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cluponka [151]
3 years ago
8

How do i solve for y and x?

Mathematics
2 answers:
yawa3891 [41]3 years ago
6 0
Use the 30-60-90 rule

X= 18 and y=6sqrt3
Akimi4 [234]3 years ago
4 0

Answer:

You use the facts about a 30, 60, 90 triangle.

Step-by-step explanation:

The angles in a triangle add up to 180.

90+30+x=180

120+x=180

x=60

This is a 30-60-90 right triangle.

The side opposite the 30 degree angle is a.

The side opposite the 90 degree angle (hypotenuse) is 2a.

The side opposite the 60 degree angle is a\sqrt{3}

We know the side opposite the 90 degree angle.

2a = 12\sqrt{3}

Divide by 2.

a=6\sqrt{3}

a is the side opposite the 30 degree angle (y)

Because we know a, we can find a\sqrt{3}.

6\sqrt{3 } (\sqrt{3})

The square roots cancel, leaving 3.

6 times 3 is 18.

Therefore, the side opposite the 60 degree angle (x) is 18.

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Evaluate C_n.xP^xQn-x For the given n=7, x=2, p=1/2
r-ruslan [8.4K]

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

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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}

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