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stellarik [79]
2 years ago
11

Math expert in polygon angle​

Mathematics
1 answer:
soldier1979 [14.2K]2 years ago
3 0

Answer:

50°

Step-by-step explanation:

The internal angles on a quadrilateral sum to 360.

So x+10+2x+x+3x=360

7x+10=360

7x=350

x=50

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2 years ago
Marcus is designing a structure to hold up scenery for the school play. The structure must be a right triangle.
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Step-by-step explanation:

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3 years ago
What is the answer for 2/5x5/8 ?
Maru [420]
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2 years ago
. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

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 \frac{\sigma}{\sqrt{n}}

Normal probability distribution

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.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
The temperature in Armand’s town in the morning was – 3.6°F. The temperature in the afternoon was 0°F. What was the overall chan
OLEGan [10]

The overall temperature would be 3.6°F as this is how much it was changed by

Answer=3.6°F

7 0
2 years ago
Read 2 more answers
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