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Marina86 [1]
3 years ago
7

If one wanted to use the best method to get storage into long-term memory, one would use _________. maintenance rehearsal rote r

ehearsal elaborative rehearsal sleep learning
Computers and Technology
1 answer:
anzhelika [568]3 years ago
7 0

Answer:

elaborative rehearsal

Explanation:

when it comes to storage of information or data into long term memory then elaborative rehearsal plays an important role.

Elaborative rehearsal is a technique which focuses on thinking about the piece of information or data's meaning which is to be stored in long term memory and linking it with the information or data which is already present or stored.

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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

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F1. Therefore, F1 will have a remaining space of 85MB from (100 - 15).
  3. P3 will be allocated F5. Therefore, F5 will have a remaining space of 115MB from (300 - 185).
  4. 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).
  5. P5 will be allocated to F6. Therefore, F6 will have a remaining space of 10MB from (185 - 175).
  6. 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

  1. P1 will be allocated to F4. Therefore, F4 will have a remaining space of 5MB from (205 - 200).
  2. P2 will be allocated to F3. Therefore, F3 will have a remaining space of 25MB from (40 - 15).
  3. P3 will be allocated to F6. Therefore, F6 will have no remaining space as it is entirely occupied by P3.
  4. P4 will be allocated to F1. Therefore, F1 will have a remaining space of of 25MB from (100 - 75).
  5. P5 will be allocated to F5. Therefore, F5 will have a remaining space of 125MB from (300 - 175).
  6. 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

  1. P1 will be allocated to F5. Therefore, F5 will have a remaining space of 100MB from (300 - 200).
  2. P2 will be allocated to F4. Therefore, F4 will have a remaining space of 190MB from (205 - 15).
  3. P3 will be allocated to part of F4 remaining space. Therefore, F4 will have a remaining space of 5MB from (190 - 185).
  4. P4 will be allocated to F6. Therefore, the remaining space of F6 will be 110MB from (185 - 75).
  5. P5 will not be allocated to any of the available space because none can contain it.
  6. 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
3 years ago
The this reference . a) can be used implicitly b) must be used implicitly c) must not be used implicitly d) must not be used 25)
Gnom [1K]

Answer:

yeet

Explanation:

because computers

6 0
3 years ago
Click to review the online content. Then answer the question(s) below, using complete sentences. Scroll down to view additional
lbvjy [14]

Answer:  you can say  or talket to them why or how

Explanation:

4 0
3 years ago
Read 2 more answers
You are troubleshooting a computer that is in the design phase. The problem you see is that the CPU is not receiving information
iogann1982 [59]

Answer:

Control bus

Explanation:

A control bus is a PC bus that is utilized by the CPU to speak with gadgets that are contained inside the PC. This happens through physical associations, for example, links or printed circuits.

The CPU transmits an assortment of control sign to parts and gadgets to transmit control sign to the CPU utilizing the control bus. One of the principle targets of a transport is to limit the lines that are required for communication

An individual bus licenses communication between gadgets utilizing one information channel. The control transport is bidirectional and helps the CPU in synchronizing control sign to inside gadgets and outer segments. It is included interfere with lines, byte empower lines, read/compose sign and status lines.

8 0
3 years ago
Think of an example in your life where a number could be described as data, information, and knowledge
zhannawk [14.2K]

Answer:

how many event you have been too in the last month (well non during this time but as an example)

Explanation:

4 0
3 years ago
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