Companies in the U.S. car rental market vary greatly in terms of the size of the fleet, the number of locations, and annual reve

Answer:

a) Taylor Industries puede reducir eficazmente sus costos laborales en los almacenes de inventario al contabilizar solo los artículos de alto valor. Estos artículos se determinan por sus valores de uso anual. Los valores de uso anual de los artículos deben ser utilizados como el grupo de costos de actividad para acumular y asignar los costos laborales en los almacenes de inventario. Taylor Industries puede establecer un punto de referencia o un límite, de modo que solo se cuenten los artículos que cumplan con este parámetro. Por ejemplo, los artículos que tengan un valor de uso anual superior a $5,000 deberían ser considerados en los conteos. Esta estrategia disminuirá la cantidad de artículos que se deben contar, reduciendo así los costos laborales.

b) Dado que el ítem 15 es crucial para las operaciones de Taylor Industries, debe clasificarse como un costo de materiales directos y no como un costo indirecto.

Explanation:

a) Datos y Cálculos:

una muestra aleatoria de 20 de los artículos de Taylor:

NUMERO DE ARTICULO USO ANUAL NUMERO DE ARTICULO USO ANUAL

1 $ 1,500 11 $ 13,000

2 12,000 12 600

3 2,200 13 42,000

4 50,000 14 9,900

5 9,600 15 1,200

6 750 16 10,200

7 2,000 17 4,000

8 11,000 18 61,000

9 800 19 3,500

10 15,000 20 2,900

Valor promedio de uso anual = $12,657.50

Answer:

a. The break-even quantity in units for manual process= 26,786 bags

b. $80,358

c. 36,000 bags

d. $108,000

e. $53,000

f. $58,000

g. 26,786 bags

h. mechanized process, manual process.

Explanation:

a. The calculation for break-even quantity is carried out using the formula

Fixed Costs ÷ (Revenue per Unit – Variable Cost per Unit)

= $ 37, 500 ÷ $3 - $1.60

= 26,786

b. The break-even point in sales dollars is found using the formula

Break-Even Point (sales dollars) = Fixed Costs ÷ Contribution Margin

with Contribution Margin defined as Price of Product – Variable Costs

= $37,500 ÷ ($3 - $1.60)

= 26,786 x $3= $80,358

c. The break-even quantity for the mechanized process can be calculated by

Fixed Costs ÷ (Revenue per Unit – Variable Cost per Unit)

= $72,000 ÷ $3 - $1 = 36,000 bags

d. To determine the revenue for the mechanized process at the break-even quantity, apply the formula

Fixed Costs ÷ Contribution Margin

where Contribution Margin = Price of Product – Variable Costs

= $72,000 ÷ ($3 - $1) = 36,000 bags x $3= $108,000

e. For 65,000 bags sold using the manual method, the calculation is 65,000 bags x $3 - $1.60 = $91,000-$37,500 (manual process fixed cost)

= $53,000

f. Utilizing the mechanized method for 65,000 bags, the equation would be 65,000 bags x $3 - $1 = $130,000 - $72,000 (manual process fixed cost)

= $58,000

h. The mechanized process option should be picked as it leads to greater returns if demand goes beyond the point of​ indifference, while the manual process should be favored if demand remains beneath that point.

The stockholder's equity statement includes common stock and retained earnings, with the adjusted ending balances detailed in the attached spreadsheet. For retained earnings, the ending balance is calculated as the beginning balance plus net income minus dividends: $52,000 + $34,000 - $12,000 = $74,000. In addition, the ending balance for common stock is calculated as the beginning balance plus new shares issued: $71,000 + $20,000 = $91,000.

The smallest largest element can be 10 when the subset is {1, 2,..., 10}. The probability of this occurrence is (1/2)^10 = 1/1024. (Each element from 1 to 10 can either be included in the subset or omitted.)

The largest element can also take on the value of 11. In this case, every number in the subset must be from 1 to 10, necessitating the exclusion of one number, yielding the probability that the largest element is 11 as C(10,1)*1/1024.

The maximum element could also be 12. Here, all subset numbers must be selected from 1 to 11, requiring the omission of two numbers, resulting in the probability for the largest element of 12 as C(11,2)*1/1024.

This reasoning can be extended similarly for additional cases:

If the largest element is 13, then the probability is C(12,3)*1/1024.

If it's 14, the probability equals C(13,4)*1/1024.

If it's 15, it corresponds to C(14,5)*1/1024.

If it's 16, then we have C(15,6)*1/1024.

If the largest element is 17, it corresponds to C(16,7)*1/1024.

If it’s 18, it translates to C(17,8)*1/1024.

If it’s 19, it equals C(18,9)*1/1024.

If the largest element is 20, it concludes with C(19,10)*1/1024.

Summing these probabilities leads to (1 + C(10,1) + C(11,2) +... + C(19,10))*1/1024. Since 1 can be treated as C(9,0), we can state (C(9,0) + C(10,1) + C(11,2) +... + C(19,10))*1/1024.

Using the Hockey Stick Identity, it holds that C(9,0) + C(10,1) + C(11,2) +... + C(19,10) equals C(20,10), so the expected value for the largest element is 1/11*C(20,10)*1/1024 = 4199/256.