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Attackers frequently use ACK scans to circumvent a firewall or other filtering tools. During a NULL scan, all packet flags are enabled. The most recent versions of Nessus Server and Client are compatible with Windows, Mac OS X, FreeBSD, and the vast majority of Linux variants.
<h3>What is ack scan ?</h3>
- ACK scans are used to identify hosts or ports that have been blocked or are resistant to other types of scanning. An attacker uses TCP ACK segments to learn about firewall or ACL configuration.
- Attackers probe our router or send unsolicited SYN, ACK, and FIN requests to specific UDP/TCP ports.
- TCP ACK Scan sends an ACK message to the target port to determine whether or not it is filtered.
- On unfiltered ports, a RST reply packet will be sent for both open and closed ports. Filtered ports will either generate no response or generate an ICMP reply packet with an unreachable destination.
- The TCP ACK scanning technique attempts to determine whether a port is filtered by using packets with the ACK flag set.
To learn more about ask scan refer to:
brainly.com/question/13055134
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Answer:
Sorted array will be:-
3 5 5 9
Explanation:
In Selection Sort algorithm it sorts the array by repeatedly finds the minimum in the array form the array that is unsorted and swaps it with the value at the first position.
The unsorted array:- 5 3 9 5
In first iteration
minimum is 3.
3 will be swapped with 5.
Now the array is 3 5 9 5
sorted array 3
In second iteration
unsorted array is 5 9 5
sorted array 3 5
and the minimum from it is 5
No swapping will occur.
In third iteration
unsorted array is 9 5
minimum is 5
sorted array 3 5 5
swap it with 9.
Now the array is sorted.
3 5 5 9
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.
