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2 years ago

(3ab-5a-3) - (3ab-1)

Mathematics

1 answer:

ipn [44]2 years ago

8 0

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2 years ago

On a recent quiz, the class mean was 72 with a standard deviation of 4.2. Calculate the z-score (to 2 decimal places) for a pers

barxatty [35]

Answer:

The Z score of the person is -3.09.

Step-by-step explanation:

The standard normal deviate (Z) for a normal distribution is given by

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

where,

X = the value of observation

$\bar{X}$ is mean of the observations and

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Applying values we get

$Z=\frac{59-72}{4.2}=-3.09$

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3 years ago

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Lyrx [107]

Answer:

i think the answer is A

Step-by-step explanation:

again "i think". so i could be wrong

5 0

2 years ago

Read 2 more answers

The answer to this problem?

LenaWriter [7]

The simplified expression of $\sqrt{\frac{162x^9}{2x^{27}}}$ is $\frac{9}{x^9}$

<h3>How to simplify the expression?</h3>

The expression is given as:

$\sqrt{\frac{162x^9}{2x^{27}}}$

Divide 162 and 2 by 2

$\sqrt{\frac{81x^9}{x^{27}}}$

Take the square root of 81

$9\sqrt{\frac{x^9}{x^{27}}}$

Apply the quotient rule of indices

$9\sqrt{\frac{1}{x^{27-9}}}$

Evaluate the difference

$9\sqrt{\frac{1}{x^{18}}}$

Take the square root of x^18

$\frac{9}{x^9}$

Hence, the simplified expression of $\sqrt{\frac{162x^9}{2x^{27}}}$ is $\frac{9}{x^9}$