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3 years ago

_____ relies on lies and misrepresentation to trick an authorized user into providing information or access to an attacker.

Computers and Technology

1 answer:

kap26 [50]3 years ago

6 0

Answer:

"Social engineering" is the correct answer for the above question.

Explanation:

Social engineering is a process, which gives the idea for the hacker or unauthorized user to get inside the system or hack the system.
It is used to read the humans data or psychologically hack human data.
It teaches the processor by an attacker to be the master of the authorized user. By this, an attacker can able to know the authorized term with the help of an authorized person.
The above question asks about the term, which is used to help the hackers to hack the system. This term is known as Social Engineering.

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The _____ establishes that the destination device is present on the network, verifies active service, and informs the destinatio

masya89 [10]

Answer: Three way handshake

Explanation:

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The three way shake process basically uses in the transmission control protocol(TCP) and it also inform to the destination device about the communication session.

Three way handshake process are properly understand in the TCP segment header by the various values in the two host exchange.

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3 years ago

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Alexus [3.1K]

Answer:

C. testing

Explanation:

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Serhud [2]

C. The insertion poin

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2 years ago

If directory server a trusts directory server b, directory server b trusts directory server c, and directory server a trusts dir

topjm [15]

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3 years ago

A direct-mapped cache holds 64KB of useful data (not including tag or control bits). Assuming that the block size is 32-byte and

Cerrena [4.2K]

Answer:

A) Number of bits for byte = 6 bits

B) number of bits for index = 17 bits

C) number of bits for tag = 15 bits

Explanation:

Given data :

cache size = 64 kB

block size = 32 -byte

block address = 32 -bit

number of blocks in cache memory

cache size / block size = 64 kb / 32 b = 2^11 hence the number of blocks in cache memory = 11 bits = block offset

A) Number of bits for byte

$log _{2} (6)^2$ = 6 bits

B) number of bits for index

block offset + byte number

= 11 + 6 = 17 bits

c ) number of bits for tag

= 32 - number of bits for index

= 32 - 17 = 15 bits

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3 years ago