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3 years ago

Help me solve this question please.

Mathematics

2 answers:

AleksandrR [38]3 years ago

6 0

The answer is D. 9(y+3) and 27 + 9y
You would distribute the 9 to both the 3 and the y.

Igoryamba3 years ago

4 0

Ah, so without giving you the answer, I'll help you. Because the question says "with an equivalent of 'y'", you're pretty much going to substitute any number you want in for y, and solve. The answer that has both sides equal to the same number, is correct.

For example:
If I had 3y+7 and 4y + 6

You're going to substitute any number you want. I chose 4 for example.
3(4) + 7= 19
4(4) +3=19

This would be your answer because both expressions have the same number as the outcome.

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Answer:

The value of given expression is $\frac{21}{128}$ .

Step-by-step explanation:

Given information: n=7, x=2, p=1/2

$q=1-p=1-\frac{1}{2}=\frac{1}{2}$

The given expression is

$C(n,x)p^xq^{n-x}$

It can be written as

$^nC_xp^xq^{n-x}$

Substitute n=7, x=2, p=1/2 and q=1/2 in the above formula.

$^7C_2(\frac{1}{2})^2(\frac{1}{2})^{7-2}$

$\frac{7!}{2!(7-2)!}(\frac{1}{2})^2(\frac{1}{2})^{5}$

$\frac{7!}{2!5!}(\frac{1}{2})^{2+5}$

$\frac{7\times 6\times 5!}{2\times 5!}(\frac{1}{2})^{2+5}$

$21(\frac{1}{2})^{7}$

$\frac{21}{128}$

Therefore the value of given expression is $\frac{21}{128}$ .

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3 years ago

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I think 15 is the answer I’m not sure tho

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