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2 years ago

What should be added to a²-2ab+b² to get a²-ab-2b²

Mathematics

1 answer:

dybincka [34]2 years ago

5 0

SOLUTION:

what should be added to a²-2ab+b² to get a²-ab-2b²

let the added no is P

$\pmb{P+(a^2-2ab+b^2)=(a^2-ab-2b^2) }$

$\pmb{P=(a^2-ab-2b^2)-(a^2-2ab+b^2) }$

$\pmb{P=a^2-ab-2b^2-a^2+2ab-b^2 }$

$\pmb{P=ab-3b^2 }$

$\therefore{\pmb{ab-3b^2 }}$ should be added to a²-2ab+b² to get a²-ab-2b²

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Step-by-step explanation:

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:

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In this problem, we have that:

$\mu = 500, \sigma = 100$

a. Greater than 600

This is 1 subtracted by the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 500}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

1 - 0.8413 = 0.1587

15.87% of the scores are expected to be greater than 600.

b. Greater than 700

This is 1 subtracted by the pvalue of Z when X = 700. So

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

$Z = \frac{700 - 500}{100}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

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2.28% of the scores are expected to be greater than 700.

c. Less than 450

Pvalue of Z when X = 450. So

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

$Z = \frac{450 - 500}{100}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

30.85% of the scores are expected to be less than 450.

d. Between 450 and 600

pvalue of Z when X = 600 subtracted by the pvalue of Z when X = 450. So

X = 600

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

$Z = \frac{600 - 500}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 450

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

$Z = \frac{450 - 500}{100}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

0.8413 - 0.3085 = 0.5328

53.28% of the scores are expected to be between 450 and 600.

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2 years ago

What should be added to a²-2ab+b² to get a²-ab-2b² ​

What should be added to a²-2ab+b² to get a²-ab-2b²