<u>Answer:</u>
<u>Cloud computing</u><em> allow the user to access software or any document from the remote place.</em>
<u>Explanation:</u>
Let us understand what does the word cloud actually means. In simple terms, we get rain from cloud, but we don’t know actually which cloud burst to give rain.
In a similar way, <em>the cloud computing is the concept of storing files in multiple servers and multiple location</em> and it provide access when the <em>client needs the source. </em>
Cloud computing enable user to work on <em>software online</em> or to download document or <em>edit / create documents online</em>. Certain services are <em>free and few other are paid.</em>
Answer:
C.find the email address of someone you know
D.maintain an address book of your contacts
E.organize your emails In folders
Answer:
Explanation:
a. In this scenario, the best solution would have an Object of Traditional Books, CD, Music, Bookstore and Customer.
b. All five objects would be able to be called by the main program loop and the Customer Object would call upon and use either the Books or CD object, While the Bookstore object would call upon all of the other objects.
c. Both the Bookstore object and Customer object will "have" other objects as the Bookstore needs to hold information on every Book or CD in the Inventory. While the Customer object would call upon the Book and CD object that they are purchasing.
d. The Music Object will extend the CD object and use information on the CD object as its parent class.
e. Since the Music Object extends the CD object it is also considered a CD since it is in CD format like the Books on CD and therefore is both objects.
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
