A help system refers to a documentation component that usually accompany every software. It is meant to help the users when they encounter problems in the course of using the software. The most difficult part of working with the help system IS PHRASING YOUR QUESTION. A user has to use the correct words and phrases in order to get the needed help from the help system.
Answer:
Hi, for this exercise we have two laws to bear in mind:
Morgan's laws
NOT(А).NOT(В) = NOT(A) + NOT (B)
NOT(A) + NOT (B) = NOT(А).NOT(В)
And the table of the Nand
INPUT OUTPUT
A B A NAND B
0 0 1
0 1 1
1 0 1
1 1 0
Let's start!
a.
Input OUTPUT
A A A NAND A
1 1 0
0 0 1
b.
Input OUTPUT
A B (A NAND B ) NAND (A NAND B )
0 0 0
0 1 0
1 0 0
1 1 1
C.
Input OUTPUT
A B (A NAND A ) NAND (B NAND B )
0 0 0
0 1 1
1 0 1
1 1 1
Explanation:
In the first one, we only need one input in this case A and comparing with the truth table we have the not gate
In the second case, we have to negate the AND an as we know how to build a not, we only have to make a nand in the two inputs (A, B) and the make another nand with that output.
In the third case we have that the OR is A + B and we know in base of the morgan's law that:
A + B = NOT(NOT(А).NOT(В))
So, we have to negate the two inputs and after make nand with the two inputs negated.
I hope it's help you.
Organization and access method.
Hope this helped! :)
- Jujufire
Complete Question:
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?
Answer:
a) 51.2 msec. b) 5.12 msec
Explanation:
If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:
The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:
a) BW = 1 Mbps = 10⁶ bps
⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec
b) BW = 10 Mbps = 10⁷ bps
⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec
Answer:
ICMP.
Explanation:
Hackers might use ICMP for gain knowledge about current system addresses and specific procedures as part of a reconnaissance process.
ICMP is a failure informing system used by network protocols such as routers to produce warning to that of the source Address while network issues interrupt IP packet transmission.
