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Yanka [14]
3 years ago
7

Choose the correct function to yield the shown result.

Computers and Technology
1 answer:
HACTEHA [7]3 years ago
8 0

Answer:

int

Explanation:

Correct on edg test 2021

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Answer:

Programming & Software Development

Explanation:

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7 0
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Que compone una maquina Rube Goldberg (operadores mecánicos- maquinas simples – mecanismos)
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dominoes, fans, PVC pipe, magnets, duct tape, marbles, cups or bowls, miniature toy cars, paper towel tubes, string.

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dominó, abanicos, tubos de PVC, imanes, cinta adhesiva, canicas, tazas o cuencos, carros de juguete en miniatura, tubos de toallas de papel, cuerdas.

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3 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

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The answer is:
You can use one budget to advertise on the Search Network and Display Network
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