Answer:
A trial balance will not balance if both sides do not equal, and the reason has to be explored and corrected.
Explanation:
The debit side and the credit side must balance, meaning the value of the debits should equal the value of the credit
Answer:
to rank tasks from most to least important
Explanation:
Prioritize means to choose priority, obviously and priority is the thing is the thing which, among other things, have the biggest importance.
Every day, especially in business, one finds himself swimming in tasks up to his neck. Obviously, not all of them can be successfully finished, or at least not without sacrificing one's personal life or sleep.
Prioritizing, therefore, serves as a helpful organising tool. After writing down all tasks that need to be done, a person should rank them by priority, which means that only urgent and important tasks will be dealt with immediately. Tasks of lower priority will be postponed, delegated or simply deleted.
<span> In a network that uses WPA2-PSK encryption you can bypass the encryption by using the weakness in the WPA2-PSK system and that is the following:the encrypted password is shared in what is known as the 4-way handshake. </span> When a client authenticates to the access point, the client and
the AP go through a 4-step process of authentication.
The same thing as if you do it once
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
