Which statement most aptly defines the term networking?

Someone help me I don’t know what to do /COMPUTER SCIENCE

nydimaria [60]

It kind of depends on the intended language, but I can see the following mistakes:

- lines not terminated with a semicolon (but not all languages require this)
- while statement missing parenthesis, ie., while (num1 <= num2), also not required in all languages
- statements below the while statement must be grouped by curly braces
- num1+5 should be num1=num1+5 or num1 += 5

Output will be:

10
15
20
25
30

5 0

4 years ago

The letters G, H, T, Y, B, and N are what type of keys?

andrey2020 [161]

Lower row keys. Your welcome

5 0

3 years ago

Read 2 more answers

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

The Apple II is an IBM-compatible PC "clone.

lutik1710 [3]

Is this supposed to be a true or false question?

4 0

3 years ago

A type of VPN connection, where a single computer logs into a remote network and becomes, for all intents and purposes, a member

elena-s [515]

Answer:

Client To Site.

Explanation:

The Client-to-Site is a virtual private network which used to provide the connection of the device like a laptop to the organization network. In the Client To Site a single computer is connected to the remotely network, for all the purpose and the intent of the network.

The single laptop is connected with a remote machine in the firewall. The "IPSec" protocol is used in the Client To Site connection.

3 0

4 years ago