Answer:
B. Combine like terms
Step-by-step explanation:
Inside the parenthesis, they combined -2 and 6 to get a total of 4x
It depends on how much you made per hour of work. Missing information.
Answer:
a) Drawing for grant charts that illustrates execution of the process is in the image attached
b) To find turn around time we use: completion time - arrival time
•Turn around time for FCFS scheduling algorithm will be:
P1 = 2-0= 2
P2= 3-0 = 3
P3 = 11-0 = 11
P4 = 15-0 = 15
P5 = 20-0 = 20
• Turn around time for SJF scheduling algorithm
P1 = 3-0= 3
P2 = 1-0 = 1
P3= 20-0 = 20
P4= 7-0 = 7
P5 = 12-0 = 12
• Turnaround time for non-preemptive algorithm
P1 = 15-0 = 15
P2 = 20-0 = 20
P3 = 8-0 = 8
P4 = 19-0 = 19
P5 = 13 - 0 = 13
• Turnaround time for RR
P1 = 2-0=2
P2= 3-0= 3
P3= 20 - 0 = 20
P4 = 19-0 = 19
P5 = 18-0 = 18
c) To find waiting time we use (turnaround time - burst time)
•Waiting time for FCFS
P1= 2-2 = 0
P2 = 3-1 = 2
P3 = 11-8 = 3
P4 = 15-4 = 11
P5 = 20-5= 15
• Waiting time for SJF
P1= 3-2 = 1
P2 = 1-1 = 0
P3 = 20-8 = 12
P4 = 7-4 = 3
P5 = 12-5 = 7
• Waiting time for non-preemptive
P1= 15-2 = 13
P2 = 20-1 = 19
P3 = 8-8 = 0
P4 = 19-4 = 15
P5 = 13 - 5 = 8
• Waiting time for RR
P1 = 2-2 = 0
P2 = 3-1 = 2
P3 = 20-8 =12
P4 = 13-4= 9
P5= 18-5=13
d) Average waiting time
•'For FCFS
= (0+2+3+11+15)/5
= 31/5
= 6.2milliseconds
•average waiting time For SJF
(1+0+12+3+7)/5
=23/5
= 4.6milliseconds
• Average waiting time For non-preemptive
(13+9+0+15+8)/5
=55/5
=11milliseconds
• average waiting time For RR
(0+2+12+9+13)/5
=7.2milliseconds
Since the SJF algorithm has 4.6milliseconds, it results in the minimum average waiting time.
The A, the C and the D because we have the vectors YX and YW which start from Y and the angle XYZ because the angle is around Y
Answer:
A
Step-by-step explanation:
You can use the Law of Cosines if you know all three sides of the triangle.
Note, however, that you can also use the Law of Cosines if you know the lengths of two of the sides and the angle between them.