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3 years ago

What is the difference between manual and computer typesetting?

Computers and Technology

1 answer:

Burka [1]3 years ago

6 0

Answer:

Manual typesetting: The form was placed in a press, inked, and an impression made on paper. During typesetting, individual sorts are picked from a type case with the right hand, and set into a composing stick held in the left hand from left to right, and as viewed by the setter upside down.

Computer typesetting: Computerized typesetting, method of typesetting in which characters are generated by computer and transferred to light-sensitive paper or film by means of either pulses from a laser beam or moving rays of light from a stroboscopic source or a cathode-ray tube (CRT).

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Give an example of an outdated memory or storage device what do you think they are outdated

andreev551 [17]

Answer:

HDD technology

Explanation:

You will find that there is a big difference between the SSD and the HDD. The SSD lists out the technical advantages as well as the disadvantages related to HDD, and this has led to the wide use of the SSD rather than the HDD, which is supposed by many being an outdated product. You will find that the SSDs last longer, and they are faster. However, the SSD is more prone to damages as compared to HDD. However, SSD looks like being more advanced considering the above-mentioned advantages, and fewer number of disadvantages. And hence, many people shifted to the SSD from HDD. Remember, SSD stands for solid state drives. And HDD means hard disk drive.

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3 years ago

If your vehicle leaves the pavement for any reason, remember to take your foot off the gas pedal, hold the wheel firmly, and____

aleksandr82 [10.1K]

A- and steer in a straight line

8 0

3 years ago

Read 2 more answers

Given six memory partitions of 100 MB, 170 MB, 40 MB, 205 MB, 300 MB, and 185 MB (in order), how would the first-fit, best-fit,

nlexa [21]

Answer:

We have six memory partitions, let label them:

100MB (F1), 170MB (F2), 40MB (F3), 205MB (F4), 300MB (F5) and 185MB (F6).

We also have six processes, let label them:

200MB (P1), 15MB (P2), 185MB (P3), 75MB (P4), 175MB (P5) and 80MB (P6).

Using First-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
P4 will be allocated to the remaining space of F1. Since F1 has a remaining space of 85MB, if P4 is assigned there, the remaining space of F1 will be 10MB from (85 - 75).
P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using First-fit include: F1 having 10MB, F2 having 90MB, F3 having 40MB as it was not use at all, F4 having 5MB, F5 having 115MB and F6 having 10MB.

Using Best-fit

P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
P6 will be allocated to the part of the remaining space of F5. Therefore, F5 will have a remaining space of 45MB from (125 - 80).

The remaining free space while using Best-fit include: F1 having 25MB, F2 having 170MB as it was not use at all, F3 having 25MB, F4 having 5MB, F5 having 45MB and F6 having no space remaining.

Using Worst-fit

P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
P5 will not be allocated to any of the available space because none can contain it.
P6 will be allocated to F2. Therefore, F2 will have a remaining space of 90MB from (170 - 80).

The remaining free space while using Worst-fit include: F1 having 100MB, F2 having 90MB, F3 having 40MB, F4 having 5MB, F5 having 100MB and F6 having 110MB.

Explanation:

First-fit allocate process to the very first available memory that can contain the process.

Best-fit allocate process to the memory that exactly contain the process while trying to minimize creation of smaller partition that might lead to wastage.

Worst-fit allocate process to the largest available memory.

From the answer given; best-fit perform well as all process are allocated to memory and it reduces wastage in the form of smaller partition. Worst-fit is indeed the worst as some process could not be assigned to any memory partition.

8 0

4 years ago

Olivia wants to change some settings in Outlook. What are the steps she will use to get to that function? open Outlook → File →

Rainbow [258]

Answer:

Explanation:

edg 2020

3 0

3 years ago

Read 2 more answers

You must keep track of some data. Your options are: (1) A linked-list maintained in sorted order. (2) A linked-list of unsorted

ladessa [460]

Answer:

Question was incomplete and continued the question

For each of the following scenarios, which of these choices would be best? Explain your answer.

BST

Sorted Array

Un-sorted Array

a) The records are guaranteed to arrive already sorted from lowest to highest (i.e., whenever a record is inserted, its key value will always be greater than that of the last record inserted). A total of 1000 inserts will be interspersed with 1000 searches.

b) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1,000,000 insertions are performed, followed by 10 searches.

Explanation:

Answer for a: Un-sorted array or Un-sorted linked list : as mentioned in the question itself that the records are arriving in the sorted order and search will not be O(log n) and insert will be not be O(n).

Answer for b : Un-sorted array or Un-sorted linkedlist : Number of the items to be inserted is already known which is 1,000,000 but it is very high and at the same time search is low. Unsorted array or Unsorted linked list will be best option here.

7 0

3 years ago