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OLga [1]
3 years ago
9

A student is curious about how a Web site appears on his computer screen. On a piece of paper,

Computers and Technology
1 answer:
nata0808 [166]3 years ago
5 0

A student is curious about how a Web site appears on his computer screen. There is a communication between the client and the server in the Application Layer.

Explanation:

  • When we choose to open a webpage (in any browser) the seventh layer of the OSI model - called Application layer will help to do that.
  • What happens after we write the webpage address in address bar is that the Application layer protocol (also called HTTP) formats and sends the request from the client's browser (Internet Explorer, Mozilla Firefox , Opera, Safari etc.) to the server.
  • It also formats and sends the server's respond back to client's browser. This process happens very fast and all the OSI model layers are processing in it (not only the Application layer).
  • All the layers are working together and each of them is responsible for some particular job, but all together they work as one. The layers communicate with each other and in case of error they will retry and fix the error or if they are unable to do it, the responsible layer will inform the user about the source of the problem.
  • What happens when page is requested and received? If we will remove the graphical and visual image and look at the process that computer does. We will see a set of commands, mathematical algorithms, symbols, letters and not understandable codes and processes.
  • When data is sent from "A" to "B", a transport layer is responsible to send and deliver it correctly and exactly the same, what was requested.
  • If the request, processed by "A" (sender) is too long, the transport layer will divide it in segments (called segmentation process) to understand well and not make a mistake while sending the data to "B" (recipient). After this process data is travelling through the network to the "B" (recipient), if the sent data is segmented or divided, the transport layer is responsible for reassemble it again and "B" (recipient) receives its requested data (It can be web page or other data).
  • If the transport layer will not do the segmentation process, then the next - network layer - will check the data and if the requested message is too long it will fragment it (called fragmentation process) and will provide the same as transport layer had to do.
  • All the layers processes are connected to each other and work cooperatively.
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Which role is delegated to personnel of the IT department and is responsible for maintaining the integrity and security of the d
enyata [817]

Correct question.

Which role is delegated to personnel of the IT department and how is responsible for maintaining the integrity and security of the data?

Answer:

<u>data custodian</u>

<u>Explanation:</u>

Remember, every organization generates data, and large organizations generate even larger data.

Hence, the role of a data custodian in an organization requires He implements measures to protect the organization's data, store and backup the data, as well as granting access to the data to authorized persons when needed.

3 0
2 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

8 0
3 years ago
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