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Archy [21]
3 years ago
9

SOMEBODY PLEASE HELP ME WITH THIS!!!

Mathematics
2 answers:
mamaluj [8]3 years ago
5 0

More than means addition. Same as means equal to.


The equation would be: g + 312 = 200

inn [45]3 years ago
3 0

g+312 = 200 is how the equation is written


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garik1379 [7]

Answer:

the answer for the 2nd question is 1/4

Step-by-step explanation:

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2 years ago
The area of a circle is 706.9 square inches. Find the radius
antoniya [11.8K]

Answer:

15 inches

Step-by-step explanation:

The area of a circle is pi r^2. We basically have to work backwards. I divided it fist by 3.14 which is the value of pi. I got 225.127, but I rounded it to 225. I found the square root of this which is 15. If we check the work. 15^2 is 225, and 225 times pi is 706.9.  

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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.

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9:{1,3 3,9}
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Twelve eggs cost $2.04 how much would 18 eggs cost?
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Answer:

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Step-by-step explanation:

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Read 2 more answers
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