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babymother [125]
3 years ago
15

Suppose that a worker in Caninia can produce either 2 blankets or 8 meals per day, and a worker in Felinia can produce either 5

blankets or 1 meal per day. Each nation has 10 workers. For many years, the two countries traded, each completely specializing according to their respective comparative advantages. Now war has broken out between them and all trade has stopped. Without trade, Caninia produces and consumes 10 blankets and 40 meals per day and Felinia produces and consumes 25 blankets and 5 meals per day. The war has caused the combined daily output of the two countries to decline by:
Business
1 answer:
emmasim [6.3K]3 years ago
5 0

Answer:

15 blankets; 35 meals

Explanation:

First, we compute Opportunity Cost (OC).

In Caninia,

OC of blanket = 8/2 = 4 meals

OC of meals = 2/8 = 0.25 blanket

In Felinia,

OC of blanket = 1/5 = 0.2 meals

OC of meals = 5/1 = 5 blanket

Since Felinia can produce blankets at lower OC (0.2 < 4), so

Felinia has comparative advantage and specializing in blankets.

Total blankets produced with trade = 5 x 10

                                                           = 50

Since Caninia can produce meals at lower OC (0.25 < 5), so

Caninia has comparative advantage and specializing in meals.

Total meals produced with trade = 8 x 10

                                                       = 80

After trade,

Total blankets produced = 10 + 25

                                         = 35

Decrease in blanket output = 50 - 35

                                              = 15

Total meals produced = 40 + 5

                                     = 45

Decrease in meals output = 80 - 45

                                            = 35

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Answer:

$35,000

Explanation:

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A firm in China sells toys to a U.S. department store chain. Other things the same, these sales:a. decrease U.S. net exports and
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Minchanka [31]

Answer:

(a) P(X\:>\:5.40)=0.9938

(b) P(X\:

(c) X=4.975 percent

Explanation:

(a) Find the z-value that corresponds to 5.40 percent

.Z=\frac{X-\mu}{\sigma}

Z=\frac{5.40-4.15}{0.5}

Z=\frac{1.25}{0.5}=2.5

Hence the net interest margin of 5.40 percent is 2.5 standard deviation above the mean.

The area to the left of 2.5 from the standard normal distribution table is 0.9938.The probability that a randomly selected U.S. bank will have a net interest margin that exceeds 5.40 percent is 1-0.9938=0.0062

(b) The z-value that corresponds to 4.40 percent is Z=\frac{4.40-4.15}{0.5}=0.5The net interest margin of 4.40 percent is 0.5 standard deviation above the mean.

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Therefore the probability that a randomly selected U.S. bank will have a net interest margin less than 4.40 percent is 0.6915

(c)  The z-value that corresponds to 95% which is 1.65

We substitute the 1.65 into the formula and solve for X.1.65=\frac{X-4.15}{0.5}

1.65\times 0.5=X-4.150.825=X-4.15

0.825+4.15=X

4.975=X

A bank that wants its net interest margin to be less than the net interest margins of 95 percent of all U.S. banks should set its net interest margin to 4.975 percent.

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