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kykrilka [37]
4 years ago
15

Can someone plzzzzzz help? I'll give points and mark as brainliest!

Mathematics
1 answer:
SVEN [57.7K]4 years ago
5 0

Answer:

<em>x = 19° ; Write 19° in the box ( don't forgot ° or answer will be incorrect )</em>

Step-by-step explanation:

If XW ║YZ, we can tell that YW is a transversal, as YW ∩ XW, YX;

By alternate interior angles, ∠ ZYW ≅ ∠ YWX, so that ⇒

m∠ ZYW = m∠ YWX,

m∠ YWX = 2x, ⇒ given m∠ ZYW = 2x

By ∑ of angles in triangle Theorem;

m∠ YXW + m∠ XYM + m∠ YWZ = 180° ⇒ through substitution,

( 3x - 5 ) + 90 + ( 2x ) = 180 ⇒ now apply algebra,

3x - 5 + 90 + 2x = 180 ⇒ combine like terms,

3x + 2x - 5 + 90 = 180 ⇒ add,

5x + 85 = 180 ⇒ subtract 85 from either side of equation,

5x = 95, divide 5 on either side,

x = 19°

<em>Solution; x = 19°</em>

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PLEASE HELP! The table shows the number of championships won by the baseball and softball leagues of three youth baseball divisi
Irina18 [472]

Answer:

Question 1: P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16}}{ \frac{4}{16}} = \frac{1}{2}

Question 2:

A. P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}

B. P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}

C. P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}

Step-by-step explanation:

Conditional probability is defined by

P(A|B)= \frac{P(A and B)}{P(B)}

with P(A and B) beeing the probability of both events occurring simultaneously.

Question 1:

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

P ( B | Y ) = \frac{ P ( B and Y)}{ P (Y)} = \frac{ \frac{2}{16} }{ \frac{4}{16} }  = \frac{1}{2}

Question 2.A:

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

then

P( B and Y)= \frac{ 2 }{ 16 }[/tex]

By definition,

P ( Y | B ) = \frac{ P(Y and B) }{ P(B) } = \frac{ \frac{2}{16} }{ \frac{6}{16} } = \frac{1}{3}

Question 2.B:

Z: Championships won by the 13 - 15 years old, beeing

P ( Z)= \frac{ 1 }{ 16 }

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

then

P( Z and B)= \frac{ 1 }{ 16 }[/tex]

By definition,

P( Z | B ) = \frac{ P ( Z and B)}{ P (B)}= \frac{ \frac{1}{16} }{ \frac{6}{16} } = \frac{1}{6}

Question 3.B

Y: Championships won by the 10 - 12 years old, beeing

P ( Y)= \frac{ 4 }{ 16 }

Z: Championships won by the 13 - 15 years old, beeing

P ( Z)= \frac{ 1 }{ 16 }

then

P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}

B: Baseball League Championships won, beeing

P ( B ) = \frac{ 6 }{16}

so

P((YorZ) and B)= \frac{3}{16}

By definition,

P((Y or Z)|B) = \frac{ P ((Y or Z) and B)}{P(B)}= \frac{ \frac{3}{16}}{ \frac{6}{16}}= \frac{1}{2}

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ay just stating facts: if two of yo homies about to bang and u pick sides then u where never real in the first place
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Answer:

true!

Step-by-step explanation:

wait bang like fight

or bang like you know  f u   c k

btw mark brainliest plz

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