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2 years ago

7+8/5-2 work it out in BIDMAS

Mathematics

1 answer:

cricket20 [7]2 years ago

4 0

Answer:

bodmas

Step-by-step explanation:

before answering read question right it bodmas

and its 7+8=15

5-2=3

--- is 3 wholes

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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

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

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

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

South Africa broke the world record for largest pizza in 1990. It weighed `26833` pounds

viva [34]

8,944 hope this helps

5 0

2 years ago

James has $32 and earns $10 per week for his allowance. What is the initial value for the scenario described?

nignag [31]

32 is your initial value, because that is what he starts with

6 0

3 years ago

Read 2 more answers

I need help proving the y-intercept is 5 but I can't figure it out and will give brainliest if you can show the steps how you go

elixir [45]

So, honestly if its allowed you can just graph the equation ti prove the y intercept is five

4 0

2 years ago

A survey among American adults concluded that 20% of adults believe Elvis is still alive, 10% believe they've been abducted by a

Kitty [74]

Answer:

$P(Alive\ or\ Aliens) = 26\%$

Step-by-step explanation:

Given

$P(Alive) = 20\%$

$P(Aliens) = 10\%$

$P(Alive\ and\ Aliens) = 4\%$

Required

Determine $P(Alive\ or\ Aliens)$

From laws of probabilities:

$P(A\ or\ B) = P(A) + P(B) - P(A\ and\ B)$

In this case:

$P(Alive\ or\ Aliens) = P(Alive) + P(Aliens) - P(Alive\ and\ Aliens)$

$P(Alive\ or\ Aliens) = 20\% + 10\% - 4\%$

$P(Alive\ or\ Aliens) = 26\%$

7 0

3 years ago