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Ann [662]
3 years ago
5

Consider two markets: the market for cat food and the market for dog food. The initial equilibrium for both markets is the same,

the equilibrium price is $1.50 , and the equilibrium quantity is 21.0 . When the price is $8.75 , the quantity supplied of cat food is 57.0 and the quantity supplied of dog food is 107.0 . For simplicity of analysis, the demand for both goods is the same. Using the midpoint formula, calculate the elasticity of supply for dog food. Please round to two decimal places.
Business
1 answer:
Yakvenalex [24]3 years ago
6 0

Answer:

Elasticity of supply for dog food = 0.95

Explanation:

From the question, we have:

New quantity supplied of dog food = 107.0

Old quantity supplied of dog food = Initial equilibrium quantity = 21.0

New price = $8.75

Old price = Initial equilibrium price = $1.50

Generally, the formula for calculating the elasticity of supply is as

follows:

Elasticity of supply = Percentage change in quantity supplied / Percentage change in price ................ (1)

Where, based on the midpoint formula, we have:

Percentage change in quantity supplied of dog food = {(New quantity supplied of dog food - Old quantity supplied of dog food) / [(New quantity supplied of dog food + Old quantity supplied of dog food) / 2]} * 100 = {(107.0 - 21.0) / [(107.0 + 21.0) / 2]} * 100 = 134.375%

Percentage change in price = {(New price - Old price) / [(New price + Old price) / 2]} * 100 = {(8.75 - 1.50) / [(8.75 + 1.50) / 2]} * 100 = 141.463414634146%

Substituting the values into equation (1), we have:

Elasticity of supply for dog food = 134.375% / 141.463414634146% = 0.94989224137931

Approximated to 2 decimal places, we have:

Elasticity of supply for dog food = 0.95

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<u>Explanation:</u>

The problem is that of Multiple-server Queuing Model.

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Service rate, \mu= 10 boats per hour.

Probability of zero boats in the system,\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\} = 0.5385

<u>Average number of boats waiting in line for service:</u>

Lq =[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0

= [\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385 = 0.0593 boats.

The average time a boat will spend waiting for service, Wq  =  0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.

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Explain why you can invest more stocks the younger you are?
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Explanation:

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Read 2 more answers
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Answer:

B. $9,957.

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g Marlboro Construction enters into a contract with a customer to build a warehouse for $725,000 on April 15, 2021 with a comple
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Answer:

B. $725,000

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Marlboro constructions expected value of the contract is 725,000 based on the given probability estimates of contract completion.

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