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3 years ago

Which type of interview is conducted in a format where the interviewee is questioned and presented to a panel of individuals?

2 answers:

5 0

A type of interview where the interviewee is questioned and presented to a panel of individuals is called a panel interview. These types of interviews are fairly common in specific situations where it is advantageous for the group of people to cross-interview the person in real-time. Something this happens in a job offer case for example.

tester [92]3 years ago

3 0

The <span>type of interview is conducted in a format where the interviewee is questioned and presented to a panel of individuals is called "Panel Interview." The interviewee is surrounded with 2 or more panelists that will ask questions that you need to answer.</span>

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This project involves writing a java program to simulate a blackjack card game. You will use a simple console-based user interfa

coldgirl [10]

Answer:i dont know sorry

Explanation:

8 0

4 years ago

If a user's input string matches a known text message abbreviation, output the unabbreviated form, else output: Unknown. Support

Molodets [167]

Answer:

Here is the JAVA program. I have added a few more abbreviations apart from LOL and IDK.

import java.util.Scanner; // to take input from user

public class MessageAbbreviation {

public static void main(String[] args) { //start of main() function body

Scanner input = new Scanner(System.in); // create object of Scanner

String abbreviation= ""; //stores the text message abbreviation

/* In the following lines the abbreviation string type variables contain the un-abbreviated forms */

String LOL = "laughing out loud";

String IDK = "i don't know";

String BFF = "best friends forever";

String IMHO = "in my humble opinion";

String TMI = "too much information";

String TYT = "take your time";

String IDC = "I don't care";

String FYI = "for your information";

String BTW = "by the way";

String OMG = "oh my God";

//prompts the user to enter an abbreviation for example LOL

System.out.println("Input an abbreviation:" + " ");

if(input.hasNext()) { // reads the abbreviation form the user

abbreviation= input.next(); }

// scans and reads the abbreviation from user in the abbreviation variable

/*switch statement checks the input abbreviation with the cases and prints the un abbreviated form in output accordingly. */

switch(abbreviation) {

case "LOL" : System.out.println(LOL);

break;

case "IDK" : System.out.println(IDK);

break;

case "BFF" : System.out.println(BFF);

break;

case "IMHO": System.out.println(IMHO);

break;

case "TMI": System.out.println(TMI);

break;

case "TYT": System.out.println(TYT);

break;

case "IDC": System.out.println(IDC);

break;

case "FYI": System.out.println(FYI);

break;

case "BTW": System.out.println(BTW);

break;

case "OMG": System.out.println(OMG);

break;

default : System.out.println("Unknown"); } } }

Explanation:

The program first prompts the user to enter an abbreviation. Lets say the user enters LOL. Now the switch statement has some cases which are the conditions that are checked against the input abbreviation. For example if the user enters LOL then the following statement is executed:

case "LOL" : System.out.println(LOL);

This means if the case is LOL then the message ( un abbreviated form) stored in the LOL variable is displayed on output screen So the output is laughing out loud.

Anything else entered other than the given abbreviations will display the message: Unknown

The output is attached as screen shot.

5 0

3 years ago

Read 2 more answers

Many documents use a specific format for a person's name. Write a program whose input is: firstName middleName lastName, and who

never [62]

Answer:

Python Program:

if __name__ == '__main__':

a = input("Enter Name: ")

b = a.split()

print(b[2] + ',' , b[0], b[1][0])

Explanation:

The input function will be used to prompt user to enter the name of the person. The person's name will be stored in variable a.

Let us assume that the user entered a name mentioned in the example, which is Pat Silly Doe, Now variable a = 'Pat Silly Doe'.

The function a.split() is used to split a string into a list. The splitting (by default) is done on the basis of white-space character. Therefore, a.split() will give a list

['Pat', 'Silly', 'Doe']

which will be later on stored in variable b.

We can use the subscript operator [] to access the values in a list. Suppose if we have a list a, then a[i] will give us ith element of a, if i < length of a.

Finally, we print the answer. b[2] will give us the last name of the person. We append a comma using '+' operator. b[0] will give us the first name. b[1] will give us the middle name, but since we only need one character from the middle name, we can use another subscript operator b[1][0] to give us the first character of the middle name.

Note: Characters of strings can be accessed using subscript operator too.

4 0

3 years ago

Program for bit stuffing...?

Olegator [25]

Answer: Program for bit stuffing in C

#include<stdio.h>

int main()

{

int i=0,count=0;

char data[50];

printf("Enter the Bits: ");

scanf("%s",data); //entering the bits ie. 0,1

printf("Data Bits Before Bit Stuffing:%s",databits);

printf("\nData Bits After Bit stuffing :");

for(i=0; i<strlen(data); i++)

{

if(data[i]=='1')

count++;

else

count=0;

printf("%c",data[i]);

if(count==4)

{

printf("0");

count=0;

}

return 0;

}

Explanation:

bit stuffing is the insertion of non-information bits during transmission of frames between sender and receiver. In the above program we are stuffing 0 bit after 4 consecutive 1's. So to count the number of 1's we have used a count variable. We have used a char array to store the data bits . We use a for loop to iterate through the data bits to stuff a 0 after 4 consecutive 1's.

4 0

3 years ago

Suppose a variable is passed By Value to a parameter of a Sub procedure, and the parameter has its value changed inside the Sub

algol [13]

Answer:

"The value of the variable will remain the same which is already have when the sub-processor is called".

Explanation:

The above question said that:-

void fun(int a)

{

a=a+1;

}

void main()

{

int a=5;

fun(a);

}

//what will be the value of a in the main function after the fun function is excuted.

Then the answer is: the value of a will be 5 in the main function.
It is because when the fun function is called, then a variable that is defined in the fun function is a local variable for fun function. That scope after the fun function is null.
The a variable inside the fun function is a different variable and the main function a variable is also a different variable.
So when the user prints the value of a variable inside the fun function, it will give the result as 6.
But when he prints the value of a variable inside the main function, then it will give the value as 5.

4 0

4 years ago