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3 years ago

What is the measure of an angle if the measures of its two adjacent supplementary angles add up to 100°?

Mathematics

1 answer:

Sindrei [870]3 years ago

8 0

<h3>Answer: 130</h3>

Explanation:

Let x be the unknown angle we want to find.

Let y be adjacent and supplementary to x. This means x+y = 180

Let z also be adjacent and supplementary to x. So x+z = 180 also

Subtracting the two equations leads to y-z = 0 and y = z. So effectively we've proven the vertical angle theorem.

Since the supplementary angles to x add to 100, we know that y+z = 100. Plug in y = z and solve for z

y+z = 100

z+z = 100

2z = 100

z = 100/2

z = 50

Therefore,

x+z = 180

x+50 = 180

x = 180-50

x = 130

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choli [55]

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3 years ago

The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 591, \sigma = 42$

What is the probability that a randomly selected application will report a GMAT score of less than 600?

This is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 591}{42}$

$Z = 0.21$

$Z = 0.21$ has a pvalue of 0.5832

58.32% probability that a randomly selected application will report a GMAT score of less than 600

What is the probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{50}} = 5.94$

This is the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 591}{5.94}$

$Z = 1.515$

$Z = 1.515$ has a pvalue of 0.9351

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

What is the probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600?

Now we have $n = 50, s = \frac{42}{\sqrt{100}} = 4.2$

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

$Z = \frac{600 - 591}{4.2}$

$Z = 2.14$

$Z = 2.14$ has a pvalue of 0.9838

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

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3 years ago

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xxTIMURxx [149]

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3 years ago

As instructor grades exams, 30%; term paper, 20%; and final exam, 50%. A student had grades of 75,80, and 85, respectively, for

Leni [432]

Okay, so 75 counts for 30% (exams), 80 counts for 20% (term paper), and 85 counts for 50% (final exam)

So you would multiply 30% by 75, 20% by 80, and 50% by 85. That would be 22.5, 16, and 42.5. Add these numbers up, and put them over the total percentage, 100%. So you would have 81/100. The student's final average is 81%.

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3 years ago

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melamori03 [73]

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