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3 years ago

A cell phone company has three different production sites.

Mathematics

1 answer:

dlinn [17]3 years ago

4 0

Answer:

3/11

Step-by-step explanation:

given that a cell phone company has three different production sites

Site name 1 2 3 Total

Production 60% 30% 10% 100%

Recalled 5% 7% 9%

Prob for recall 0.003 0.021 0.009 0.033

Total probability for recall = 0.033

Prob for recall and came from site 3 = 0.009

So required probability

= If a randomly selected cell phone has been recalled, what is the probability that it came from Site 3

= $\frac{0.009}{0.033} \\=\frac{3}{11}$

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Step-by-step explanation:

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3 years ago

8. The Last Supper

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The number of possible seats is an illustration of permutation

There are 1728 possible sitting arrangements

<h3>How to determine the number of seats</h3>

From the question, we have the following highlights:

Chris can only take 1 seat (i.e. the central seat)
Jo can take 2 seats (i.e. the seats adjacent the central seat)
Alex, Barb and Dave can take 3! number of seats
Eddie, Fred, and Gareth can take 3! number of seats on the right of Chris.
The remaining 4 adults do not have preference, then they can seat in 4! ways

So, the number of sitting arrangement is:

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brainly.com/question/12468032

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2 years ago

The Cost of 625 bolts is $100.What is the coast of 10 bolts?

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3 years ago

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,

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

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3 years ago

What would 18^72+27(36)+36 be

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Answer:

u can use a calculator

Step-by-step explanation:

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2 years ago

Read 2 more answers