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vodka [1.7K]
3 years ago
14

Solve Y/2 +22=38 for . do u mind answering for some points?

Mathematics
1 answer:
Rasek [7]3 years ago
3 0

Answer:

Step-by-step explanation:

Y = 32

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Need help with math!!!!!!
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3

Step-by-step explanation:

the third option

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What are the solutions to the equation 3(X -4)(X +5) =0
cestrela7 [59]
Answer should be: X = 4, -5
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it takes 120 newspapers to fill 8 recyceling bins what is the unit rate in newspapers to recyceling bins
barxatty [35]
15 newspaper per recycling bin
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What is 2x-4. I really don’t understand
ddd [48]

Answer:

2 * -4 = -8

Step-by-step explanation:

2 * -4 = -8

It is just like multiplying two numbers like... 2 * 4 = 8

You must need to know that:

+a * -b = -c

-a * -b = +c

+a * +b = +c

In your question;

+a * -b = -c

a = 2

b = -4

c = -8

therefore:

2 * -4 = -8

3 0
3 years ago
A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and
Hitman42 [59]

Answer:

0.13% probability that this selected sample has an average thickness greater than 53

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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n 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 = 50, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

What is the probability that this selected sample has an average thickness greater than 53?

This is 1 subtracted by the pvalue of Z when X = 53. So

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

By the Central Limit Theorem

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

Z = \frac{53 - 50}{1}

Z = 3

Z = 3 has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.13% probability that this selected sample has an average thickness greater than 53

5 0
3 years ago
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