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3 years ago

Substitute for y and rewrite the system of equations as a single equation which can be used to solve for x.

Mathematics

1 answer:

Vlada [557]3 years ago

7 0

Answer:

x = 3 and y = 1

Step-by-step explanation:

Given equations

y=3x-8 and y=4-x

We will substitute y from second equation in the first and get:

4-x = 3x-8 first we will add (+x) to both sides and get

4-x+x = 3x+x-8 => 4 = 4x-8 now we will add (+8) to the both sides and get

4+8 = 4x-8+8 => 12 = 4x now we will change the sides of equation and get

4x = 12 => x = 12/4 = 3 => x=3 now we will replace x=3 in the second equation

y= 4-x = 4-3=1 => y=1

Finally we have (x , y) = ( 3 , 1)

God with you!!!

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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]

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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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