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Bad White [126]
3 years ago
13

The area of a rectangle is the product of its length and width. Both length and width are whole numbers. A rectangular poster ha

s an area of 88 square inches. The width of the poster is greater than 10 inches and is prime.
Mathematics
1 answer:
lesya [120]3 years ago
3 0

Answer: a rectangular poster has an area of 260 square inches.the width of the poster is greater than 10 inches and is a prime number

Step-by-step explanation: a rectangular poster has an area of 260 square inches.the width of the poster is greater than 10 inches and is a prime number.what is the width

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70% of the coins are quarters

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What is the problem of 200 and 400
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200+400=600 u just add 2 and 6 togather and u get 6 and then annd the two zeros on too it 
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Phone
Daniel [21]

Answer:

B : 850

Step-by-step explanation:

8 0
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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
ILL GIVE YOU BRAINLIST !! *have to get it right ! *
Marianna [84]

Answer:

-1/2

Step-by-step explanation:

Point L is at (-2,3)

Point M is at (2,1)

We can find the slope using

m =(y2-y1)/(x2-x1)

   = (1-3)/(2 - -2)

    =(1-3)/(2+2)

   =-2/4

  =-1/2

5 0
3 years ago
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