Answer:
C. $12,000 under the cash method and $4,000 under the accrual method
Explanation:
Under the cash basis of accounting, whenever an amount is received or paid, it is recognized in the books of accounts
Whereas, on the accrual basis of accounting, the cash received / payment or not it is recognized in the books of accounts. It is recorded when it is earned not when it is received or paid.
So, by this above information
The cash method would recognize $12,000 ($1,000 × 12 months)
Whereas
The accrual method would recognize $4,000 ($1,000 × 4 months)
We assume the books are closed on December, 31
Answer:
(C) Pepin The Short
Explanation:
In 741AD, Pepin took over from his father as Mayor of the Palace. He ruled alongside his elder brother.
In 743AD, Pepin and his brother chose Childeric to be the <em>apparent</em> King of the Franks. Both brothers still wielded the functional power to the throne. Childeric was just to 'appear to be' the King (unknown to him though).
In 747AD, Pepin's brother stepped down (intentionally and on his own accord). Pepin then became the only ruler of the entire Frankish territory.
In 751AD, Pepin, without full support from his clan, lured Childeric into monastery in order to remove him as the 'face of Francia'.
Pope Zacharias helped Pepin to be proclaimed King of the Franks, against all opposition.
Answer:
18.52%
Explanation:
Calculation for the what would be the equity weight
Using this formula
Equity weight =E÷E+P+D
Let plug in the formula
Equity weight=$2,000,000×$27÷$2,000,000×$27+$1,000,000×$14.50+$10,000×.98×$1,000
Equity weight=$14,500,000÷$78,300,000
Equity weight=.1852×100
Equity weight=18.52%
Therefore what would be the equity weight is 18.52%
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.
