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kicyunya [14]
3 years ago
11

Can someone me with this please?

Mathematics
1 answer:
Phoenix [80]3 years ago
5 0
Number 1 = -6, -1, 3, 7, 11
Number 2 = -4,  -3, -2, 1, 2
Number 3 = 3, 2, 1, 0, -1
Number 4 = -1, 2, 5, 8, 11 
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Forty-five percent of what number is 22.5? A. 10.125 B. 100 C. 50 D. 30
Mnenie [13.5K]
Look at the answers and be logical, try rounding up so instead of 45% try 50% now 22.5 would be half the number, have a look at the answers and see what is closest to double 22.5. C looks reasonable. Hope I helped, have a nice day :)
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Prime factorisation of 49, 56,42 plz answer
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Factor the expression using the GCF. 24y+88x =
Ne4ueva [31]

Answer:

4(6y + 22x)

Step-by-step explanation:

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4 0
3 years ago
The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard
Pie

Answer:

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

Step-by-step explanation:

To solve this question, we have 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 = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

Z = \frac{30732 - 30393}{472.81}

Z = 0.72

Z = 0.72 has a pvalue of 0.7642.

X = 30054

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

Z = \frac{30054 - 30393}{472.81}

Z = -0.72

Z = -0.72 has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

4 0
3 years ago
The grid below shows figure Q and its image figure Q' after a transformation:
Masteriza [31]
<h3>Answer: choice B) counterclockwise rotation of 90 degrees around the origin</h3>

To go from figure Q to figure Q', we rotate one of two ways

* 270 degrees clockwise

* 90 degrees counterclockwise

Since "270 clockwise" isn't listed, this means "90 counterclockwise" is the only possibility.

4 0
3 years ago
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