A technician is evaluating malware that was found on the enterprise network. After reviewing samples of the malware binaries, the technician finds each has a different hash associated with it. Which of the following types of malware is MOST likely present in the environment?

a. Trojan
b. Polymorphic worm
c. Rootkit
d. Logic bomb
e. Armored virus

Answer:

Bulk Ethernet Cable - Category 5e or CAT5e

Bulk RJ45 Crimpable Connectors for CAT-5e

Explanation:

Answer:

The program written in Python is as follows

nums = []

for i in range(0,68):

-->nums.append(i)

print(nums)

Explanation:

Please note that --> is used to denote indentation

The first line creates an empty list

nums = []

This line creates an iteration from 0 to 67, using iterating variable i

for i in range(0,68):

This line saves the current value of variable i into the empty list

nums.append(i)

At this point, the list has been completely filled with 0 to 67

This line prints the list

print(nums)

Answer:

According to the principles of design, Repetition refers to the recurrence of elements of the design

One of the advantages of this is that it affords uniformity. Another is that it keeps the user of such a system familiar or with the interface of the operating system.

One major drawback of this principle especially as used in the question is that it creates a familiar route for hackers.

Another drawback is that creates what is called "repetition blindness". This normally occurs with perceptual identification tasks.

The  phenomenon may be  due to a failure in sensory analysis to process the same shape, figures or objects.

Cheers!

Answer:

1. Divide each bonus by regular bonus apply this to all the data

2. In cell C11 write, "Average" press tab key on the keyboard and then select the range of the cells either by typing "C7:C10" or by selecting it through the mouse.

Explanation:

The average bonus multiplier can be found by dividing each bonus with the regular bonus applying this to all the data and then putting the average formula and applying it to the cells C7:C10.  

After dividing the bonus with regular bonus, in cell C11 write, "Average" press tab key on the keyboard and then select the range of the cells either by typing "C7:C10" or by selecting it through the mouse.

Answer:

Here is the sentinel-controlled while loop:

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

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

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

int test_score; // declare an integer variable for test scores

//prompts user to enter test scores and type-99 to stop entering scores

cout << "Enter the test scores (enter -99 to stop): ";

cin >> test_score; // reads test scores from user

while (test_score != -99){ // while loop keeps executing until user enters -99

cin >> test_score; } } // keeps taking and reading test scores from user

Explanation:

while loop in the above chunk of code keeps taking input scores from user until -99 is entered. Sentinel-controlled loops keep repeating until a sentinel value is entered. This sentinel value indicates the end of the data entry such as here the sentinel value is -99 which stops the while loop from iterating and taking the test score input from user.

The complete question is that the code should then report how many scores were  entered and the average of these scores. Do not count the end sentinel -99 as a score.

So the program that takes input scores and computes the number of scores entered and average of these scores is given below.

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

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

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

double sum = 0.0;  // declares sum variable to hold the sum of test scores

int test_score,count =0;  

/* declares test_scores variable to hold the test scores entered by user and count variable to count how many test scores input by user */

cout << "Enter the test scores (or -99 to stop: ";

//prompts user to enter test scores and type-99 to stop entering scores

cin >> test_score;  // reads test scores from user

while (test_score != -99){  // while loop keeps executing until user enters -99

count++;  /* increments count variable each time a test cores is input by user to count the number of times user entered test scores */

sum = sum + test_score;  // adds the test scores

cin >> test_score;}  // reads test scores from user

if (count == 0)  // if user enters no test score displays the following message

cout << "No score entered by the user" << endl;

else  //if user enters test scores

//displays the numbers of times test scores are entered by user

cout<<"The number of test scores entered: "<<count;

/* displays average of test scores by dividing the sum of input test scores with the total number of input test scores */

cout << "\n The average of " << count << " test scores: " <<sum / count << endl;}

The program along with its output is attached.

Answer:

Here is the JAVA program to calculate volume of cuboid:

import java.util.Scanner; // Scanner class is used to take input from user

public class CuboidVol { // class to calculate volume of cuboid

public static void main(String[] args) { // start of main() function body

Scanner input= new Scanner(System.in); //create Scanner class object

// prompts user to enter length of cuboid

System.out.println("Enter the cuboid length:");

double length=input.nextDouble(); //reads the input length value from user

// prompts user to enter width of cuboid

System.out.println("Enter the cuboid width:");

double width=input.nextDouble(); //reads the input width from user

// prompts user to enter height of cuboid

System.out.println("Enter the cuboid height:");

double height=input.nextDouble(); //reads the input height from user

/* the following formula is to calculate volume of a cuboid by multiplying its length width and height and a double type variable cuboidVolume is defined to store the value of the resultant volume to it */

double  cuboidVolume= length*width*height; //calculates cuboid volume

//displays volume of cuboid and result is displayed up to 2 decimal places

System.out.printf("Volume of the cuboid (length " + length + "/ height " + height + "/ width" +width +" is: " + "%.2f",cuboidVolume);   } }

Explanation:

The formula for the volume of a cuboid is as following:

Volume = Length × Width ×  Height

So in order to calculate the volume of cuboid three variable are required for length, width and height and one more variable cuboidVolume to hold the resultant volume of the cuboid.

The program is well explained in the comments added to each statement of the program. The program prompts the user to enter the value of height width and length of cuboid and the nextDouble() method is used to take the double type input values of height length and width. Then the program declares a double type variable cuboidVolume to hold the result of the volume of cuboid. Then the last printf statement is used to display the volume of cuboid in the format format "Volume of the cuboid (length  / height  / width ) is" and the result is displayed up to 2 decimal places.

The screenshot of the program along with its output is attached.

Answer:

Here is the modified DemoSalesperson2

public class DemoSalesperson2{ // class DemoSalesperson2

public static void main(String[] args) { //start of main() function body

SalesPerson[] sp = new SalesPerson[10]; // creates an array of SalesPerson object named sp

int start_id =111; // assign 111 to starting id number

double start_sal=25000; //assign 25000 to starting salary means salary will start from amount 25000

 for (int i =0; i<sp.length; i++) { //loop iterates until the value of loop variable i remains less then length of the array object sp

  sp[i] =new SalesPerson(start_id+i,start_sal+5000*(i)); /*at each iteration  ID number from 111 through 120 and a sales value that ranges from $25,000 to $70,000, increasing by $5,000 for each successive is displayed using object array sp[] which calls the SalesPerson constructor to access id and sales */

 System.out.println(sp[i].getId()+"   "+sp[i].getAnnualSales() );  } } } /*uses object sp of class SalesPerson to access the methods getId which returns the id and getAnnualSales method which returns the annual sales, to display id and annual sales */

Explanation:

The question first requires a SalesPerson class with these requisites:

Class name: Salesperson.

Data fields for Salesperson: int ID number , double annual sales amount.

Methods: constructor SalesPerson() that requires values for both data fields, as well as get and set methods for each of the data fields.

So according to the complete question's requirement, I have implemented SalesPerson class:

public class SalesPerson {   // class name

//data members: integer type id variable and double type sales variable

   private int id;  

   private double sales;  

   public SalesPerson(int id_no, double annual_sales) {   //constructor that requires values for id and sales

       id = id_no;  

       sales = annual_sales;     }  

   public int getId() {   //accessor getId() method to get or access the value of id data field

       return id;     }  

   public void setId(int id_no) {   //mutator setId to set the value of id data field

       id = id_no;     }  

   public double getAnnualSales() {   //accessor getAnnualSales() method to get or access the value of sales data field

       return sales;     }  

   public void setAnnualSales(double annual_sales) { //mutator setAnnualSales to set the value of sales data field

       sales = annual_sales;     }  }

However you can use my DemoSalesperson2 application for your own SalesPerson class.

Now I will explain the working of for loop used in DemoSalesperson2

Iteration 1:

i=0

i<sp.length is true because length of sp[] is 10 so i<10

This means the body of for loop will execute.

The statement inside the body of for loop is :

 sp[i] =new SalesPerson(start_id+i,start_sal+5000*(i));

class SalesPerson constructor is used to access the data members of SalesPerson class

start_id = 111

start_sal=25000

start_id+i = 111 + 0 = 111

start_sal+5000*(i) = 25000 + 5000 (0) = 25000

The values 111 and 25000 are stored to sp[0] means at the first index of the sp array.

System.out.println(sp[i].getId()+"   "+sp[i].getAnnualSales() );  

This statement uses sp object to get access to the getId and getAnnualSales methods to print the id number and sales amount.

So the output at 1st iteration is:

111   25000.0

Iteration 1:

The value of i is incremented to 1 so now i = 1

i<sp.length is true because length of sp[] is 10 so i<10

This means the body of for loop will execute.

The statement inside the body of for loop is :

 sp[i] =new SalesPerson(start_id+i,start_sal+5000*(i));

class SalesPerson constructor is used to access the data members of SalesPerson class

start_id = 111

start_sal=25000

start_id+i = 111 + 1 = 112

start_sal+5000*(i) = 25000 + 5000 (1) = 25000 + 5000 = 30000

The values 111 and 25000 are stored to sp[1] means at the first index of the sp array.

System.out.println(sp[i].getId()+"   "+sp[i].getAnnualSales() );  

This statement uses sp object to get access to the getId and getAnnualSales methods to print the id number and sales amount.

So the output at 1st iteration is:

112   30000.0

The loop keeps iterating and the value of id is incremented by 1 and value of sales is increased by 5000 at every iteration.

This loop continues to execute until the value of i exceeds the length of sp array i.e. 10.

The values from iteration 1 to 8 following the above explanation are:

Iteration 1:

111    25000.0

Iteration 2:

112    30000.0

Iteration 3:

113    35000.0

Iteration 4:

114   40000.0  

Iteration 5:

115   45000.0  

Iteration 6:

116   50000.0                                                          

Iteration 7:

117   55000.0                                                                                                      

Iteration 8:                                                                                      

118   60000.0      

Iteration 9:                                                                                    

119   65000.0

Iteration 10:                                                                                    

120   70000.0

At iteration 11 the for loop body will not execute as value of i at 11th iteration is i = 10 and i<sp.length evaluates to false as i = sp.length because length of sp array is 10 and now the value of i is also 10. So the loop ends. Screenshot of the program and its output is attached.

Answer:

packet filtering

Explanation:

We can use a packet filtering firewall, for something like this, reasons because when visiting a site these types of firewalls should block all incoming traffic and analyze each packet, before sending it to the user. So if the packet is coming from a malicious origin, we can then drop that packet and be on our day ;D

If my computer has 4 gigabytes of ram memory then I have 4e+9 bytes of memory.  I think-

Answer: That isn't a software problem, its a hardware problem. There is a difference between 12gb of ram versus 100 gb of HDD. Ram is temporary storage, you can not turn SSD or HDD storage into RAM. If you are a gamer, STEER AWAY FROM LAPTOPS, DO NOT BUY A GAMING LAPTOP. Either buy a DESKTOP (or build) or a console. They give more performance for the price and they are easily upgradable. Not many people can open up a laptop and put in more RAM because the hardware is incredibly small.

Answer:

Linked lists and arrays are both linear data structures but while an array is a collection of items that can be accessed randomly, a linked list can be accessed sequentially.

A sparse matrix contains very few non-zero elements. For example;

_                        _

|  0   0   3  0  6    |

|   0   5   0  0  4   |

|   2   0   0  0  0   |

|_ 0   0   0  0  0 _|

In the implementation of a sparse matrix, the following are some of the pros and cons of using a linked list over an array;

PROS

i.  Linked lists are dynamic in nature and are readily flexible - they can expand and contract without having to allocate and/or de-allocated memory compared to an array where an initial size might need to be set and controlled almost manually. This makes it easy to store and remove elements from the sparse matrix.

ii. No memory wastage. Since the size of a linked list can grow or shrink at run time, there's no memory wastage as it adjusts depending on the number of items it wants to store. This is in contrast with arrays where you might have unallocated slots. Also, because the zeros of the sparse matrix need not be stored when using linked lists, memory is greatly conserved.

CONS

i. One of the biggest cons of linked lists is the difficulty in traversing items. With arrays, this is just of an order of 0(1) since the only requirement is the index of the item. With linked lists, traversal is sequential which means slow access time.

ii. Storage is another bottle neck when using linked lists in sparse matrix implementation. Each node item in a linked list contains other information that needs to be stored alongside the value such as the pointer to the next or previous item.

Answer:

link encryption

Explanation:

In e-mail, SSL/TLS provides link encryption.

The SSL is an acronym for Secured Socket Layer and it is one of the secured way of authenticating and encrypting data between a computer and the mail server.

In the case of a TLS, it is an acronym for Transport Layer Security and it basically is used for providing authentication and encryption of data between two communicating systems on a network.

This ultimately implies that, SSL/TLS are standard network protocols that provides data integrity and privacy to users when communicating over the internet or networking devices as they're made to encrypt user credentials and data from unauthorized access. The SSL/TLS are an application layer protocol used for the encryption of mails sent over the internet, in order to protect user information such as username and password.

Answer:

A. LCD power inverter

B. AC adapter

C. Battery

Answer:

(b) is used in Traceroute

(d) is used by Ping

Explanation:

ICMP is the short form of Internet Control Message Protocol. It is a protocol used by networking devices such as routers to perform network diagnostics and management. Since it is a messaging protocol, it is used for sending network error messages and operations information. A typical message could be;

i. Requested service is not available

ii. Host could not be reached

ICMP does not use ports. Rather it uses types and codes. Some of the most common types are echo request and echo reply.

Traceroute - which is a diagnostic tool - uses some messages available in ICMP (such as Time Exceeded) to trace a network route.

Ping - which is an administrative tool for identifying whether a host is reachable or not - also uses ICMP. The ping sends ICMP echo request packets to the host and then waits for an ICMP echo reply from the host.

ICMP is not required to solve NAT traversal problem neither does it have a new version in IPV6.

Answer:

double ComputeGasVolume(double pressure, double temperature, double moles){

   double volume = moles*GAS_CONST*temperature/pressure;

    return volume;        

}

Explanation:

You may insert this function just before your main function.

Create a function called ComputeGasVolume that takes three parameters, pressure, temperature, and moles

Using the given formula, PV = nRT, calculate the volume (V = nRT/P), and return it.

Answer:

I believe it is A

Explanation:

Answer:

Following are the description of the given points:

Explanation:

A) The Multiple greedy approaches could exist, which could be to reach for the closest emergency diamond, and yet clearly we are going to miss some very essential routes in that case. So, it can make every argument quickly, and seek to demonstrate a reference case for that.

B) An approach to the evolutionary algorithm and its users are going to just have states M x N. But each state (i, j) indicates the absolute difference between the amount of the selected diamond and the amounts of the costs incurred.

DP(i, j) = DimondVal(i, j) + max ((DP(i, j-1)-2), (DP(i-1,j-1)-3))

DP(i, j) is a description of the state.

DimondVal(i, j) is the diamond value at (i j), 0 if there is no diamond available.

This states must calculate the states of M N  and it involves continuous-time for each State to determine. Therefore the amount of time of such an algo is going to be O(MN).

Answer:

Here is the C++ program for Employee class with fields id,name and sal.

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

#include <string>  //to manipulate and use strings

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

class Employee {  //class Employee

private:  

/* the following data members are declared as private which means they can only be accessed by the functions within Employee class */

  string name;  //name field

  int id; //id field

  double sal;   //salary field

public:    

  Employee();  // constructor that initializes an object when it is created

/* setName, setID and setSalary are the mutators which are the methods used to change data members. This means they set the values of a private fields i.e. name, id and sal */

  void setName(string n)  //mutator for name field

     { name = n; }        

  void setId(int i)  //mutator for id field

     { id = i; }        

  void setSalary(double d)  //mutator for sal field

     { sal = d; }  

/* getName, getID and getSalary are the accessors which are the methods used to read data members. This means they get or access the values of a private fields i.e. name, id and sal */

  string getName()  //accessor for name field

     { return name; }        

  int getId()  //accessor for id field

     { return id; }        

  double getSalary()  //accessor for sal field

     { return sal; }  };  

Employee::Employee() {  //default constructor where the fields are initialized

  name = "";  // name field initialized

  id = 0;  // id field initialized to 0

  sal = 0;   }   // sal field initialized to 0

void display(Employee);  

// prototype of the method display() to display the data of Employee

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

  Employee emp;  //creates an object emp of Employee class

/*set the name field to Abc Xyz which means set the value of Employee class name field to Abc Xyz  through setName() method and object emp */

  emp.setName("Abc Xyz");  

/*set the id field to 1234 which means set the value of Employee class id field to 1234  through setId() method and object emp */

  emp.setId(1234);

/*set the sal field to 1000 which means set the value of Employee class sal field to 1000  through setSalary() method and object emp */

  emp.setSalary(1000);    

  display(emp);  }   //calls display() method to display the Employee data

void display(Employee e) {  // this method displays the data in the Employee //class object passed as a parameter.

/*displays the name of the Employee . This name is read or accessed through accessor method getName() and object e of Employee class */

  cout << "Name: " << e.getName() << endl;  

/*displays the id of the Employee . This id is read or accessed by accessor method getId() and object e */

  cout << "ID: " << e.getId() << endl;

/*displays the salary of the Employee . This sal field is read or accessed by accessor method getSalary() and object e */

  cout << "Salary: " << e.getSalary() << endl;  }

Explanation:

The program is well explained in the comments mentioned with each statement of the program.

The program has a class Employee which has private data members id, name and sal, a simple default constructor Employee(), mutatator methods setName, setId and setSalary to set the fields, acccessor method getName, getId and getSalary to get the fields values.

A function display( ) is used to display the Employee data i.e. name id and salary of Employee.

main() has an object emp of Employee class in order to use data fields and access functions defined in Employee class.

The output of the program is:

Name: Abc Xyz                                                                                                      

ID: 1234                                                                                                                  

Salary: 1000

The program and its output are attached.