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Mariulka [41]
3 years ago
9

Where is information stored in the computer?​

Computers and Technology
1 answer:
m_a_m_a [10]3 years ago
4 0

Answer:

files

Explanation:

data is stored in files

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The use of computers in education is referred to as computer assisted instruction (CAI). Write a program that will help an eleme
dybincka [34]
I don't think some will write the whole program here, but which part are you having trouble with?
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4 years ago
A keyboard, mouse, and microphone are examples of ________.
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A keyboard, mouse, and microphone all are examples of <u>peripheral devices.</u>

Explanation:

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3 years ago
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What single awk command can be used to display all the login names and their associated numerical user IDs
Luden [163]

Answer:

Used the command syntax; awk -F":" '{ print "username: " $<number location> "\t\tuid:" $<number location> }' <target folder>

Explanation:

Linux operating system is a fast open-source computer platform for programmers and network administrators. Its system is arranged in a hierarchical tree structure with the root represented as "/" (for absolute path).

The passwd is a folder in the Linux OS that holds the login details of all users in the system network. The 'awk' is one of the commands used to get information from a file in a folder. It prints out the result by specifying the location of the values (like the username and user id) as a variable (with prefix '$') and then the target folder.

8 0
3 years ago
Which career qualification is unique to the Energy Transmission career pathway and not to the Energy Distribution pathway? color
Sergeeva-Olga [200]

Answer:

a, c, d

Explanation:

6 0
3 years ago
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Suppose you have a 16-bit machine with a page size of 16B. Assume that any unsigned 16-bit integer can be a memory address. Also
irinina [24]

Answer:

2^11

Explanation:

Physical Memory Size = 32 KB = 32 x 2^10 B

Virtual Address space = 216 B

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Page size = 16 B. Therefore, Frame size = 16 B

If there is a restriction, the number of bits is calculated like this:  

number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]

where

physical memory size = 32KB  which is the restriction

n bit machine = frame size = 16

Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11

7 0
4 years ago
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