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3 years ago

9a-18b+21c apply distributive property to factor out gcf of all three terms

Mathematics

2 answers:

Anni [7]3 years ago

5 0

Answer:3(3a - 6b + 7c)

hope it helped :)

horsena [70]3 years ago

3 0

The GCF of the three terms (9a, -18b and 21c) is 3

Rewrite each of the terms so 3 is a factor
9a = 3*3a
-18b = -3*6b
21c = 3*7c

So we can say...
9a - 18b + 21c = 3*3a - 3*6b + 3*7c
9a - 18b + 21c = 3(3a - 6b + 7c)

Answer: 3(3a - 6b + 7c)

If you distribute outer 3 to each of the inner terms and multiply, you'll get the original expression again.

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Answer:

The probability of a randomly selected student scoring in between 77.6 and 88.4 is 0.8185.

Solution:

Given, Scores on a test are normally distributed with a mean of 81.2

And a standard deviation of 3.6.

We have to find What is the probability of a randomly selected student scoring between 77.6 and 88.4?

For that we are going to subtract probability of getting more than 88.4 from probability of getting more than 77.6

Now probability of getting more than 88.4 = 1 - area of z – score of 88.4

$\mathrm{Now}, \mathrm{z}-\mathrm{score}=\frac{88.4-\mathrm{mean}}{\text {standard deviation}}=\frac{88.4-81.2}{3.6}=\frac{7.2}{3.6}=2$

So, probability of getting more than 88.4 = 1 – area of z- score(2)

= 1 – 0.9772 [using z table values]

= 0.0228.

Now probability of getting more than 77.6 = 1 - area of z – score of 77.6

$\mathrm{Now}, \mathrm{z}-\text { score }=\frac{77.6-\text { mean }}{\text { standard deviation }}=\frac{77.6-81.2}{3.6}=\frac{-3.6}{3.6}=-1$

So, probability of getting more than 77.6 = 1 – area of z- score(-1)

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Answer:

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Step-by-step explanation:

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