Answer:
Network scan
Explanation:
Cyber security can be defined as preventive practice of protecting computers, software programs, electronic devices, networks, servers and data from potential theft, attack, damage, or unauthorized access by using a body of technology, frameworks, processes and network engineers.
Some examples of cyber attacks are phishing, zero-day exploits, denial of service, man in the middle, cryptojacking, malware, SQL injection, spoofing etc.
Generally, a security assessment is carried out by network security experts on a regular basis to determine potential loopholes or vulnerabilities in the information and technology (IT) infrastructure. One of the techniques or approach used in security assessment is a network scan.
A network scan is a security assessment technique used for the automatic detection of host systems on a network. Although a network scan isn't capable of discovering or detecting all the weaknesses on a network, it avails users information about the computer systems that are active on the network and what services the computer systems offer or what ports are available on them.
Answer:
getline(cin, address);
Explanation:
Given
String object: address
Required
Statement that reads the entire line
The list of given options shows that the programming language is c++.
Analysing each option (a) to (e):
a. cin<<address;
The above instruction will read the string object until the first blank space.
Take for instance:
The user supplied "Lagos state" as input, only "Lagos" will be saved in address using this option.
b. cin address:
This is an incorrect syntax
c. getline(cin,address);
Using the same instance as (a) above, this reads the complete line and "Lagos state" will be saved in variable address
d. cin.get(address);
address is created as a string object and the above instruction will only work for character pointers (i.e. char*)
<em>From the above analysis, option (c) is correct.</em>
It’s A because surface website’s are available for everybody but the dark web is opposite
Answer:
a. Utilization = 0.00039
b. Throughput = 50Kbps
Explanation:
<u>Given Data:</u>
Packet Size = L = 1kb = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
RTT = 20 msec
<u>To Find </u>
a. Sender Utilization = ?
b. Throughput = ?
Solution
a. Sender Utilization
<u>As Given </u>
Packet Size = L = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec
Utilization = Transmission Time / RTT + Transmission Time
= 8 micro-sec/ 20 msec + 8 micro-sec
= 0.008 sec/ 20.008 sec
Utilization = 0.00039
b. Throughput
<u>As Given </u>
Packet Size = 1kb
RTT = 20ms = 20/100 sec = 0.02 sec
So,
Throughput = Packet Size/RTT = 1kb /0.02 = 50 kbps
So, the system has 50 kbps throughput over 1 Gbps Link.
Just press on the slow motion button I think
