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charle [14.2K]
3 years ago
12

Barney has 14 dimes and quarters worth $2.15. write the system of equations

Mathematics
1 answer:
grigory [225]3 years ago
5 0

Answer: 14 dimes and 3 quarters

Step-by-step explanation:

14x10=140+25=165+25=190+25=215 aka $2.15 meaning that there is 14 dimes and 3 quarters

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mrs_skeptik [129]

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It's 3 feet

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Because you are trying to reach the window that is 15 feet, so you would place the 18 foot ladder 3 feet from the house so then it would be at an angle and then it'll reach to the window properly.

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Mr. Gubareff gave his class 15 minutes to read. Sabira read 6 ¼ pages in that time. At what rate, in pages per hour, did Sabira
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I think its 25 (proceed with caution)

Step-by-step explanation:

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The coordinate of the vertices for square CDEF translated 3 units right and 4 units up if the vertices are C(-3,1), D(1,5), E(5,
ELEN [110]
The statement is false because not all the corresponding point have been translated correctly from the original.

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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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Answer:

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