<span>Think about the things people might have wanted private, but after death they have no control.</span>
Answer:
Check the explanation
Explanation:
The two tools that are utilized for consideration in cyber attacks are operating systems and computer systems. The operating system Kali Linux 2.6 3.x Mix (64-bi) and windows XP SPO (32-bit).
Penetration testing is a very important technique or procedure in auditing the loopholes and security in a particular operating system. The computer systems are Nmap and VMware. The author posit that VMware is a virtual machines programming that gives a virtual PC the capacity for more than one event or action of the working frameworks to keep cunning one similar server.
Nmap (“Network Mapper”) according to the authors, it is a free and open source utility for system reviewing and security. these tools are used for the purpose of finding an open port and to also gain access while another considers the vulnerabilities of the operating system.
ALL together these tools assist in building the foundation for ethical hacking. Using Nmap allows a hacker to gain access to the system thereby seeing all the flaws to actively peruse that account. All these tools work together for a successful hacking process and procedures.
Answer:
syndication
Explanation:
According to Gitlin, during the 1950s and 60s television production costs began to exceed the licensing fees the networks paid in order to broadcast their programming. But the studios could make that money back by putting a show in syndication after it produced 100 episodes that could be programmed in re-runs. Syndication is the licensing or sale of a publication material by television stations.
Answer:
The answer is A.Absolute reference.
Explanation:
Absolute reference is a cell reference whose location remains constant when the formula is copied.
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];
