Solve. f/5 - 24 = -28 f =

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,

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

3 0

3 years ago

Can someone help me with this?

bulgar [2K]

Answer:

I thinks its C , but im sorry if its wrong.

Step-by-step explanation:

6 0

3 years ago

Sophomore, junior, and senior students at a high school will be surveyed regarding a potential increase in the extracurricular s

vitfil [10]

Answer: C) 4

Step-by-step explanation:

5 0

2 years ago

Pls answer asapppddddddd

kvv77 [185]

Answer:

don't know........

......

4 0

3 years ago

Which equation best describes the relationship shown in the table?

Lerok [7]

Answer:

The answer is C. y=2x+3

Step-by-step explanation:

A. y=3x you would pick a coordinate to replace the X and Y value. -1=3(-2) which gives you -1=-6, which wouldn't work.

B. y=x+1 you would pick a coordinate to replace the X and Y value. -1=-2+1 which would give you -1=-1 which would work, but you have to check all of the coordinates to make sure it is the correct one. 3=0+1 which would give you 3=1 so this is not the correct answer.

C. y=2x+3 You would pick a coordinate to replace the X and Y value. -1=2(-2)+3 so then you would simplify that which would give you -1=-1 which would work, but you still need to check to make sure that all of the coordinates work. 3=2(0)+3 which would give you 3=3. 9=2(3)+3 which would give you 9=9. Since this one is able to have three coordinates that can be plugged into it more than likely it is the correct answer, but just to be certain lets see if D will work.

D. y=x+13 you would substitute the coordinates for the x and y values -1=-2+13 which would give you -1=11 which would not be correct.

4 0

3 years ago