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LUCKY_DIMON [66]
2 years ago
9

Lai worked on a global team for an American company, and all her work had to be completed in her second language, English. Somet

imes her teammates misinterpreted her meaning. Lai has unintentionally created of a(n) ______ barrier to communication.
A) Decoding
B) Sender
C) Feedback
D) Receiver
E) Encoding
Computers and Technology
2 answers:
STALIN [3.7K]2 years ago
5 0

Answer:

E) Encoding

Explanation:

    In a telephone conversation, a simple online conversation or a video conference conversation, for example, the communication situation is based on the exchange of messages from one point to another, provided that the message is linguistically encoded. This condition is no different in a simpler communicational situation: in any case, message coding refers to the organization of the terms that compose it in a logical system of recognizable (decodable) signs by a group of speakers. Message coding in linguistic communication is a conventional process that is pre-established among speakers of a language.

N76 [4]2 years ago
4 0

Answer:

E) Encoding.

Explanation:

Communication is important aspect of life.Communication barriers occurs at all stages of communication from when the sender sends the message to the receiver receiving the message. Lack of communication skill in that particular comes in encoding barrier.So we conclude that the answer is Encoding.

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Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the
zmey [24]

Answer:

part (a).

The program in cpp is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>0)

   {

       if(balance<payment)    

         { payment = balance;}

       else

       {  

           std::cout << balance <<"\t\t\t"<< payment << std::endl;

           balance = balance - payment;

       }

   }

   return 0;

}

part (b).

The modified program from part (a), is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   double charge=0.01;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   balance = balance +( balance*charge );

   std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>payment)

   {

           std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

           balance = balance +( balance*charge );

           balance = balance - payment;

       }

   if(balance<payment)    

         { payment = balance;}          

   std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

   return 0;

}

Explanation:

1. The variables to hold the loan balance and the monthly payment are declared as double.

2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.  

3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.

4. The computed values are displayed for each month till the loan balance becomes zero.

5. The output for both part (a) and part (b) are attached as images.

4 0
2 years ago
in a small office, there are 5 computers, a network printer, and a broadband connection to the internet. what devices are needed
Nataly [62]

Answer:

<em>Ethernet cables, Network Adapters, Modem, Routers, Switches.</em>

Explanation:

<em>The devices that are required in setting up a wired network for the 5 computer comprises of the following devices </em>

  • <em>Ethernet Cables</em>
  • <em>Network Adapters</em>
  • <em>Modem/Router</em>
  • <em>Network Switch</em>

<em>Ethernet cables: They are called network cables or RJ-45 cables used to in connecting two or more computers together. it has different categories called, the untwisted pair and twisted pair Ethernet, with a speed from 10-1000</em>

<em>Network Adapters : This adapters allows a computer device to connect and interface with a network computer</em>

<em>Modem/Routers : A router is a device that that sits in the  middle  between your local computers and modems. it takes receives information or gets information from the modem and delivers it to the computer</em>

<em>Network switch: it connects more than two computers together to a network and share data among themselves and other devices on the network</em>

6 0
2 years ago
An automobile battery, when connected to a car radio, provides 12.5 V to the radio. When connected to a set of headlights, it pr
ladessa [460]
Base on the question, and in my further computation, the possible answers would be the following and I hope you are satisfied with my answer and feel free to ask for more.

- If you want to determine the Thevenin equivalent voltage and resistance without overloading the battery, then apply some known resistance 

<span><span>RL</span><span>RL</span></span> and measure the output voltage as <span><span>VL</span><span>VL</span></span>. Measure the voltage without a load as <span><span>V<span>OC</span></span><span>V<span>OC</span></span></span>. The voltage divider equation tells us that

<span><span><span>VL</span>=<span>V<span>OC</span></span><span><span>RL</span><span><span>R<span>TH</span></span>×<span>RL</span></span></span></span><span><span>VL</span>=<span>V<span>OC</span></span><span><span>RL</span><span><span>R<span>TH</span></span>×<span>RL</span></span></span></span></span>

Solve for <span><span>R<span>TH</span></span><span>R<span>TH</span></span></span>, and you know that <span><span><span>V<span>TH</span></span>=<span>V<span>OC</span></span></span><span><span>V<span>TH</span></span>=<span>V<span>OC</span></span></span></span>.

6 0
2 years ago
Temperature Class Write a Temperature class that will hold a temperature in Fahrenheit and provide methods to get the temperatur
posledela

Answer:

Explanation:

   public class Temperature

   {

       double ftemp;

       public int Constructor(double fahrenheit)

       {

           ftemp = fahrenheit;

          return Convert.ToInt32(ftemp);

       }

       public void setFahrenheit(double fahrenheit)

       {

           ftemp = fahrenheit;

       }

       public void getFahrenheit()

       {

           ftemp = Constructor(ftemp);

       }

       public void getCelcius()

       {

           ftemp = (ftemp - 32) * 5 / 9;

       }

       public void getKelvin()

       {

           ftemp = (ftemp - 32) * 5 / 9 + 273.15;

       }

   }

8 0
2 years ago
If there are 8 opcodes and 10 registers, a. What is the minimum number of bits required to represent the OPCODE? b. What is the
Nimfa-mama [501]

Answer:  

For 32 bits Instruction Format:

OPCODE   DR               SR1                   SR2      Unused bits

a) Minimum number of bits required to represent the OPCODE = 3 bits

There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.

Ceil (log2 (8)) = 3

b) Minimum number of bits For Destination Register(DR) = 4 bits

There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value.  4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.  

Ceil (log2 (10)) = 4

c) Maximum number of UNUSED bits in Instruction encoding = 17 bits

Total number of bits used = bits used for registers + bits used for OPCODE  

     = 12 + 3 = 15  

Total  number of bits for instruction format = 32  

Maximum  No. of Unused bits = 32 – 15 = 17 bits  

OPCODE                DR              SR1             SR2              Unused bits

  3 bits              4 bits          4 bits           4 bits                17 bits

7 0
2 years ago
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