What is typically unique in each database table?

The difference between a dot matrix printer and a line printer

bezimeni [28]

Answer:

please give me brain list and follow

Explanation:

Difference Between Dot Matrix and Line Printer is that Dot-matrix printer produce printed images, they produce image when tine wire pins on a print head mechanism strike an inked ribbon. While Line printer is a type of impact printer which is high-speed and printer an entire line at a time.

4 0

3 years ago

Read 2 more answers

Write a shell (text-based) program, called first_word.py, that opens the file stuff.txt and prints out the first word of every l

postnew [5]

Answer:

See explaination

Explanation:

Please check below for the code for first_word.py file which will print first word of an input file stuff.txt. So the open function will open file in read mode. We had maintained a count variable so that we can skip printing first word of first line. Also the command line.strip() checks whether string is empty or not so that we will not get index error while calling split function over line to get first word.

#!/usr/local/bin/python

stuff = open("stuff.txt", "r");

count = 0;

for line in stuff:

count+=1;

if (count>1 and line.strip()):

print(line.split(maxsplit=1)[0])

3 0

3 years ago

Which of these ia an example of gene flow?

GalinKa [24]

In population genetics, gene flow (also known as gene migration) is the transfer of alleles or genes from one population to another. Migration into or out of a population may be responsible for a marked change in allele frequencies (the proportion of members carrying a particular variant of a gene).

7 0

3 years ago

Disk requests come in to the disk driver for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. A seek takes 6 msec per cyli

oksano4ka [1.4K]

Answer:

For FCFS = 876msec For SSTF = 360msec For SCAN(elevator) = 348msec

Explanation:

Considering FCFS algorithm.

In FCFS, the requests are addressed in the order they arrive in the disk queue, this means that the number of cylinders traveled becomes equal to the total of disk requests. With the arm initially at 20, the first request is to read cylinder 10.

Therefore the cylinders traversed for the first request = 20 – 10 = 10

For the second request i.e. a movement from cylinder 10 to cylinder 22, the number of cylinders traversed is = 22 - 10 = 12.

Similarly, for the third request seek arm will return to 20 from 22 so, cylinders traversed through would be = 22-20 = 2.

Also for the fourth request, cylinders traversed would be = 20 – 2 = 18.

For the fifth request, cylinders traversed = 40 – 2 = 38.

Now for the sixth request cylinders traversed = 40 – 6 = 34.

For the seventh and last request, cylinders traversed = 38 – 6 = 32.

So now to get the how much seek time is required for Disk scheduling algorithm

First we would add the total cylinders traversed = 10 + 12 + 2+ 18+ 38 + 34 + 32

= 146 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

= 146 X 6

= 876msec

Considering SSTF algorithm.

In SSTF (Shortest Seek Time First), requests having shortest seek time are executed first. So, the seek time of every request is calculated in advance in the queue and then they are scheduled according to their calculated seek time. What this means is that the closest disk (cylinder) next to the position of the seek arm is attended to first. With the arm at 20 initially, the first request is to read cylinder 22 (i.e. the closest cylinder to the seek arm)

Therefore the cylinders traversed for the first request = 22-20 = 2

For the second request, the disk to focus on is disk 20 and the cylinders traversed = 22-20 = 2

Similarly for the third request the seek arm will move to 10 from 20 so, cylinder traversed = 20-10 =10

For fourth request, cylinder traversed = 10 – 6 = 4

For the fifth request, cylinder traversed = 6 – 2 = 4

For sixth request, since all other disk request closer to it has been attended to the seek arm will move to disk 38 to attend to that disk request So, the cylinder traversed = 38 – 2 = 36

For the last request, cylinder traversed = 40 -38 = 2

So now to get the how much seek time is required for Disk scheduling algorithm

First we would add the total cylinders traversed = 2 + 2 +10 + 4 + 4 + 36 + 2

= 60 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

= 60 X 6 = 360msec

From Here we see that SSTF is better or an improvement to FCFS as it decrease the average response time (Average Response time is the response time of the all requests).

Considering SCAN (elevator) algorithm

In SCAN algorithm the disk arm moves into a particular direction and services the requests coming in its path and after reaching the end of disk, it reverses its direction and again services the request arriving in its path. So, this algorithm works as an elevator and hence also known as elevator algorithm. Therefore the number of cylinder traveled becomes equal to the total of disk request. With the arm at 20 initially

The first request is to read cylinder 22 i.e. the first cylinder on the upward movement

Therefore the cylinders traversed would be = 20 – 22 = 2

For the second request is to read cylinder 38, and the cylinders traversed would be = 38 – 22 =16

For the third request, seek arm will move to 40 So, the cylinders traversed would be = 40 – 38 = 2

For the fourth request, seek arm will return to 20 since from 40 since 40 is the highest in this upward elevator movement So, cylinders traversed would be = 40 -20 = 20

For the fifth request, cylinder traversed would be = 20 – 10 = 10

For the sixth request, cylinder traversed would be = 10 – 6 = 4

For the seventh and last request, cylinder traversed = 6 – 2 = 4

So now to get the how much seek time is required for Disk scheduling algorithm

First we would add the total cylinders traversed = 2 + 16 + 2 + 20 +10+ 4 + 4 = 58 cylinders

So therefore the total seek time = number of cylinders traversed X seek time per cylinder

= 58 X 6

= 348msec

From Here we see that SCAN is better or an improvement to FCFS and SSTF as it decrease the average response time (Average Response time is the response time of the all requests).

6 0

3 years ago

What does the internet engineering task force (IEFT) do?

Anvisha [2.4K]

The Internet Engineering Task Force is the premier internet standard body that defines standard internet protocols like TCP/IP. It is a large community that consists of operators, network designers, researchers and vendors drawn from the internet society’s individual and organization bodies and is concerned with the evolution and the smooth operation of the internet.

7 0

3 years ago