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kaylaamberd
10.11.2019 •
Mathematics
Acompany that makes nails finds that it would sell 42 boxes of nails per month if they are priced at $20 per box. if the company lowers the price to $10 per box, then more people would purchase nails and the company could expect to sell 52 boxes per month. find the equation of the line for the demand equation if x represents the number of boxes sold and y is the price of a box of nails.
a.) y = 62 + x
b.) y = 42 − x
c.) y = 42 + x
d.) y = 62 − x
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Ответ:
If x = number of boxes sold and y = price of nails box, then it is best represented by the equation y = 62 - x.
Checking:
If y = 20 and x = 42,
20 = 62 - 42
20 = 20
If y = 10 and x = 52.
10 = 62 - 52
10 = 10
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Ответ:
a. Utilization of machine A = 0.8
Utilization of machine B =![\frac{2}{9}](/tpl/images/0515/3838/746ea.png)
b. Throughput of the production system:
c. Average waiting time at machine A = 16 mins
d. Long run average number of jobs for the entire production line = 3.375 jobs
e. Throughput of the production system when inter arrival time is 1 =![\frac{5}{6} mins](/tpl/images/0515/3838/41b5b.png)
Step-by-step explanation:
Machines A and B in the production line are arranged in series
Processing times for machines A and B are calculated thus;
Inter arrival time is given as 5 mins
since the processing time for machine B adds up the processing time for machine A and the inter arrival time,
Inter arrival time for machine B,
a. Utilization can be defined as the proportion of time when a machine is in use, and is given by the formula![\frac{\beta }{M}](/tpl/images/0515/3838/b67ad.png)
Therefore the utilization of machine A is,
And utilization of machine B is,
b. Throughput can be defined as the number of jobs performed in a system per unit time.
Throughput of machines A and B,
Throughput of the production system is therefore the mean throughput,
c. Average waiting time according to Little's law is defined as the average queue length divided by the average arrival rate
Average queue length,![L_q = \frac{P_A^2}{1-P_A} = \frac{0.8^2}{1-0.8}=\frac{0.64}{0.2}= 3.2](/tpl/images/0515/3838/0dd2d.png)
Average waiting time =![\frac{3.2}{\frac{1}{5} }= 3.2*5=16mins](/tpl/images/0515/3838/1d630.png)
d. Since the average production time per job is 30 mins;
Probability when machine A completes in 30 mins,
And probability when machine B completes in 30 mins,
The long run average number of jobs in the entire production line can be found thus;
e. If the mean inter arrival time is changed to 1 minute
Utilization of machine A,![P_A = \frac{\beta_A }{M_A} = 4](/tpl/images/0515/3838/c7b27.png)
Utilization of machine B,![P_B = \frac{\beta_B}{M_B} = \frac{1}{3}](/tpl/images/0515/3838/b157d.png)
Throughput;