And this help me please as well.

Charlie makes pizza at a restaurant. The customers always compliment how great the pizza tastes. But Charlie takes a long time t

Oksi-84 [34.3K]

To get this u must do 8372928+(2728279298)=382627-6382638)()()(7372637 I did it right I’m the test

8 0

3 years ago

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To use Aggregate functions on subgroups within the total result set, place the Aggregate function in the Select clause and add t

densk [106]

To use Aggregate functions on subgroups within the total result set, place the Aggregate function in the Select clause and add the Group By clause to the query.

<h3>What is an aggregate function?</h3>

An aggregate function is known to be a tool that functions only by calculating one or more values and then it also returns a single value.

Conclusively, The aggregate function is a tool that is often employed with the GROUP BY clause and HAVING clause of the SELECT words.

Learn more about Aggregate functions from

brainly.com/question/15875528

8 0

2 years ago

How does having weak security on your browser represent the weakest link in a network

kozerog [31]

Answer:

1. Your location is always exposed "might not be a problem, until it is"

2. Computer Viruses as your browser allows them

3. Hackers, of course, Ransomware attacks

4. SpyWares, read terms and condition or atleast visit trusted sources

5. Phishing, intermediatte security can handle this risk, but fatal if not. as in online identity stolen, bank accounts accessed.

6...... ALWAYS STAY SECURED

YOU WERE NEVER THE TARGET, JUST THE VICTIM

5 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

When is a handrail required for stairs

ch4aika [34]

Answer: when is not able to get up the stairs like a normal person

Explanation:

5 0

3 years ago

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And this help me please as well. ​

And this help me please as well.