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Makovka662 [10]
3 years ago
13

What is the preferred procedure for putting customers on hold​

Computers and Technology
2 answers:
vesna_86 [32]3 years ago
8 0

Answer/Explanation:

Explain the reason you need to put the caller on hold.

Ask for the person's phone number, in case the conversation is cut off.

Promise to return in a minute—or your best estimate of how long the hold will last. If you think the delay will be longer than a couple of minutes, ask for a number and a time when you can call back.

If, despite your best efforts, the delay is taking longer than anticipated, check in with the caller so that he doesn't feel abandoned. Again, offer to call back, "rather than keep you holding."

When you return to the phone, apologize for the inconvenience and thank the caller for his patience.

Use the caller's name frequently to add a sense of connection.

The first step you can take when putting a caller on hold and making it a satisfying experience is a systematic one. When a caller is on hold, 60 seconds seems like an eternity, so see if you can program the phone system to play music (calming classical or jaunty jazz is best) to break this perception of an interminable wait. Aside from that, however, putting someone on hold in a professional way depends on your behavior. The golden rule to keep in mind is as follows: how would you like to be treated if you were on the other end of the line?

MAKE ME THE BRAINLIEST!!!!!!!!!!!!

Semmy [17]3 years ago
6 0

Answer: Persuading them!!

Explanation:

Maybe have a T.V near the costumers to watch while moving so they can not get bored IDK if your making a store business but this is all I have hope it helps.. plz give me BRAINLIEST!!

I'm 2 brainliest away!..

You might be interested in
Part 1: Create an application that allows you to enter student data that consists of an ID number, first name, last name, and gr
aleksley [76]

<u>PART A</u>

import java.io.*;

import java.util.Scanner;

public class StudentsStanding {

   private static final double CUT_OFF = 2.0;

   public static void main(String[] args)

   {

       try {

           PrintWriter goodFile = new PrintWriter(new FileWriter("goodStanding.txt"));

           PrintWriter probationFile = new PrintWriter(new FileWriter("probation.txt"));

           char choice;

           String firstName,lastName;

           Scanner input = new Scanner(System.in);

           int id;

           double gradePoint;

           do {

               System.out.print("Enter student Id: ");

               id = input.nextInt();

               input.nextLine();

               System.out.print("Enter First name: ");

               firstName = input.nextLine();

               System.out.print("Enter last name: ");

               lastName = input.nextLine();

               System.out.print("Enter grade point: ");

               gradePoint = input.nextDouble();

               if(gradePoint<CUT_OFF)

                   probationFile.println(id+","+firstName+","+lastName+","+gradePoint);

               else

                   goodFile.println(id+","+firstName+","+lastName+","+gradePoint);

               System.out.print("Do you want to continue..(y/n): ");

               choice = input.next().toLowerCase().charAt(0);

           }while (choice!='n');

           //close the streams

           goodFile.close();

           probationFile.close();

       }

       catch (IOException ex)

       {

           System.err.println("IO Exception occurs...");

       }

   }

}

<u>PART 2</u>

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class StudentsStanding2 {

   private static final double CUT_OFF = 2.0;

   public static void main(String[] args)

   {

       String header = "ID\t"+"First Name"+"\t"+"Last Name"+"\t"+"Grade Point"+"\t"+"Exceeds/Fall\n";

       try {

           Scanner probationFile= new Scanner(new File("probation.txt"));

           Scanner goodFile = new Scanner(new File("goodStanding.txt"));

           System.out.println("Cut Off is: "+CUT_OFF);

           System.out.println(header);

           double point ,exceed;

           while (probationFile.hasNextLine())

           {

               String[] tokens = probationFile.nextLine().split(",");

               point = Double.parseDouble(tokens[3]);

               exceed = CUT_OFF-point;

               System.out.println(String.format("%4s %7s %10s %10s %10s",tokens[0],tokens[1],tokens[2],point,String.format("%.2f",exceed)));

           }

           //Close the stream

           probationFile.close();

           while (goodFile.hasNextLine())

           {

               String[] tokens = goodFile.nextLine().split(",");

               point = Double.parseDouble(tokens[3]);

               exceed = point - CUT_OFF;

               System.out.println(String.format("%4s %7s %10s %10s %10s",tokens[0],tokens[1],tokens[2],point,String.format("%.2f",exceed)));

           }

           goodFile.close();

       }

       catch (FileNotFoundException ex)

       {

           System.err.println("File not found!!!");

       }

   }

}

6 0
3 years ago
Live.com is Microsoft's free web-based email provider.<br> A. True<br> B. False
Zigmanuir [339]
True Live.com is Microsoft's free web- based email provider
5 0
2 years ago
A server-side extension ________________. provides its services to the web server in a way that is totally transparent to the cl
damaskus [11]
A: provides its services to the Web server in a way that is totally transparent to the client browser.
8 0
3 years ago
Microsoft's ____ is one of the major web-based development environments.
finlep [7]
The answer is ".NET" (had to increase character count to let me post).
4 0
3 years ago
Write a program that generates 1,000 random integers between 0 and 9 and displays the count for each number. (Hint: Use a list o
masya89 [10]

Answer:

import random

count0, count1, count2, count3,

count4, count5, count6, count7,

count8, count9, i = [0 for _ in range(11)]

while i < 1000:

   number = random.randint(0,9)

   if number == 0:

       count0 = count0 + 1

   if number == 1:

       count1 = count1 + 1

   if number == 2:

       count2 = count2 + 1

   if number == 3:

       count3 = count3 + 1

   if number == 4:

       count4 = count4 + 1

   if number == 0:

       count5 = count5 + 1

   if number == 6:

       count6 = count6 + 1

   if number == 7:

       count7 = count7 + 1

   if number == 0:

       count8 = count8 + 1

   if number == 9:

       count9 = count9 + 1

   

   i = i+1

print("0's: "+ str(count0) + "\n"+ "1's: "+ str(count1) + "\n"+

"2's: "+ str(count2) + "\n"+ "3's: "+ str(count3) + "\n"+

"4's: "+ str(count4) + "\n"+ "5's: "+ str(count5) + "\n"+

"6's: "+ str(count6) + "\n"+ "7's: "+ str(count7) + "\n"+

"8's: "+ str(count8) + "\n"+ "9's: "+ str(count9) + "\n")

Explanation:

- Initialize variables to hold the count for each number

- Initialize <em>i</em> to control the while loop

- Inside the while loop, check for the numbers and increment the count values when matched.

- Print the result

5 0
2 years ago
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