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Eddi Din [679]
3 years ago
8

What is another way to write 9(3+11)

Mathematics
1 answer:
maksim [4K]3 years ago
7 0
10(2+11 or 12(0+11) or 12(11+0)
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Find the value of x for which the lines l and m are parallel.
Rina8888 [55]

Answer:

A) 13

Step-by-step explanation:

l // m

9x + 10 = 127

9x = 117

x = 13

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6x + 5y = 82<br> 2x + 5y = 54
vovikov84 [41]

Answer:

A) 6x + 5y = 82

B) 2x + 5y = 54  We'll multiply equation B) by -1

B) -2x -5y = -54 Then we'll add this to A)

A) 6x + 5y = 82

4x = 28

x = 7

y = 8

Step-by-step explanation:

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•<br> Evaluate 15 / (-3) + 5|1-21 / (-5)
katrin [286]

Answer:

20

Step-by-step explanation:

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Can u pls help me with this question ​
Tresset [83]
The answer is a b c e
3 0
3 years ago
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard
avanturin [10]

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that \mu = 33208, \sigma = 2503.

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when X = 33208+633 = 33841 subtracted by the pvalue of Z when X = 33208 - 633 = 32575

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57

So

X = 33841

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

Z = \frac{33841 - 33208}{357.57}

Z = 1.77

Z = 1.77 has a pvalue of 0.9616

X = 32575

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

Z = \frac{32575- 33208}{357.57}

Z = -1.77

Z = -1.77 has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

4 0
3 years ago
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