The economic term for the practice of assigning small parts of a complicated job to individual workers who specialize in doing there just smart part is called spe<span>cialization</span><span>. Specialization is achieved when the assigned personnel is master of his art and the resources are focused on the particular field. This leads to an efficiency called allocative efficiency.</span>
Po = 0.5385, Lq = 0.0593 boats, Wq = 0.5930 minutes, W = 6.5930 minutes.
<u>Explanation:</u>
The problem is that of Multiple-server Queuing Model.
Number of servers, M = 2.
Arrival rate, 
= 6 boats per hour.
Service rate, 
= 10 boats per hour.
Probability of zero boats in the system,![\mathrm{PO}=1 /\{[(1 / 0 !) \times(6 / 10) 0+(1 / 1 !) \times(6 / 10) 1]+[(6 / 10) 2 /(2 ! \times(1-(6 /(2 \times 10)))]\}](https://tex.z-dn.net/?f=%5Cmathrm%7BPO%7D%3D1%20%2F%5C%7B%5B%281%20%2F%200%20%21%29%20%5Ctimes%286%20%2F%2010%29%200%2B%281%20%2F%201%20%21%29%20%5Ctimes%286%20%2F%2010%29%201%5D%2B%5B%286%20%2F%2010%29%202%20%2F%282%20%21%20%5Ctimes%281-%286%20%2F%282%20%5Ctimes%2010%29%29%29%5D%5C%7D)
= 0.5385
<u>Average number of boats waiting in line for service:</u>
Lq =![[\lambda.\mu.( \lambda / \mu )M / {(M – 1)! (M. \mu – \lambda )2}] x P0](https://tex.z-dn.net/?f=%5B%5Clambda.%5Cmu.%28%20%5Clambda%20%2F%20%5Cmu%20%29M%20%2F%20%7B%28M%20%E2%80%93%201%29%21%20%28M.%20%5Cmu%20%E2%80%93%20%5Clambda%20%292%7D%5D%20x%20P0)
= ![[\{6 \times 10 \times(6 / 10) 2\} /\{(2-1) ! \times((2 \times 10)-6) 2\}] \times 0.5385](https://tex.z-dn.net/?f=%5B%5C%7B6%20%5Ctimes%2010%20%5Ctimes%286%20%2F%2010%29%202%5C%7D%20%2F%5C%7B%282-1%29%20%21%20%5Ctimes%28%282%20%5Ctimes%2010%29-6%29%202%5C%7D%5D%20%5Ctimes%200.5385)
= 0.0593 boats.
The average time a boat will spend waiting for service, Wq = 0.0593 divide by 6 = 0.009883 hours = 0.5930 minutes.
The average time a boat will spend at the dock, W = 0.009883 plus (1 divide 10) = 0.109883 hours = 6.5930 minutes.
Answer:
26.4%.
Explanation:
Net Profit:
= Saving of Labor & other Costs - Maintenance Cost of Machine - Depreciation On Machine (100,000/ 16 years)
= $40,000 - $10,000 - $6,250
= $23,750
Initial Investment:
= Cost of new Machine - Salvage value of old machine
= $100,000 - $10,000
= $90,000
Simple Rate of Return = Net Profit ÷ Initial Investments
= $23,750 ÷ $90,000
= 0.264 × 100
= 26.4%
Answer:
True.
Explanation:
A responsability chart is useful for describing the participation of persons completing labors and deliverables for a project or a business process. Some of the information considered in this type of chart is:
- Labor or procedure.
- Responsible.
- Supervisor.
- delivery date.
Answer:
$3,000
Explanation:
Warranty expense is an obligation on the business because business is liable to accept the claims of warranty. A estimated percentage of warranty expense is charges as an expense in each period.
Total Sales = $500 x 6,000 units = $3,000,000
Warranty Expense for the year = Sales units x 3% x warranty cost per unit
Warranty Expense for the year = 6,000 x 3% x $50 = $9,000
Recognised warranty cost in the year = 120 units x $50 = $6,000
Accrued Warranty expense = $9,000 - $6,000 = $3,000
