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Kitty [74]
2 years ago
12

These figures illustrate the production possibilities available to Kate and Sarah with eight hours of labor in their bakery. Ans

wer the questions according to these figures.
SAT
1 answer:
jeyben [28]2 years ago
8 0

Given the above production possibility curve:

  • Kate has an absolute advantage in neither goods
  • Sarah has the absolute advantage in both goods.

<h3>What is absolute advantage?</h3>

This is the ability of one person or group to do a certain economic activity more effectively than another person or group.

<h3>How do we arrive at the above answer?</h3>

From the PPC curves of both manufacturers, the attached schedule of production which shows how they use 8 hours of labor indicates that:

Kate produces 4 loaves of bread or 8 pieces of cake; while

Sarah produces 12 loaves or 9 pieces of cake.

Given the definition of Absolute Advantage above, it is thus correct to state that:

  • Kate has an absolute advantage in neither goods: while
  • Sarah has the absolute advantage in both goods.

Learn more about absolute advantage at;
brainly.com/question/1655791
#SPJ1

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Consider a set of cards that has four cards labeled 1, 2, 3, and 4. Suppose you pick two cards, without replacement, to obtain t
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Answer:

\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}

Explanation:

Given

Cards = \{1,2,3,4\}

Required

The sampling distribution

The possible selection of 2 cards without replacement is as follows:

S = \{(1,2) (1,3) (1,4) (2,1) (2,3) (2,4) (3,1) (3,2) (3,4) (4,1) (4,2) (4,3)\}

Calculate the mean

\begin{array}{cccccccccccc}{Selection} & {(1,2)} & {(1,3)} & {(1,4)} & {(2,1)} & {(2,3)} & {(2,4)}& {(3,1)} & {(3,2)} & {(3,4)} & {(4,1)} & {(4,2)} & {(4,3)} \ \\ {Mean} & {1.5} & {2} & {2.5} & {1.5} & {2.5} & {3} & {2} & {2.5} & {3.5} & {2.5} & {3} & {3.5}\ \end{array}

List out the mean and the respective frequency

1.5 \to 2

2 \to 2

2.5 \to 4

3 \to 2

3.5\to  2

Total \to 12

Calculate the probability of each mean

P(1.5) \to \frac{2}{12} \to \frac{1}{6}\\

P(2) \to \frac{2}{12} \to \frac{1}{6}\\

P(2.5) \to \frac{4}{12} \to \frac{1}{3}\\

P(3) \to \frac{2}{12} \to \frac{1}{6}\\

P(3.5) \to \frac{2}{12} \to \frac{1}{6}

So, the table of sampling distribution is:

\begin{array}{cccccc}{Mean} & {1.5} & {2} & {2.5} & {3} & {3.5}\ \\ {Probability} & {\frac{1}{6}} & {\frac{1}{6}} & {\frac{1}{3}} & {\frac{1}{6}} & {\frac{1}{6}}\ \end{array}

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3 years ago
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