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3 years ago

How does email software protect the privacy of your messages?

Computers and Technology

2 answers:

zhannawk [14.2K]3 years ago

5 0

D is the answer to the question hope it helps

kari74 [83]3 years ago

4 0

The right answer is C. Servers encrypt all incoming and outgoing messages.

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Different the need for external or secondary memory

alexira [117]

Answer:

Secondary storage is needed to keep programs and data long term. Secondary storage is non-volatile , long-term storage. Without secondary storage all programs and data would be lost the moment the computer is switched off.

External storage enables users to store data separately from a computer's main or primary storage and memory at a relatively low cost. It increases storage capacity without having to open up a system.

4 0

3 years ago

An object on the computer use to store data, information and settings is known as?

kodGreya [7K]

Answer:

hard drive

Explanation: im sure tell me if its right

6 0

3 years ago

Read 2 more answers

kenny6666 [7]

1. true
2. false
3. true
4. true
5. true

if you would like an elaboration, let me know

8 0

3 years ago

Let A and B be two stations attempting to transmit on an Ethernet. Each has a steady queue of frames ready to send; A’s frames w

Mariana [72]

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete and correct question in the attachment file.

For Point a:

For the second round,

A is selects kA(2) either 0 or 1, so for each of them, that is $\frac{1}{2}$ .

B selects $kB(2)\ from\ (0, 1, 2, 3)$ for each choice with the probability of $\frac{1}{4}$ .

If $kA(2) < kB(2)$ wins the second rear race.

$\to P[A \ wins] = P[kA(2) < kB(2)]$

$= P[kA(2) = 0] \times P[kB(2) > 0] + P[kA(2) = 1] \times P[kB(2) > 1]\\\\= \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{2}{4} \\\\=\frac{3}{8} +\frac{2}{8} \\\\= \frac{3+2}{8}\\\\= \frac{5}{8}$

For Point b:

Throughout this example, $kA(3)$ also selects to be either 0 or 1 with such a $\frac{1}{2}$ probability. So, although B chooses $kB(3)$ from $(0, 1, 2, 3, 4, 5, 6, 7),$ the probabilities each are $\frac{1}{8}$ :

$\to P[A \ wins] = P[kA(3) < kB(3)]$

$= P[kA(3) = 0] \times P[kB(3) > 0] + P[kA(3) = 1] \times P[kB(3) > 1]\\\\= \frac{1}{2} \times \frac{7}{8} + \frac{1}{2} \times \frac{6}{8}\\\\= \frac{7}{16} + \frac{6}{16}\\\\= \frac{7+6}{16} \\\\= \frac{13}{16}\\\\$

For point c:

Assume that B tries again 16 times (typical value), and it destroys. In addition, throughout the exponential background n is obtained at 10 when choosing k between 0 to 2n−1. The probability of A winning all 13 backoff events is: $P[A \text{wins remaining races}] = 16\pi i =4P[A \ wins \ i |A \ wins \ i -1 ]$

Let the k value kA(i) be A for the backoff race I select. Because A retains the breed

$=(kA(i)] \cdot P[kA(i+ 1)< kB(i+ 1)] \geq P[kA(i) + 1$

For point d:

Two stations A and B are supposed. They assume that B will try 16 times afterward. Even so, for A, 16 races were likely to also be won at a rate of 0.82 For Just higher expectations of three A, B, and C stations. For Station A, possibility to win all backoffs

3 0

3 years ago

Charles is having a lot of problems with errors in a very complicated spreadsheet that he inherited from a colleague, and he tur

Kobotan [32]

Charles is having a lot of problems with errors in a very complicated spreadsheet that he inherited from a colleague, and he turns to another co-worker, Seymour, for tips on how to trace errors in the sheet. Seymour tells Charles that all error values begin with the same symbol, so they can be easily identified as errors.

The symbol is A)#.

Explanation:

An error is something you have done which is considered to be incorrect or wrong, or which should not have been done.
Generally errors are classified into three types: systematic errors, random errors and blunders
Error values begin with the same symbol as they are easy to identify. Once an error is identified, it can be corrected.
Error correction is the process of detecting errors in transmitted messages and reconstructing the original error-free data. Error correction ensures that corrected and error-free messages are obtained at the receiver side.
Error Correction can be handled in two ways: Backward error correction: Once the error is discovered, the receiver requests the sender to retransmit the entire data unit. Forward error correction: In this case, the receiver uses the error-correcting code which automatically corrects the errors.

3 0

4 years ago