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3 years ago

If A is an n× n matrix, then det A = det AT . Verify this theorem for 2 × 2 matrices.

Mathematics

1 answer:

Bond [772]3 years ago

4 0

Answer:

Verified

Step-by-step explanation:

Let A matrix be in the form of

$\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

Then det(A) = ad - bc

Matrix A transposed would be in the form of:

$\left[\begin{array}{cc}a&c\\b&d\end{array}\right]$

Where we can also calculate its determinant:

det(AT) = ad - bc = det(A)

So the determinant of the nxn matrix is the same as its transposed version for 2x2 matrices

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The probability of pulling an orange is 2 out of 5 if I 20 times how many of them would be orange

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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

8 0

3 years ago

Each of the letters of the word ALABAMA are written on a piece of paper and then

Kruka [31]

Answer:

3/7

Step-by-step explanation:

There are 7 letters in ALABAMA. And there are 4 A's in ALABAMA. So if you're trying to not get an A then you would subtract 4 from 7. Which you would get 3/7 of the letters in ALABAMA are not A.

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2 years ago

The probability that a randomly chosen citizen-entity of Cygnus is of pension age† is approximately 0.7. What is the probability

victus00 [196]

Answer: 0.2401

Step-by-step explanation:

The binomial distribution formula is given by :-

$P(x)=^nC_xp^x(1-p)^{n-x}$

where P(x) is the probability of x successes out of n trials, p is the probability of success on a particular trial.

Given : The probability that a randomly chosen citizen-entity of Cygnus is of pension age† is approximately: p =0.7.

Number of trials : n= 4

Now, the required probability will be :

$P(x=4)=^4C_4(0.7)^4(1-0.7)^{4-4}\\\\=(1)(0.7)^4(1)=0.2401$

Thus, the probability that, in a randomly selected sample of four citizen-entities, all of them are of pension age =0.2401

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3 years ago