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In-s [12.5K]
3 years ago
9

Write a expression that represents the difference of y and 32 , plus the product of y and 4  

Mathematics
2 answers:
Sveta_85 [38]3 years ago
6 0
Lets break down the problem into bits and pieces then put them together. So when we say difference, we usually mean subtract so when we say the difference of y and 32, it means y - 32. Next, let's look at when we say the product of y and 4. When we say product, we would mean multiplication. So the result of that will be 4y. After that, the problem says we should add the two expressions. So the whole expression will be (y-32) + 4y. 
katen-ka-za [31]3 years ago
3 0
The difference of y and 32 is y-32.

The product of y and 4 is y*4, or 4y.

(y -32)+ 4y

The final answer is (y -32)+ 4y~
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Please help ASAP I am in desperate need
Vlad1618 [11]

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x = 9

TH = 12

Step-by-step explanation:

since TM is half of HM, the equation is:

2(21 - x) = 3x - 3

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8 0
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
3 years ago
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