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Rom4ik [11]
3 years ago
8

The following is a recipe for Gazpacho, a cold vegetable soup. Write how much of each item in the

Mathematics
1 answer:
denis23 [38]3 years ago
8 0

9514 1404 393

Answer:

  replace recipe quantities:

  1/4 ⇒ 5/8; 1/2 ⇒ 1 1/4; 1 ⇒ 2 1/2; 1 1/2 ⇒ 3 3/4; 2 ⇒ 5

Step-by-step explanation:

The given recipe serves 4, so must be multiplied by 10/4 = 5/2 to make it make 10 servings.

The numbers in the recipe (ignoring units or ingredients) are ...

  1/4, 1/2, 1, 1 1/2, 2

Each of these numbers needs to be multiplied by 5/2 to get the number for the larger recipe.

  1/4 × 5/2 = 5/8

  1/2 × 5/2 = 5/4 = 1 1/4

  1 × 5/2 = 5/2 = 2 1/2

  (1 1/2) × 5/2 = 3/2 × 5/2 = 15/4 = 3 3/4

  2 × 5/2 = 5

Then, to make the larger recipe, rewrite it with the quantities replaced as follows:

  old value ⇒ new value

  1/4 ⇒ 5/8

  1/2 ⇒ 1 1/4

  1 ⇒ 2 1/2

  1 1/2 ⇒ 3 3/4

  2 ⇒ 5

__

For example, 1 1/2 lbs of fresh tomatoes ⇒ 3 3/4 lbs of fresh tomatoes

_____

<em>Additional comment</em>

If you actually want to create the recipe, you may find it convenient to use a spreadsheet to list quantities, units, and ingredient names. Then you can add a column for the quantities for a different number of servings, and let the spreadsheet figure the new amounts. (A spreadsheet will compute quantities in decimal, so you will need to be familiar with the conversions to fractions--or use metric quantities.)

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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 }

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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 }

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P (Y or Z) = P(Y) + P(Z) = \frac{6}{16}

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

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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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