Categorize the statements according to whether they promote economic growth or inhibit economic growth. Promote economic growth
1 answer:
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Answer:
The incomplete part of the question is "Using a cap-and-trade system of tradable emission allowances will eliminate half of the sulfur dioxide pollution at a cost of $1 million per year. If the permits are not tradable, what will be the cost of eliminating half of the pollution? If permits cannot be traded, then the cost of the pollution reduction will be $1 million per year." The full question is attched as picture as well
1) Tradable permit system
Then lower MAC firm will abate the all pollution units
Then as MAC1 = $250, MAC2 = $275
Firm 1 = Consolidated electric
Firm 2 = Commonwealth utility
Then 1 will sell all permits to 2, at a price between $250 & $275.
So total cost of abatement of 20 units = MAC1 * 20
= $250 * 20 Unit
= $5,000
2) Non-tradable permits
Total cost = MC1*10 + MC2*10
= $2,500 + $2,750
= $5,250
Answer:
730 items
Explanation:
The objective of the given information is to determine the number of hamburgers UAHH should order for the following conditions:
Average daily demand 600
Standard deviation of demand 100
Desired service probability 99%
Hamburger inventory 800
The formula for a given order quantity in a fixed period of time can be expressed as :
where;
= order quantity = ???
= daily demand average = 600
L = lead time in days = 1
T = time taken = 1
z = no of standard deviation = ???
= standard deviation of usage in lead time and time taken = ???
I = present inventory level = 800
=
× standard deviation of daily demand
=
= 1.4142 * 100
= 141.42 items
From the Desired service probability 99% = 0.99; we can deduce the no of standard deviation by using the excel function (=NORMSINV (0.99))
z = 2.33
From
q = 729.5086 items
q ≅ 730 items
Therefore; the number of hamburgers UAHH should order from the following given conditions = 730 items