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JulsSmile [24]
3 years ago
9

A security administrator is reviewing the following information from a file that was found on a compromised host: Which of the f

ollowing types of malware is MOST likely installed on the compromised host?
A. Keylogger
В. Spyware
C. Trojan
D. Backdoor
E. Rootkit
Computers and Technology
1 answer:
Roman55 [17]3 years ago
8 0

Answer:

C. Trojan

Explanation:

In Cybersecurity, vulnerability can be defined as any weakness, flaw or defect found in a software application or network and are exploitable by an attacker or hacker to gain an unauthorized access or privileges to sensitive data in a computer system.

This ultimately implies that, vulnerability in a network avail attackers or any threat agent the opportunity to leverage on the flaws, errors, weaknesses or defects found in order to compromise the security of the network.

In this scenario, a security administrator is reviewing the following information from a file that was found on a compromised host: "cat suspiciousfile.txt."

Some of the ways to prevent vulnerability in a network are;

1. Ensure you use a very strong password with complexity through the use of alphanumerics.

2. You should use a two-way authentication service.

3. You should use encrypting software applications or services.

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Suppose a host has a 1-MB file that is to be sent to another host. The file takes 1 second of CPU time to compress 50%, or 2 sec
kozerog [31]

Answer: bandwidth = 0.10 MB/s

Explanation:

Given

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.50 MB / Bandwidth)

Transfer Size = 0.50 MB

Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)

Total Time = 1 s + RTT + (0.50 MB / Bandwidth)

Compression Time = 1 sec

Situation B:

Total Time = Compression Time + Transmission Time

Transmission Time = RTT + (1 / Bandwidth) xTransferSize

Transmission Time = RTT + (0.40 MB / Bandwidth)

Transfer Size = 0.40 MB

Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)

Total Time = 2 s + RTT + (0.40 MB / Bandwidth)

Compression Time = 2 sec

Setting the total times equal:

1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)

As the equation is simplified, the RTT term drops out(which will be discussed later):

1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)

Like terms are collected:

(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s

0.10 MB / Bandwidth = 1 s

Algebra is applied:

0.10 MB / 1 s = Bandwidth

Simplify:

0.10 MB/s = Bandwidth

The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.

(2) . Assume the RTT for the network connection is 200 ms.

For situtation 1:  

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB

Total Time = 1.2 sec + 5 sec

Total Time = 6.2 sec

For situation 2:

Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize

Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB

Total Time = 2.2 sec + 4 sec

Total Time = 6.2 sec

Thus, latency is not a factor.

5 0
3 years ago
Exercise 4: Bring in program grades.cpp and grades.txt from the Lab 10 folder. Fill in the code in bold so that the data is prop
Alex73 [517]

Answer:

Here is the C++ program:

#include <fstream>  //to create, write and read a file

#include <iostream>  // to use input output functions

using namespace std;  //to access objects like cin cout

const int MAXNAME = 20;  //MAXNAME is set to 20 as a constant

int main(){  //start of main() function

ifstream inData;  //creates an object of ifstream class

inData.open("grades.txt");  //uses that object to access and opens the text file using open() method

char name[MAXNAME + 1];   // holds the names

float average;   //stores the average

inData.get(name,MAXNAME+1);  //Extracts names characters from the file and stores them as a c-string until MAXNAME+1 characters have been extracted

while (inData){  //iterates through the file

 inData >> average;  //reads averages from file using the stream extraction operator

 cout << name << " has a(n) " << average << " average." << endl; //prints the names along with their averages

 inData.ignore(50,'\n');  //ignores 50 characters and resumes when  new line character is reached. It is used to clear characters from input buffer

 inData.get(name,MAXNAME+1);}  //keeps extracting names from file

return 0; }

Explanation:  

The program is well explained in the comments added to each line of the code. The program uses fstream object inData to access the grades.txt file. It gets and extracts the contents of the file using get() method, reads and extracts averages from file using the stream extraction operator. Then program displays the names along with their averages from grades.txt on output screen.

The grades.txt file, program and its output is attached.

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