You are writing a paragraph in a word processor. You
want to use the same text that appears elsewhere in a
different document, but you want to keep it in the other
document. To get the other text into the new document
most efficiently, you should:
Cut and paste
Ctrl + S
Copy and paste
Print the text and retype it so you don't make any mistakes.

Answer:

d. Delivers increased bandwidth using fewer devices and energy.

Explanation:

Alcatel-Lucent, formed in 1919 was a French global telecommunications equipment manufacturing company with its headquarter in Paris, France.

They provide services such as telecommunications and hybrid networking solutions deployed both in the cloud and properties.

Alcatel-Lucent's High Leverage Network (HLN) increases bandwidth and network capabilities while reducing the negative impact on the environment. This works because the High Leverage Network (HLN) delivers increased bandwidth using fewer devices and energy on Internet Protocol (IP) networks.

The Alcatel-Lucent's High Leverage Network (HLN) provides reduced cost of transmitting data as fewer network equipments are used with less adverse effects on the environment.

The High Leverage Network (HLN) when successfully implemented helps telecom firms to improve their operational efficiency, maintenance costs, and enhance network performance and capacity to meet the bandwidth demands of their end users.

Answer: Provided in the explanation section

Explanation:

(a). BIND is a type of DNS server used on the internet, actually it is reported to be most widely used DNS server on the internet. Also BIND is the de-facto standard on Unix like operating systems and Linux. Thus it can be used to locate computers on a network using domain names instead of their IP addresses. BIND was originally programmed at the University of California, Berkeley in the 1980s, this was made possible by a grant from US based DARPA program.

Originally, BIND stood for Berkeley Internet name daemon, but nowadays it is called Berkeley Internet name domain.

(b). In some of the clients in windows and Windows servers we will install LDP. By the LDP.exe we can test tDNS, Active directory woking , even we can add or modify. LDP is used to test and verify the Active Directory objects. We will install in windows machine then open the software, then from the menu we will select connect and type the IP adress of the Active Directory then we will select or type the objects of Active directory. For Linux system we have to openldap and can search the AD.

From Linux or Unix we can use the nmap tool which can test the network connectivity between all the computers whether it is windows or Linux. By nmap we can test by ICMP protocol to verify the network connectivity. By nmap we can scan all the port which are required to open from client to the server. We can also use nslookup to test DNS server from the client. By nslookup we can check that client can connect the server.

Answer:

The algorithm takes 346 ms to run on an input array with 1500 rows and 4096 columns.

Explanation:

For an input array with 1000 rows and 512 columns, the algorithm takes 173 ms.

We want to find out how long will the algorithm take to run on an input array with 1500 rows and 4096 columns?

Let the algorithm take x ms to run 1500 rows and 4096 columns.

For an input of n rows and m columns, it takes

[tex]n \: log_{2} \:m[/tex]

So,

[tex]1000 \: log_{2} \:512 = 173 \\\\1000 \: log_{2} \:2^{9} = 173 \:\:\: ( 2^{9} = 512) \\\\1000 \times 9 = 173 \:\:\:\:\: \: \: eg. 1[/tex]

and

[tex]1500 \: log_{2} \:4096 = x \\\\1500 \: log_{2} \:2^{12} = x \:\:\: ( 2^{12} = 4096) \\\\1500 \times 12 = x \:\:\:\:\: \: \: eg. 2[/tex]

Now divide the eq. 2 by eq. 1 to find the value of x

[tex]\frac{1500 \times 12}{1000 \times 9} = \frac{x}{173} \\\\\frac{18000 }{9000} = \frac{x}{173} \\\\2 = \frac{x}{173} \\\\x = 2 \times 173 \\\\x = 346 \: ms[/tex]

Therefore, the algorithm takes 346 ms to run on an input array with 1500 rows and 4096 columns.

The bitmap is 1110 0000 0001 0000 after some use. The system always looks for free blocks beginning using a contiguous allocation technique.

After some use, the new bitmap is 1110 0000 0001 0000. The system employs a contiguous allocation mechanism and starts searching for free blocks at the position.

(a) File B is made up of five blocks, 1111 1111 1111 0000.

b) The batch number 1000 0001 1111 0000 deletes File A.

c) Eight blocks are used for writing in file C.

1111 1111 1111 1100

File B is deleted (d). 1111 1110 0000 1100

Free-space bitmaps are one method used by some file systems to keep track of allocated sectors. Even though the most fundamental implementation of free-space bitmaps is incredibly wasteful, some modern file systems use complex or hybrid versions of them.

Therefore, after some use, the bitmap is 1110 0000 0001 0000. The contiguous allocation strategy is always used by the system to look for free blocks to start with.

To learn more about bitmap, refer to the link:

https://brainly.com/question/26230407

#SPJ2

Answer:

The pseudocode is as given below while the flowchart is attached.

Explanation:

The pseudocode is as follows

input customer name, number of texts  

Set Basic bill=5 $;

if the number of texts is less than or equal to 60

Extra Charge=0;

If the number of texts is greater than 60 and less than 200

number of texts b/w 60 and 200 =number of texts-60;

Extra Charge=0.1*(number of texts b/w 60 and 200);

else If the number of texts is greater than 200

number of texts beyond 200 =number of texts-200;

Extra Charge=0.25*(number of texts beyond 200)+0.1*(200-60);

Display Customer Name

Total Bill=Basic bill+Extra Charge;

Total Bill after Tax=Total Bill*1.12;

Answer:

Explanation:

The objective of this question and what we are meant to do is to compute a short program in C++ that will display the information given in the question.

Let get started!.

C ++ Program Code.

#include <iostream>

#include<string>

using namespace std;

int main()

{

string name,email,hometown,highschool,previousColleges,mathOrCSCource,lastMathOrCSCource,CompType,ExtraCur,FavBook

,planAfterGrad,phone;

//Reading Data

cout<<"Name:";

getline(cin,name);

cout<<"Phone:";

getline(cin,phone);

cout<<"Email:";

getline(cin,email);

cout<<"HomeTown:";

getline(cin,hometown);

cout<<"High School(s):";

getline(cin,highschool);

cout<<"Previous Colleges:";

getline(cin,previousColleges);

cout<<"List of math/CS Cources:";

getline(cin,mathOrCSCource);

cout<<"When,Where and what was your last math or CS course:";

getline(cin,lastMathOrCSCource);

cout<<"Type(s) of computer you are confident working with:";

getline(cin,CompType);

cout<<"ExtraCurricular Activities (job,club,sports,etc.):";

getline(cin,ExtraCur);

cout<<"Favourite books,movies,music:";

getline(cin,FavBook);

cout<<"What you plan to do after Graduation:";

getline(cin,planAfterGrad);

//Printing data

cout<<"--------------------Printing Data---------------------------"<<endl;

cout<<"Name:"<<name<<endl;

cout<<"Phone:"<<phone<<endl;

cout<<"Email:"<<email<<endl;

cout<<"HomeTown:"<<hometown<<endl;

cout<<"High School(s):"<<highschool<<endl;

cout<<"Previous Colleges:"<<previousColleges<<endl;

cout<<"List of math/CS Cources:"<<mathOrCSCource<<endl;

cout<<"When,Where and what was your last math or CS course:"<<lastMathOrCSCource<<endl;

cout<<"Type(s) of computer you are confident working with:"<<CompType<<endl;

cout<<"ExtraCurricular Activities (job,club,sports,etc.):"<<ExtraCur<<endl;

cout<<"Favourite books,movies,music:"<<FavBook<<endl;

cout<<"What you plan to do after Graduation:"<<planAfterGrad<<endl;

return 0;

}

Answer:

Following are the algorithm to this question:

Find_Heavier_Coins(Coin[9]):

   i) Let Coin[] Array represent all Coins.  

   ii) Divide coin[] into 3 parallel bundles and each has three coins, example 0-2, 3-5, 6-8 indicate a1 a2 a3

   iii) Randomly select any two bundles and place them in balance [say a1 and a2]

   iv) If Weigh of a1 and a2 has same:

           // checking that if a3 has heavier coin

           Choose any two 6-8 coins and place them onto balance [say 6 and 8]

           If those who weigh has the same weight:

               Third coin is coin heavier [say 7]

           else:  

               The coin [6 or 8] is the one which is heavier on the balance

       else:

           //The coin has the package which would be heavier on the balance [say a1]

           Select any two coins on balance from of the heavier package [say 0 and 1]

   If they weigh the same weight:

       Third coin is coin heavier [say 2]

   else:

       The coin that is heavier on the balance is the [or 0]

Explanation:

In the above-given algorithm code, a method Find_Heavier_Coins is declared which passes a coin[] array variable in its parameters. In the next step, if conditional block is used that checks the values which can be described as follows:

Answer:

a. starting balance

Explanation:

A program that calculates the current balance of a savings should ask the user for the starting balance and then ask for the annual interest rate. A loop should then iterate once for every month of the savings period in order to perform other tasks such as asking the user for the total amount deposited into the account, the total amount withdrawn from the account, and calculate the interest rate. The program will then proceed to display the result at the end of the savings period.

Answer:

Numeric Addition

Explanation:

Literally, the + sign is an arithmetic operator used to perform addition operation between numeric datatypes (integer, float, decimal, double, long integer, etc...)

Its default built in function of the plus is sign is to add variables, constants and/or expressions.

The plus sign is used as follows;

a = 5;

b = 3;

c = a + b

-------------

a + b = 5 + 3 = 8

8 will be saved in variable c. This is only possible using the plus sign.

The plus sign also has other function such as string concatenation but this function is language dependent (i.e. some programming languages do not use + for string concatenation).

Conclusively, the built in function of the plus sign is for numeric addition

Answer:

Option (d) is correct

Explanation:

Previously saved capture files can be read using Wireshark. For this, select the File then open the menu. Then Wireshark will pop up the “File Open” dialog box.

Wireshark capture files, like the DemoCapturepcap file found in this lab, have a .pcapng extension, which stands for packet capture, next generation.

Answer:

None

Explanation:

These are all stereotypes. Sure, there are definitely some people who fit their stereotypes, but not all. It's just a generalization at the end of the day. I can't really agree with any of them due to the fact that it's all stereotyping.

Perhaps you feel differently, and believe that some of these example are true. I can't though, sorry. Hope this take helps.

external hard drive

external hard drive

Answer:

Amswer would be D

Explanation:

Answer:

Output device

Explanation:

Im pretty sure that correct

Explanation:

A transaction is a very small system unit that can contain many low-level tasks. A transaction in a database system should maintain, Atomicity, Consistency, Isolation, and Durability − these are commonly known as ACID properties − in order to ensure accuracy, completeness, and integrity of the data.

An example of a simple transaction is as below,  Suppose a bank employee transfers Rs 500 from A's account to B's account.

A’s Account

Open_Account(A)

Old_Balance = A.balance

New_Balance = Old_Balance - 500

A.balance = New_Balance

Close_Account(A)

B’s Account

Open_Account(B)

Old_Balance = B.balance

New_Balance = Old_Balance + 500

B.balance = New_Balance

Close_Account(B)

Answer:

15 minutes

Explanation:

this meeting is normally timeboxed to a maximum duration of 15 minutes.

Answer:

A.  15 minutes

B.  45 minutes

C.  10 minutes

D.  30 minutes

The Answer is A): 15 minutes

explanation:

PLZ Mark Me As Brainliest

Answer:

#main function start

def main():

#list to store names entered by user

name_list=[]

#variable to take input from user to keep rotating the loop and take input from user again and again

ch='y'

fname=""

lname=""

while ch=='y' or ch=='Y':#keep the loop taking input from user till user do not want to enter more

#take input from user

name=[]

fname=input("Enter first name: ")

lname=input("Enter last name: ")

name.append(fname)

name.append(lname)

#append the previously input first and last name in main list

name_list.append(name)

ch=input("Do you want to continue: (y/Y for yes)")

#main function end

#get file name from user

filename=input("Enter output file name")

#open file in write mode

file=open(filename,"w")

#keep loop add data in file till main list 'name_list' have

for i in name_list:

#Write data in file

file.write("%s, %s\r\n" % (i[0],i[1]))

#close the file

file.close()

#call main function

main()

Explanation:

input the above code and see output

Answer:

import java.util.Scanner;

public class UnitTesting {

   // Function returns origNum cubed

   public static int cubeNum(int origNum) {

       return origNum * origNum * origNum;

   }

   public static void main (String [] args) {

       System.out.println("Testing started");

       System.out.println("2, expecting 8, got: " + cubeNum(2));

       System.out.println("3, expecting 27, got: " + cubeNum(3));

       System.out.println("-1, expecting -1, got: " + cubeNum(-1));

       System.out.println("Testing completed");

       return;

   }

}

Explanation:

Added statements are highlighted.

Since the cubeNum function calculates the cube of the given number and we are asked to write two statements to test inputs 3 and -1, pass the parameters 3 and -1 to the function called cubeNum and print the results using print statements

Answer:

The java program is as follows.

import java.lang.*;

public class Numbers

{

   //variables to hold two numbers

   static int num1=-8, num2=-2;

   //method to return integer with higher magnitude

   public static int maxMagnitude(int userVal1, int userVal2)

   {

       if(Math.abs(userVal1) > Math.abs(userVal2))

           return userVal1;

       else    

           return userVal2;

   }

public static void main(String[] args) {

    int max = maxMagnitude(num1, num2);

 System.out.println("The number with higher magnitude among "+num1+ " and "+num2+ " is "+max);

}

}

OUTPUT

The number with higher magnitude among -8 and -2 is -8

Explanation:

1. Two integer variables are declared to hold the two numbers. The variables are declared at class level (outside main()) and declared static.

2. The variables are initialized with any random value.

3. The method, maxMagnitude(), takes the two integer parameters and returns the number with higher magnitude. Hence, return type of the method is int.

public static int maxMagnitude(int userVal1, int userVal2)

4. This method finds the number with higher magnitude using the Math.abs() method inside if-else statement.

               if(Math.abs(userVal1) > Math.abs(userVal2))

                   return userVal1;

        else    

                   return userVal2;

5. Inside main(), the method, maxMagnitude(), is called. This method takes the two variables as input parameters.

6. The integer returned by the method, maxMagnitude() is assigned to a local integer variable, max, as shown.

      int max = maxMagnitude(num1, num2);

7. The result is displayed to the user as shown.

System.out.println("The number with higher magnitude among "+num1+ " and "+num2+ " is "+max);

8. The class variables and the method, maxMagnitude(), are declared static so that they can be used and called inside main().

9. The variable, max, is called local variable since it can be used only within the main() where it is declared. It is not declared static.

10. The class is declared public since it has the main() method and execution begins with main().

11. The program is saved with the same name as the name of the public class having main(). In this case, the program is saved as Numbers.java.

12. No user input is taken for the two input parameter values. The static variables can be assigned any integer value and the program can be tested accordingly.

Answer:

public static void PrintShampooInstructions(int numberOfCycles){

       if (numberOfCycles<1){

           System.out.println("Too Few");

       }

       else if(numberOfCycles>4){

           System.out.println("Too many");

       }

       else

           for(int i = 1; i<=numberOfCycles; i++){

               System.out.println(i +": Lather and rinse");

           }

       System.out.println("Done");

   }

Explanation:

I have used Java Programming language to solve this

Use if...elseif and else statement to determine and print "Too Few" or "Too Many".

If within range use a for loop to print the number of times

Answer:

#In Python

def shampoo_instructions(num_cycles):

   if num_cycles < 1: ///

       print ('Too few.') ///

   elif num_cycles > 4:

       print ('Too many.')

   else:

       i = 0

       while i<num_cycles:

           print (i+1,": Lather and rinse.")

           i = i + 1

       print('Done.')

user_cycles = int(input())

shampoo_instructions(user_cycles)

Explanation:

def shampoo_instructions(num_cycles): #def function with loop

   if num_cycles < 1:  #using 1st if statement

       print('Too few.')

   elif num_cycles > 4:

       print ('Too many.')

   else:

       i = 0

       while i<num_cycles:

           print (i+1,": Lather and rinse.")

           i = i + 1

       print('Done.')

user_cycles = int(input())

shampoo_instructions(user_cycles)

Find the given attachments

Answer:

This program is written in python programming language

Comments are used for explanatory purpose;

Take note of indentations(See Attachment)

Program starts here

#Initialize Sum to 0

sum = 0

#Prompt user for start value

start= int(input("Start Value: "))

#Prompt user for stop value

stop= int(input("Stop Value: "))

#Check if start is less than stop

if(start<=stop):

       #Iterate from start to stop

       for i in range(start, stop+1, 1):

              sum+=i**3

else:

print("Start must be less than or equal to Stop")

#Display Result

print("Expected Output: ",sum)

#End of Program

Answer:

yooooooooooooooooooooo

Explanation:

Answer:

The correct answer will be "Project Gutenberg".

Explanation:

So that the above is the right answer.

Answer:

130 bits

Explanation:

If the length of a key = N bits

Total number of key combinations = [tex]\frac{2^{N} }{2}[/tex]

For a key length of 100, the total number of key combinations will be  [tex]\frac{2^{100} }{2} = 2^{99}[/tex] combinations. This is a whole lot and it almost seems impossible to be hacked.

The addition of  one bit to the length of the symmetric key in 1 year is sufficient to make the key strong, because even at double its speed, the processor can still not crack the number of combinations that will be formed.

Therefore, if a bit is added to the key length per year, after 30 years, the length of the symmetric session key will be 130 bits.

Answer: provided in the explanation section

Explanation:

// code to copy

Car.java

public class Car {

  public void m1()

  {

  System.out.println("car 1");

}

  public void m2() {

      System.out.println("car 2");

}

  public String toString()

  {

  return "vroom";

}

}

Truck.java

public class Truck extends Car{

  public void m1()

  {

  System.out.println("truck 1");

  }

  public void m2()

  {

  super.m1();

  }

  public String toString()

  {

      return super.toString() + super.toString();

  }

}

MonsterTruck​​​​​​​.java

public class MonsterTruck extends Truck

{

  public void m1() {

System.out.println("monster 1");

}

public void m2() {

super.m1();

super.m2();

}

public String toString() {

return "monster " + super.toString();

}

public static void main(String[] args) {

  MonsterTruck mt=new MonsterTruck();

  mt.m1();

  mt.m2();

  System.out.println(mt);

}

}

cheers i hope this helped !!!

Filtering is a useful way to see only the data that you want displayed in Access databases.

Answer:

The database can include the fields related to the Author such as name and address, related to Publisher such as publisher's name and address, related to book such as title and ISBN of the book and related to contract such as payment schedule, contract start and end.

Explanation:

The following fields can be needed for this database:

Answer:

def output_ints_less_than_or_equal_to_threshold(user_values, upper_threshold):

  for value in user_values:

      if value < upper_threshold:

          print(value)  

def get_user_values():

  n = int(input())

  lst = []

  for i in range(n):

      lst.append(int(input()))

  return lst  

if __name__ == '__main__':

  userValues = get_user_values()

  upperThreshold = int(input())

  output_ints_less_than_or_equal_to_threshold(userValues, upperThreshold)

Explanation:

Answer: provided in the explanation section

Explanation:

The question says:

Write a short assembly language program in either our 8088 SCO DOSBox or 80386+ MASM Visual Studio 2017 environment that demonstrates data storage and retrieval from memory. As an example consider some value which is either 16 or 32 bits, after instantiating as an immediate value transfer to and retrieve from memory using assembly language instructions.

The Answer:

multi-segment executable file template. data segment ; add your data here! pkey db "press any key...$" ends stack segment dw 128 dup(0) ends code segment start: ; set segment registers: mov ax, data mov ds, ax mov es, ax ; add your code here mov cx,4 input: mov ah,1 int 21h push ax loop input mov dx,13d mov ah,2 int 21h mov dx,10d mov ah,2 int 21h mov cx,4 output: pop bx mov dl,bl mov ah,2 int 21h loop output exit: lea dx, pkey mov ah, 9 int 21h ; output string at ds:dx ; wait for any key....

mov ah, 1 int 21h mov ax, 4c00h ; exit to operating system. int 21h ends end start ;

set entry point and stop the assembler.

Cheers I hope this helps!!!

Answer:

The step by step explanation

Explanation:

Before doing anything, the first thing to do is to select the 100BaseTX network adapter and reconnect the Cat5e cable. The 100BaseTX network adapter supports Fast

Ethernet which is all that is required.

Now do the following steps:

Step one

In Office 2, switch to the motherboard view of the computer (turning off the workstation as necessary).

Step two

On the Shelf, expand the Network Adapters category.

Step three

Identify the network adapter that supports Fast Ethernet 100BaseTX. Drag the network adapter from the Shelf to a

free PCI slot on the computer.

Step four

To connect the computer to the network, switch to the back view of the computer.

Step five

Drag the Cat5 cable connector from the motherboard's NIC to the port of the 100BaseTX network adapter.

Step six

To verify the connection to the local network and the Internet, switch to the front view of the computer.

Step seven

Click the power button on the computer case.

Step eight

After the workstation's operating system is loaded, click the networking icon in the notification area and click

Open Network and Sharing Center. The diagram should indicate an active connection to the network and the

Internet.

For one to be able to confirm the speed of the connection, it can be done by clicking the Local Area Connection link in the Network and

Sharing Center.