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3 years ago

n an open economy, why is the supply curve for dollars in the foreign-currency exchange market vertical?

Business

1 answer:

Zarrin [17]3 years ago

4 0

Answer:

In an open economy, the supply curve for dollars in the foreing-currency exhange market is vertical, because the supply does not depend on the exchange currency rate.

The supply of dollars in an open economy depends on the interest rate, which is determined by the difference between imports and exports (which is the same as the difference between purchases and sales of foreign capital).

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Conventional wisdom says one should measure a manager's investment performance over an entire market cycle. What arguments suppo

White raven [17]

Answer:

In every form of analysis, it is always safer to take a macro or holistic view of the situation. This is true for the investment performance of a manager. One investment decision that went right does not suffice to classify an investment portfolio manager as proficient, neither is one that went south enough to tag him deficient.

The forecasting ability of managers, on the balance of probability, will vary for different cases, with a helicopter view of providing a more accurate measure of their performance.

However, if it was possible to analyse the market for volatility and adjust our forecasts it becomes unnecessary to look at and analyse all the information from a 12-month cycle before coming to terms about the performance of the manager.

Cheers!

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3 years ago

Suppose that the price of basketball tickets at your college is determined by market forces. Currently, the demand and supply sc

belka [17]

Answer:

a) see attached graph. There is nothing unusual with the supply curve, it is simply fixed. This happens to most services, e.g. there is a fixed number of hotel rooms available for rent, in the short run you cannot add more rooms per night if the demand increases. In order to increase the quantity supplied, you would need to build a larger hotel, or in this case, a larger stadium.

b) the equilibrium price is $8 and the equilibrium quantity is 8,000 tickets

c) if the college plans to increase enrollment, the demand might increase, leading to a higher equilibrium price, but the supply will remain the same until the stadium is expanded.

Explanation:

Price Quantity Demanded (Qd) Quantity Supplied (Qs)

$4 10,000 8,000

$8 8,000 8,000

$12 6,000 8,000

$16 4,000 8,000

$20 2,000 8,000

3 0

2 years ago

Get master dude . Its worth it just answer the question by saying MASTER

Brums [2.3K]

Answer:

MASTER

Explanation:

Apparently it says to write it so that's is what I did is there anything wrong about that bye

8 0

3 years ago

Read 2 more answers

Paying attention to the trends that might imact your future is called

MArishka [77]

Career Resilience: Major Trends That Will Impact Your Future !

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2 years ago

A rectangular storage container with an open top is to have a volume of 10 m^3. The length of its base is twice the width. Mater

DiKsa [7]

Answer:

The dimension that minimizes the container is width = 1; base = 2 and height = 5

The minimum cost is $36

Explanation:

Let the width be x

So:

$Width = x$

$Base = 2 * Width$

$Base = 2 * x$

$Base = 2x$

$Height = h$

Volume of the box is:

$Volume = 10m^3$ ---- Given

Volume is calculated as:

$Volume = Base * Width * Height$

$Volume = 2x * x * h$

$Volume = 2x^2h$

Substitute 10 for Volume

$10 = 2x^2h$

Make h the subject

$h = \frac{10}{2x^2}$

$h = \frac{5}{x^2}$

Next, we calculate area of the sides.

$Area = Base + Sides$

Because it has an open top, the area is:

$Base\ Area = Base * Width$

$Sides\ Area = 2[(Width * Height) + (Base * Height)]$

$Base\ Area = 2x * x$

$Base\ Area = 2x^2$

$Sides\ Area = 2[(Width * Height) + (Base * Height)]$

$Side\ Area = 2[(x * h) + (2x * h)]$

$Side\ Area = 2[(xh) + (2xh)]$

$Side\ Area = 2[3xh]$

$Side\ Area = 6xh$

The base area costs $6 per m²

So, the cost of 2x² would be:

$Cost = 6 * 2x^2$

$Cost = 12x^2$

The side cost area costs $0.8 per m²

So, 6xh would cost

$Cost = 0.8 * 6xh$

$Cost = 4.8xh$

Total Cost (C) is:

$C = 12x^2 + 4.8xh$

Recall that $h = \frac{5}{x^2}$

So:

$C = 12x^2 + 4.8x *\frac{5}{x^2}$

$C = 12x^2 + 4.8 *\frac{5}{x}$

$C = 12x^2 + \frac{4.8 *5}{x}$

$C = 12x^2 + \frac{24}{x}$

Take derivative of C

$C^{-1} = 24x - \frac{24}{x^2}$

Take LCM

$C^{-1} = \frac{24x^3 - 24}{x^2}$

Equate $C^{-1}$ to 0

$0 = \frac{24x^3 - 24}{x^2}$

Cross multiply

$24x^3 - 24 = 0 * x^2$

$24x^3 - 24 = 0$

Add 24 to both sides

$24x^3 = 24$

Divide through by 24

$x^3 = 1$

Take cube roots of both sides

$x = 1$

Recall that

$Width = x$

$Base = 2x$

$Height = h$ and $h = \frac{5}{x^2}$

Solve for these dimensions:

$Width = 1$

$Base = 2 * 1$

$Base = 2$

$h = \frac{5}{1^2}$

$h = \frac{5}{1}$

$h = 5$

i.e.

$Height = 5$

Hence, the dimension that minimizes the container is width = 1; base = 2 and height = 5

Recall that

$C = 12x^2 + \frac{24}{x}$

Substitute 1 for x

$C = 12(1^2) + \frac{24}{1}$

$C = 12 + 24$

$C = 36$

The minimum cost is $36

8 0

3 years ago