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k0ka [10]
3 years ago
14

A free-frame list Select one: a. is a set of all frames that are used for stack and heap memory. b. is a set of all frames that

are currently unallocated to any process. c. is a set of all frames that are filled with all zeros. d. is a set of all frames that are currently being shared by at least two processes.
Computers and Technology
1 answer:
zhuklara [117]3 years ago
7 0

Answer:

b. is a set of all frames that are currently unallocated to any process

Explanation:

The free frame list is the list that used for all kind of the frames that presently non-allocated to any kind or process

Therefore as per the given situation, the correct option is b as it fits to the current situation

Hence, all the other options are wrong

So, only option b is correct

The same is to be considered

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Which term refers to a type of an attack in which an attacker makes his data look like it is coming from a different source addr
Andrew [12]

<u>Man-in-the-middle attack</u> refers to a type of an attack in which an attacker makes his data look like it is coming from a different source address, and is able to intercept information transferred between two computers.

<u>Explanation:</u>

A man-in-the-middle attack (MITM) is an assault where the aggressor furtively transfers and potentially changes the correspondences between two gatherings who accept that they are straightforwardly speaking with one another. This happens when the assailant catches a segment of a correspondence between two gatherings and retransmits it sometime in the future. The assailant would then be able to screen and perhaps change the substance of messages. The utilization of such encoded burrows makes extra secure layers when you get to your organization's secret systems over connections like Wi-Fi.

8 0
3 years ago
What is up what everbody up to
andriy [413]

Answer: Eating chicken. drinking cool aid, and playing basketball

Explanation: look at my skin color and u will understand

5 0
3 years ago
Read 2 more answers
The measure of the maximum amount of data that can travel through a computer’s communications path in a given amount of time is
STatiana [176]

Answer:

Bandwidth

Explanation:

Bandwidth is the rate of transfer of data in the given time. Its unit is Bit/sec.

It is used to measure the transfer rate of bit in a network.

4 0
4 years ago
13) What are the benefits and detriments of each of the following? Consider both the systems and the programmers’ levels. a. Sym
dsp73

Answer:

Automatic and Explicit Buffering.

In the case of explicit buffering, the length of the queue is provided while in automatic buffering the queue size needs to be indefinite. In automatic buffering there is no need to block the sender while coping the message. While in explicit buffering the sender is blocked for the space in queue.

No memory is being wasted in explicit buffering.

Send by Copy and Send by Reference.

By using the send by copy method, the receiver is not able to change the state of parameter while send by reference allow. The advantage of using the send by reference method is that it allows to change a centralized application to its distributed version.

Fixed-sized and Variable-sized Messages.

In fixed size messaging refers, the buffer size is fixed. This means only a particular number of messages can only be supported by the fixed size buffer. The drawback of the fixed size messages is that they must be a part of fixed size buffer. These are not a part of variable size buffer. The advantage of variable size message is that the length of the message is variable means not fixed. The buffer length is unknown. The shared memory is being used by the variable size messages.

Explanation:

5 0
3 years ago
Suppose you are an ISP that owns a / 22 IPv4 address block. Can you accommodate requests from six customers who need addresses f
lisabon 2012 [21]

Answer:

It is not possible.

Explanation:

In this example, we need to accommodate 473 computers for six clients that are 473 IP addresses.

For this request just we have /22 IPv4 address blocks, this mean

22 red bits 11111111111111111111110000000000 <--- 10 host bits

We must increase red bits to 25, we need these 3 bits to create 6 sub red, in this case, 2^3 = 8 sub red.

Why did we ask 3 bits? Because if we ask only 2, 2^2 = 4, and we need 6 sub red.

25 red bits 11111111111111111111111110000000 7 host bits

In this case, we need more than 260 computers, but just we have 7 bits, this means.

2^7 = 128 and just one customer needs 260, for that is impossible.

7 0
3 years ago
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