular shape, we have four sides, where two sides are 2x + 6 feet and two sides are x + 7 feet. To find the perimeter, we add up the lengths of all four sides.

2x + 6 + 2x + 6 + x + 7 + x + 7

or

2(2x + 6) + 2(x + 7)

6x + 26

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3 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 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 = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
4 years ago
Question 11: please help. I will give brainliest to correct answer.
harkovskaia [24]

Answer:

3

Step-by-step explanation:

8 0
3 years ago
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