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.
You need to go into excel and make it there
The answer to this is probably something like the human factor or a human error. That is because no matter how great the conditions are, you can't prevent people from making a mistake eventually and this can include even simple things like falling down stairs. What you can do is make sure you get good employees, but even those banal injuries still count as a workplace accident and the worker needs to be paid and covered to mend it.
Answer:the quantities of some factors of production are fixed; the quantities of all factors of production can be varied - D
Explanation:
In the short run, some factors of production are fixed, which is usually the capital. Therefore for a company to increase output, it would need employ more workers, but would not increase capital.
Therefore in the short run, we can get diminishing marginal returns, which may cause marginal costs to start increasing quickly.
Also, in the short run, prices and wages fall out of equilibrium because a sudden rise in demand may lead to higher prices, and companies may not have the the capacity to respond and increase supply.
Long run
In the long run, usually greater than 6 months, all main factors of production are variable. The company has time to build a bigger one making it respond to changes in demand which means that a sudden rise in demand, would have a complimentary increase in supply to meet the demands and prices can be adjusted.
.
Answer:
39: $182.46
Explanation:
In this problem, the daily production level at the lowest average cost per player and the average cost (in $) can be estimated by considering the graph (average cost (y) against production level (x)) in the previous question. By drawing the graph up to scale, it can be deduced from the graph that the lowest average cost is approximately $182.46 and the production level at that point is 39.
