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LUCKY_DIMON [66]
3 years ago
14

Which of the following is benefit of information communication technology?

Computers and Technology
2 answers:
nekit [7.7K]3 years ago
7 0

Answer:

cost reduction

Explanation:

apex

igor_vitrenko [27]3 years ago
5 0

Answer:

C. Cost reduction

Explanation:

ICT covers all the products that can store, retrieve, transmit or receive and manipulate the information via the electronic medium in a binary or digital mode. As an example, we can consider the Personal computer, digital television, robots or email as well as the digital communication technologies that allow the organizations and people to talk and share the information between them in a digital manner. And this is cost-effective, there is no employee benefit, increased use of postal mail or less technology. The only advantage is cost reduction. And hence C, Cost reduction is the right answer.

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1.1 what is the osi security architecture? 1.2 what is the difference between passive and active security threats? 1.3 list and
ioda

The well-designed standard architecture for security features in computer networking is defined by the OSI Security Architecture.

The OSI architecture is widely accepted because it establishes the process for ensuring safety in an organization. In a live attack, the victim is made aware of it. The victim of a passive attack is not made aware of the attack. System resources can be modified during an active attack. System resources are not changing while under passive attack. Monitoring a system is the focus of passive assaults, which do not require changing any data on the target system. On the system under attack, active threats will alter data.

Learn more about system here-

brainly.com/question/14253652

#SPJ4

6 0
1 year ago
Write a method so that the main() code below can be replaced by the simpler code that calls method mphAndMinutesToMiles(). Origi
defon

ANSWER

The JAVA program after simplification is as below.

import java.util.Scanner;

public class CalcMiles {

   

   // variables declaration

   static double milesPerHour;

   static double minutesTraveled;      

   static double hoursTraveled;

   static double milesTraveled;

   

   // method declared static

   public static void mphAndMinutesToMiles(double speed, double mins)

   {

       // computations to calculate distance travelled

       hoursTraveled = mins / 60.0;

       milesTraveled = hoursTraveled * speed;

       

       // result displayed on the screen  

     System.out.println("Miles: " + milesTraveled);

   }

   

   // Scanner object created inside main()

   // user input taken inside main()

   public static void main(String [] args)

   {

       Scanner scnr = new Scanner(System.in);

       System.out.println("Enter miles travelled per hour ");

       milesPerHour = scnr.nextDouble();

       System.out.println("Enter minutes required to travel ");

       minutesTraveled = scnr.nextDouble();

       

       mphAndMinutesToMiles(milesPerHour, minutesTraveled);

       

   }

}

OUTPUT

Enter miles travelled per hour  

2.3

Enter minutes required to travel  

1234

Miles: 47.30333333333333

EXPLANATION

The program is simplified as explained below.

1. User input is taken using the object of Scanner class.

2. This object of Scanner class can only be created inside main() method hence, user input can only be taken inside main().

3. The code to calculate the result is separated in another method, mphAndMinutesToMiles(). This method is declared with return type void since no value is returned.

4. After user input is taken in main() for miles travelled per hour and minutes required to travel, these values are passed as parameters to the mphAndMinutesToMiles() method.

5. These parameters are used in calculation part. The total miles travelled in total hours (obtained from minutes), is calculated. The code to display the result is added to this method.

6. Inside the main method, only the code to create Scanner object, code to take user input for two variables and code to call the mphAndMinutesToMiles() is included.

7. The variables are declared as static and declared at class level.

8. No variable is declared inside any of the two methods.

5 0
3 years ago
Read 2 more answers
45 points!!
pav-90 [236]

Answer:

ENQUIRE database system

Explanation:

5 0
3 years ago
Read 2 more answers
Suppose there are 69 packets entering a queue at the same time. Each packet is of size 7 MiB. The link transmission rate is 1.7
svp [43]

Answer:

69.08265412 milliseconds

Explanation:

Lets first convert 7 MiB to bits

7*1024*1024*8=58720256 bits

Now convert bits to Gbits

58720256/10^{9}  =0.058720256 Gbits

Queuing Delay = Total size/transmission link rate

Queuing Delay= \frac{0.058720256}{1.7} =0.03454132706 seconds

Delay of packet number 3 = 0.03454132706*2=0.06908265412 seconds

or 0.06908265412= 69.08265412 milliseconds

7 0
3 years ago
What is the processing speed for the second generation of computers​
Helga [31]

10mbp

I hope it helps you

5 0
3 years ago
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