Answer:
Explanation:
The journal entry is shown below:
Cash A/c Dr $768,000
Service Charge Expense A/c $32,000
To Accounts Receivable A/c $800,000
(Being the cash is received and the remaining balance is debited to the cash account )
The computation of the service charge expense is shown below:
= Accounts Receivable × service charge percentage
= $800,000 × 4%
= $32,000
Deadweight losses occur when the quantity of an output produced is: ... Less than or greater than the competitive equilibrium quantity. Such that the marginal benefit of the output is just equal to the marginal cost.
The answer to this question is "OUTCOME FAIRNESS". Such as in addition to compensation, the customers expect OUTCOME FAIRNESS. In other words, the customers expect fairness in terms of policies, rules, guidelines, and timeless of the complaint process. Therefore, the answer is the last item in the choices which is outcome fairness.
Answer:
Diluted earnings per share is $1.7 per share
Explanation:
The number of diluted shares from the options is calculated thus
Total number of shares from options 34,500
Actual number of shares that can be purchased
(options shares*option price/share market price)
(34,500*$11/$15) (25,300)
Diluted shares 9,200
Diluted earnings per share=net income/(outstanding common stock + diluted common stock)
net income is $331,840
outstanding common stock is 186,000
diluted common stock is 9200
diluted earnings per share=$331,840/(186,000+9200)
=$1.7 per share
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.
