Answer:
Two advantages of the agile methods are:
- The agile method require less documentation process as compared to waterfall model and saves maximum time and money. It basically reduces the efforts and the amount of work.
- In agile method, there is always high customer satisfaction present. In agile method, it is easy to modify in the data as compared to the waterfall model. The customers and developers always interact with each other as the interaction is very important for the good results in the project.
Two disadvantages of the agile methods are:
- For the large and complex projects, sometimes it is difficult to determine the requirements in the software development as projects are easily go off track.
- There is less predictability and the projects are easily messed up if the projects requirement are not clear by the customer end.
Video and Sound are the 2 most basic categories of film
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:
boolean isEven = false;
if (x.length % 2 == 0)
isEven = true;
Comparable currentMax;
int currentMaxIndex;
for (int i = x.length - 1; i >= 1; i--)
{
currentMax = x[i];
currentMaxIndex = i;
for (int j = i - 1; j >= 0; j--)
{
if (((Comparable)currentMax).compareTo(x[j]) < 0)
{
currentMax = x[j];
currentMaxIndex = j;
}
}
x[currentMaxIndex] = x[i];
x[i] = currentMax;
}
Comparable a = null;
Comparable b = null;
if (isEven == true)
{
a = x[x.length/2];
b = x[(x.length/2) - 1];
if ((a).compareTo(b) > 0)
m = a;
else
m = b;
}
else
m = x[x.length/2];
