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Molodets [167]
2 years ago
14

which of the following protocols allows hosts to exchange messages to indicate problems with packet delivery?

Computers and Technology
1 answer:
Sloan [31]2 years ago
4 0

The protocols that allows the exchange of messages by host as an indication for packet delivery problem is Internet control message protocol (ICMP).

  • Internet Control Message Protocol (ICMP) can be regarded as control channel that us used in reporting errors as well as  performing network diagnostics.

  • In this ICMP process, a message will be sent back to the sender from the receiver, in a case whereby the data doesn't  appear as it  should.

Some types of errors that can be handled by this protocol are;

  1. destination unreachable
  2. time exceeded
  3. source quench
  4. parameter problems

Therefore, ICMP is been utilized by  network devices in  send error messages as well as operational information

learn more at: brainly.com/question/10124043?referrer=searchResults

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svp [43]

Answer:

examples are:

com or .edu is a top-level domain name (TLD)

cornell.edu is a second-level domain name (SLD)

bigred.cornell.edu is a third-level or three-part domain name

project.bigred.cornell.edu is a fourth-level or four-part domain name

hope this helps you.

4 0
2 years ago
Bertha was recording a drama club practice. during editing, she noticed that she missed recording the beginning of a shot in the
Ray Of Light [21]

Bertha should use end roll technique so that she doesn't miss the initial action of a drama.

b. end roll

<u>Explanation:</u>

End roll is a simple method to decide whether the film is pushing ahead or not. When you utilize the film advance to wind the film, you essentially need to check if the handle on the left (that you use to rewind the film) is turning.

In the event that it turns, great, it implies that the film is appropriately locked in. So Bertha should utilize the end move strategy with the goal that she doesn't miss the underlying activity of a dramatization.

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2 years ago
Computer security experts devote their time and energy to the protection of sensitive data and the prevention of an outside atta
Alik [6]

Answer:

Information security policy are used for the prevention of intruders hacking a network when an organization start getting IT related attacks.

Explanation:

Information security policy are used for the prevention of intruders hacking a network when an organization start getting IT related attacks.

An information security policy are set of rules/policies designed to guide employees for the protection of the security of company information and IT systems. The reasons for these policies are:

  1. It defines what is required from organization’s employees for the security of the IT systems
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4 0
3 years ago
A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st
Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

^nC_r = \frac{n!}{(n-r)!r!}

Where n = 15\ and\ r = 4

^{15}C_4 = \frac{15!}{(15-4)!4!}

^{15}C_4 = \frac{15!}{11!4!}

^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}

^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}

^{15}C_4 = \frac{32760}{24}

^{15}C_4 = 1365

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

p = 3/15

p = 0.2

q = 1 - p

q = 1 - 0.2

q = 0.8

Solving further using binomial;

(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

Probability =  ^nC_3pq^3

Substitute 4 for n, 0.2 for p and 0.8 for q

Probability =  ^4C_3 * 0.2 * 0.8^3

Probability =  \frac{4!}{3!1!} * 0.2 * 0.8^3

Probability = 4 * 0.2 * 0.8^3

Probability = 0.4096

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

Probability =  q^n

Substitute 4 for n and 0.8 for q

Probability =  0.8^4

Probability = 0.4096

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

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2 years ago
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g100num [7]

Answer:

i dont really like music so sorry i cant help i hope someone can help you with this.

8 0
3 years ago
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