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rewona [7]
3 years ago
9

If e^9xe^2xe^-8=e^x what is the value of x

Mathematics
1 answer:
Inessa [10]3 years ago
6 0

Answer:ur daaddy

Step-by-step explanation:

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Attachment geometry please help
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Center of circle = (5,-6)
equation =  (x-5)^2+ (x+6)^2 = 3^2
8 0
4 years ago
What is the equation of a line that passes through points A<br> (1, 2) and B (4, 3)?
Tpy6a [65]

Answer:

3y=x+5

Step-by-step explanation:

The slope of the line is (1/3). The equation of the line is y=(1/3)x+5/3 or 3y=x+5

8 0
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

8 0
3 years ago
Hi sorry im slow could anyone help ?
vivado [14]

Step-by-step explanation:

your answer is 60

hope it helps :)

6 0
3 years ago
Read 2 more answers
Pat said that the amount of money he has in his pocket is greater than or equal to $5.75 but less than $13.00. Which number line
blsea [12.9K]
Draw a line that starts at 5.75 and ends at 12.99
7 0
3 years ago
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