Answer:
An HTML tag is a special word or letter surrounded by angle brackets, < and >. You use tags to create HTML elements , such as paragraphs or links.
Explanation:
Many elements have an opening tag and a closing tag — for example, a p (paragraph) element has a <p> tag, followed by the paragraph text, followed by a closing </p> tag. HTML Editors
Step 1: Open Notepad (PC) Windows 8 or later: ...
Step 1: Open TextEdit (Mac) Open Finder > Applications > TextEdit. ...
Step 2: Write Some HTML. Write or copy the following HTML code into Notepad: ...
Step 3: Save the HTML Page. Save the file on your computer. ...
Step 4: View the HTML Page in Your Browser. hope this helps you :) god loves you :)
Answer:
The answer is "907,200".
Explanation:
In the word CALIFORNIA, there is a total of 10 letters in which letter A and Letter I is repeated 2 times. To solve this problem we use the multinomial principle.
The multinomial distribution model shows the results of n trials where the results of each study are prescriptive. It is also known as a distribution that makes the binomial distribution generalizable.
Ex:
<h2>
</h2><h2>
</h2><h2>
</h2><h2>
</h2><h2>
</h2>
Answer: A. passive attacks
Explanation:
Passive attack is a type of attack where by a system is tracked and scanned to determine how vulnerable it is and to also detect open port. The major purpose of a passive attack on a network is to gain access to the information of the target without altering the data.
A passive attack on a network may not be necessarily malicious, it is a type of network attack that is very difficult to detect.
Methods of carrying out passive attacks.
•War driving: scanning of networks is carried out to detect Wi-Fi that are vulnerable.
•Dumpster driving: Attackers or intruders search for information from discarded computers and
other gadgets, those information can help to attack a network.
Answer:
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
H(X)=5.4224 bits per symb
H(X|Y="not C")=0.54902 bits per symb
Explanation:
P(B)=2P(C)
P(A)=2P(B)
But
P(A)+P(B)+P(C)=1
4P(C)+2P(C)+P(C)=1
P(C)=1/7
Then
P(A)=4/7
P(B)=2/7
Probability Distribution={(A, 4/7), (B, 2/7), (C, 1/7)}
iii
If X={A,B,C}
and P(Xi)={4/7,2/7,1/7}
where Id =logarithm to base 2
Entropy, H(X)=-{P(A) Id P(A) +P(B) Id P(B) + P(C) Id P(C)}
=-{(1/7)Id1/7 +(2/7)Id(2/7) +(4/7)Id(4/7)}
=5.4224 bits per symb
if P(C) =0
P(A)=2P(B)
P(B)=1/3
P(A)=2/3
H(X|Y="not C")= -(1/3)Id(I/3) -(2/3)Id(2/3)
=0.54902 bits per symb
I think that it is 4 megabytes