Answer:
d.a surplus of apples must have existed.
Explanation:
If the supply of apples increased while there is no change in the demand for Apples, there would be a surplus and price would fall.
If the supply of apples decreased while there is no change in the demand for Apples, there would be a shortage of apples and price would rise.
If the demand for apples recently increased while supply remains unchanged, there would be a shortage and the price of apples would rise.
I hope my answer helps you
Answer:
A sunk cost is the correct answer to this question.
Explanation:
Sunk cost:- Sunk costs are those expenses that have been accumulated in the past and are thus in some way unrelated to judgment-making.
In the question referred to above, the company has already made $14 to produce. This cost will be inconsequential even if the company makes the units as it is or procedures them further.
As a result, $14 is a sunk expense.
Other options are incorrect because they are not related to the given scenario.
Answer:
Explanation:
The policy of tax cut will be less effective in country B than in country A since the value of the tax multiplier is lower in country B.
The multiplier effect refers to the increase in final income arising from any new injections.
Calculating the Multiplier Effect for a simple economy
k = 1/MPS
A = 1/0.1 =10
B= 1/.5=2
Explanation:
1. The journal entry is as follows:
On March 1
Prepaid insurance A/c Dr $36,000
To Cash A/c $36,000
(Being the prepaid insurance is recorded for cash)
For recording the advance purchase of insurance, we debited the prepaid insurance and credited the cash account. Both the accounts are recorded at $36,000 so that the proper posting could be done.
Answer:
They should operate Mine 1 for 1 hour and Mine 2 for 3 hours to meet the contractual obligations and minimize cost.
Explanation:
The formulation of the linear programming is:
Objective function:
Restrictions:
- High-grade ore: 
- Medium-grade ore: 
- Low-grade ore: 
- No negative hours: 
We start graphing the restrictions in a M1-M2 plane.
In the figure attached, we have the feasible region, where all the restrictions are validated, and the four points of intersection of 2 restrictions.
In one of this four points lies the minimum cost.
Graphically, we can graph the cost function over this feasible region, with different cost levels. When the line cost intersects one of the four points with the lowest level of cost, this is the optimum combination.
(NOTE: it is best to start with a low guessing of the cost and going up until it reaches one point in the feasible region).
The solution is for the point (M1=1, M2=3), with a cost of C=$680.
The cost function graph is attached.
