You can access various sites on WWW by using hyperlinks or by?
Answer is: A following directions on-screen
Answer: Geotagging was enabled on her smartphone
Explanation:
The security weakness that is the most likely cause of the security breach is that geotagging was enabled on the vice president's smartphone.
Geotagging, occurs when geographical identification metadata are added to websites, photograph, video, etc. Geotagging can be used to get the location of particular place.
In this case, since geotagging was enabled on her smartphone, it was easy for the attacker to locate her house.
Answer:
<u>d. Statement A is true and Statement B is false</u>
Explanation:
Indeed, when using duplex transmission either node can transmit while the other node can receive data from the network. Also, in half-duplex transmission, both the nodes can transmit as well as receive data.
However, in half-duplex transmission, the nodes <em>cannot </em>transmit and receive data at the same time. Hence, this makes Statement B false, while Statement A is true.
Answer:NULL Scan
Explanation:RFC 793(Request for comments) is a type of RFC command labeled with the number 793 which can operate with the TCP protocol.They are sort of document form which is from IETF( Internet Engineering Task Force ).
The null scan is scanning protocol used by legal as well as illegal hackers for working in the transfer control protocol architecture. It is used for the identification of the the ports and holes in TCP servers.they can also have the negative impact if used by the illegal hackers.
Answer:
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
Explanation:
The Worst case will happen when f(a) > 2*f(b) ; f(b) > 2*f(c) ; f(c) > 2*f(d) ; f(d) > 2*f(e) and f(e) > 2*f(f).
Where f(x) is frequency of character x.
Lets consider the scenario when
f(a) = 0.555, f(b) = 0.25, f(c) = 0.12, f(d) = 0.05, f(e) = 0.02 and f(f) = 0.005
Please see attachment for image showing the steps of construction of Huffman tree:- see attachment
From the Huffman tree created, we can see that endcoding e() of each character are as follows:-
e(a) = 0
e(b) = 10
e(c) = 110
e(d) = 1110
e(e) = 11110
e(f) = 11111
So we can see that maximum length of encoding is 5 in this case.
