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nikitadnepr [17]
2 years ago
15

What is the answer for 6k-9=-15

Mathematics
2 answers:
Svetach [21]2 years ago
8 0
6k -9 = -15
     +9   +9
6k = 24
 /6     /6
k = 4
docker41 [41]2 years ago
5 0
Simplifying 6k + -9 = -15 Reorder the terms: -9 + 6k = -15 Solving -9 + 6k = -15 Solving for variable 'k'. Move all terms containing k to the left, all other terms to the right. Add '9' to each side of the equation. -9 + 9 + 6k = -15 + 9 Combine like terms: -9 + 9 = 0 0 + 6k = -15 + 9 6k = -15 + 9 Combine like terms: -15 + 9 = -6 6k = -6 Divide each side by '6'. k = -1
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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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