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2 years ago

A mesh network topology allows all computers to be directly connected. True or False

Computers and Technology

1 answer:

pav-90 [236]2 years ago

4 0

Answer:

true

Explanation:

mesh technology is a network set up where every device and computer are interconnected with each other.

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Define reverse engineering in hacker context

bogdanovich [222]

Taking the product and taking it apart to insert a malicious process into it, then implement it into the world.

Hope this helps.

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3 years ago

Which of the following inputs help robotic vacuum cleaners maneuver around obstacles?

Alja [10]

Answer:

c lasers and sensors

Explanation:

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3 years ago

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Una pregunta cuales son los ataques de Sindel que por favor y me lo diga gracias

Len [333]

I do not know spanish please say in english

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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

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3 years ago

Which of the instances below is an example of a Trojan horse? Question 2 options: a) An attachment in an email that is really a

sergij07 [2.7K]

Game-theif Trojans is the malware that makes a backdoor into your email account by which you are playing a game or has downloaded it and thus by this all information of user is exposed.

So option D is the answer.

Option A cannot be answer because Trojan does not use files for its purpose.

Option B cannot be a good option because Trojan works in online could.

Option C cannot be a option because Trojan does not need executable to attack your account.

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3 years ago

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