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Wittaler [7]
3 years ago
15

A communication medium which allows receivers to observe multiple cues, such as body language and tone of voice, and allows send

ers to get feedback, is called a _____________ medium.
Computers and Technology
1 answer:
castortr0y [4]3 years ago
4 0

Answer: Rich medium

Explanation: A communication is said to be rich id it provides the services like observing the body language, immediate communication, instant judging of the voice tone etc. These factors are commonly found in the face to face interaction which is considered as the rich source of communication. It is considered as rich medium because  it has the capability of receiving the output immediately .

You might be interested in
How many bits does it take to store a 3-minute song using an audio encoding method that samples at the rate of 40,000 bits/secon
blondinia [14]

Answer:

115200000 bits

Explanation:

Given Data:

Total Time = 3 minute = 3 x 60 sec = 180 sec

Sampling rate = 40,000 bits / sample

Each sample contain bits = 16 bits /sample

Total bits of song = ?

Solution

Total bits of song = Total Time x Sampling rate x Each sample contain bits

                             = 180 sec x 40,000 bits / sec x 16 bits /sample

                             = 115200000 bits

                             =  115200000/8 bits

                             = 14400000 bytes

                             = 144 MB

There are total samples in one second are 40000. Total time of the song is 180 seconds and one sample contains 16 bits. so the total bits in the song are 144 MB.

5 0
3 years ago
In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. Whe
never [62]

Answer:

The equivalent program in C++:

#include<iostream>

#include <sstream>

using namespace std;

int main(){

   string Score, Rank;

   cout<<"Enter student score and class rank: ";

   cin>>Score>>Rank;

   int testScore = 0, classRank = 0;

   stringstream sstream(Score);

   sstream>>testScore;

   

   stringstream tream(Rank);

   tream>>classRank;

   

   if (testScore >= 90){

       if(classRank >=25){cout<<"Accept";}

       else{cout<<"Reject";}

   }

   else if(testScore >= 80){

       if(classRank >=50){cout<<"Accept";}

       else{cout<<"Reject";}

   }

   else if(testScore >= 70){

       if(classRank >=75){cout<<"Accept";}

       else{cout<<"Reject";}

   }

   else{cout<<"Reject";}

   return 0;

}

Explanation:

This declares Score and Rank as string variables

   string Score, Rank;

This prompts the user for score and class rank

   cout<<"Enter student score and class rank: ";

This gets the user input

   cin>>Score>>Rank;

This declarees testScore and classRank as integer; and also initializes them to 0

   int testScore = 0, classRank = 0;

The following converts string Score to integer testScore

<em>    stringstream sstream(Score);</em>

<em>    sstream>>testScore;</em>

The following converts string Rank to integer classRank

   stringstream tream(Rank);

   tream>>classRank;

The following conditions implement the conditions as given in the question.    

If testScore >= 90

<em>    if (testScore >= 90){</em>

If classRank >=25

<em>        if(classRank >=25){cout<<"Accept";}</em>

If otherwise

<em>        else{cout<<"Reject";}</em>

<em>    } ---</em>

If testScore >= 80

<em>    else if(testScore >= 80){</em>

If classRank >=50

<em>        if(classRank >=50){cout<<"Accept";}</em>

If otherwise

<em>        else{cout<<"Reject";}</em>

<em>    }</em>

If testScore >= 70

<em>    else if(testScore >= 70){</em>

If classRank >=75

<em>        if(classRank >=75){cout<<"Accept";}</em>

If otherwise

<em>        else{cout<<"Reject";}</em>

<em>    }</em>

For testScore less than 70

<em>    else{cout<<"Reject";}</em>

<em />

3 0
2 years ago
Binary is a base-2 number system instead of the decimal (base-10) system we are familiar with. Write a recursive function PrintI
Paladinen [302]

Answer:

In C++:

int PrintInBinary(int num){

if (num == 0)  

 return 0;  

else

 return (num % 2 + 10 * PrintInBinary(num / 2));

}

Explanation:

This defines the PrintInBinary function

int PrintInBinary(int num){

This returns 0 is num is 0 or num has been reduced to 0

<em> if (num == 0)  </em>

<em>  return 0;  </em>

If otherwise, see below for further explanation

<em> else </em>

<em>  return (num % 2 + 10 * PrintInBinary(num / 2)); </em>

}

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

num % 2 + 10 * PrintInBinary(num / 2)

The above can be split into:

num % 2 and + 10 * PrintInBinary(num / 2)

Assume num is 35.

num % 2 = 1

10 * PrintInBinary(num / 2) => 10 * PrintInBinary(17)

17 will be passed to the function (recursively).

This process will continue until num is 0

7 0
3 years ago
Please tell fast plzzzzzz.​
lana [24]

Answer:

True

Explanation:

4 0
2 years ago
Given parameters b and h which stand for the base and the height of an isosceles triangle (i.e., a triangle that has two equal s
schepotkina [342]

Answer:

The area of the triangle is calculated as thus:

Area = 0.5 * b * h

To calculate the perimeter of the triangle, the measurement of the slant height has to be derived;

Let s represent the slant height;

Dividing the triangle into 2 gives a right angled triangle;

The slant height, s is calculated using Pythagoras theorem as thus

s = \sqrt{b^2 + h^2}

The perimeter of the triangle is then calculated as thus;

Perimeter = s + s + b

Perimeter = \sqrt{b^2 + h^2} + \sqrt{b^2 + h^2} +b

Perimeter = 2\sqrt{b^2 + h^2} + b

For the volume of the cone,

when the triangle is spin, the base of the triangle forms the diameter of the cone;

Volume = \frac{1}{3} \pi * r^2 * h

Where r = \frac{1}{2} * diameter

So, r = \frac{1}{2}b

So, Volume = \frac{1}{3} \pi * (\frac{b}{2})^2 * h

Base on the above illustrations, the program is as follows;

#include<iostream>

#include<cmath>

using namespace std;

void CalcArea(double b, double h)

{

//Calculate Area

double Area = 0.5 * b * h;

//Print Area

cout<<"Area = "<<Area<<endl;

}

void CalcPerimeter(double b, double h)

{

//Calculate Perimeter

double Perimeter = 2 * sqrt(pow(h,2)+pow((0.5 * b),2)) + b;

//Print Perimeter

cout<<"Perimeter = "<<Perimeter<<endl;

}

void CalcVolume(double b, double h)

{

//Calculate Volume

double Volume = (1.0/3.0) * (22.0/7.0) * pow((0.5 * b),2) * h;

//Print Volume

cout<<"Volume = "<<Volume<<endl;

}

int main()

{

double b, h;

//Prompt User for input

cout<<"Base: ";

cin>>b;

cout<<"Height: ";

cin>>h;

//Call CalcVolume function

CalcVolume(b,h);

//Call CalcArea function

CalcArea(b,h);

//Call CalcPerimeter function

CalcPerimeter(b,h);

 

return 0;

}

3 0
3 years ago
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