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ruslelena [56]
1 year ago
5

What are the missing sides in the triangle written as integers or as decimals rounded to the

Mathematics
1 answer:
Alecsey [184]1 year ago
5 0

The value of x is 12.7 and the value of y is 9. The correct option is the third option x = 12.7; y = 9

<h3>Trigonometry </h3>

From the question, we are to determine the values of the missing sides

From the diagram,

Opposite = 9

Adjacent = y

Hypotenuse = x

Included angle = 45°

Using SOH CAH TOA, we can write that

tan 45^\circ= \frac{9}{y}

1 = \frac{9}{y}

∴ y = 9

Since the triangle is a right triangle, we can write that

x² = y² + 9² (<em>Pythagorean theorem</em>)

x² = 9² + 9²

x² = 81 + 81

x² = 162

x = √162

x = 12.7

Hence, the value of x is 12.7 and the value of y is 9. The correct option is the third option x = 12.7; y = 9

Learn more on Trigonometry here: brainly.com/question/24259483

#SPJ1

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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
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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}

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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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