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V125BC [204]
3 years ago
6

He puts 1/4 of the treasure to aside for himself he gives 1/2 of what is left to his friend bill he shares 2/3 of the remaining

treasure equally between the rest of the crew he buries the remaining 3 bars on a secret island how many bars did the pirates have
Mathematics
1 answer:
Andrej [43]3 years ago
4 0

Answer:

24 bars of treasure

Step-by-step explanation:

You just need to do the opposite in order to find the total value since you divided the treasure up we need to multiply the treasure “back”.

3 buried bars is equal to 1/3 of the remaining treasure. Multiply by 3 to get 3/3 or 100% of buried and crew portion. So we now have 9 bars. Then multiply by 2 to get 100% of what was given to bill, crew and buried. We now have 18 bars. Then we know that we took 1/4 for ourselves so that means 18 bars represents 3/4 of the treasure. So when we look at factors of 18 we find that 3 x 6 = 18 and 4 x 6 = 24 so we conclude that we had 24 bars of treasure before the split.

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The perimeter of this rectangle is 156 cm.
erastova [34]

Answer:

x = 15

Therefore, the length and width are 31 and 47.

Step-by-step explanation:

Perimeter = (2x + 1) + (2x + 1) + (3x + 2) + (3x + 2) = 156

Simplified,

10x + 6 = 156

Subtract 6 from both sides, then divide by 10.

10x = 150

x = 15

To find the length and width, substitute 15 (the value of x) into the individual equations.

2(15) + 1 = 31

3(15) + 2 = 47

5 0
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}

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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kupik [55]

Answer:

x ∈ R,a=1

Step-by-step explanation:

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