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stepan [7]
3 years ago
11

What is the most efficient way to control the type of information that is included in the .msg file when a user forwards a

Computers and Technology
1 answer:
Jet001 [13]3 years ago
8 0

Answer: Use the "As an Outlook Contact option.

Explanation:

The most efficient way to control the type of information which will be included in the .msg file when a contact is forwarded to another user by a user is to use the Use the "As an Outlook Contact option.

After clicking on the contact that you want to forward it to, then click on forward, click on contact and click on As an Outlook Contact. You can then complete the email message, after which you'll click on send.

You might be interested in
The sum of these 9 numbers is 36. 2, 2, 6, 2, 1, 8, 7, 5, 3 What is the mean of these 9 numbers?
fiasKO [112]

Answer:

4

Explanation:

The mean is the sum divided by the count of numbers, so 36/9 = 4

6 0
2 years ago
Read 2 more answers
"Write an iterative function iterPower(base, exp) that calculates the exponential baseexp by simply using successive multiplicat
sp2606 [1]

Answer:

I am writing the function using Python. Let me know if you want the program in some other programming language.            

def iterPower(base, exp):    

    baseexp = 1

    while exp > 0:

        baseexp = baseexp*base

        exp= exp - 1

    return baseexp

base = 3

exp = 2

print(iterPower(base, exp))

Explanation:

  • The function name is iterPower which takes two parameters base and exp. base variable here is the number which is being multiplied and this number is multiplied exponential times which is specified in exp variable.
  • baseexp is a variable that stores the result and then returns the result of successive multiplication.
  • while loop body keeps executing until the value of exp is greater than 0. So it will keep doing successive multiplication of the base, exp times until value of exp becomes 0.
  • The baseexp keeps storing the multiplication of the base and exp keeps decrements by 1 at each iteration until it becomes 0 which will break the loop and the result of successive multiplication stored in baseexp will be displayed in the output.
  • Here we gave the value of 3 to base and 2 to exp and then print(iterPower(base, exp)) statement calls the iterPower function which calculates the exponential of these given values.
  • Lets see how each iteration works:
  • 1st iteration

baseexp = 1

exp>0 True because exp=2 which is greater than 0

baseexp = baseexp*base

               = 1*3 = 3

So baseexp = 3

exp = exp - 1

      = 2 - 1 = 1    

exp = 1

  • 2nd iteration

baseexp = 3

exp>0 True because exp=1 which is greater than 0

baseexp = baseexp*base

               = 3*3 = 9

So baseexp = 9

exp = exp - 1

      = 1-1 = 0    

exp = 0

  • Here the loop will break now when it reaches third iteration because value of exp is 0 and the loop condition evaluates to false now.
  • return baseexp statement will return the value stored in baseexp which is 9
  • So the output of the above program is 9.
5 0
3 years ago
___________ are the constant values that are used in a program. ​
77julia77 [94]

Answer: Data

Explanation:

3 0
4 years ago
1. The following programs require using arrays. For each, the input comes from standard input and consists of N real numbers bet
Mamont248 [21]

Answer:

import java.util.*;

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.Arrays;

class GFG

{

  // Function for calculating mean

  public static double findMean(double a[], int n)

  {

      int sum = 0;

      for (int i = 0; i < n; i++)

          sum += a[i];

 

      return (double)sum / (double)n;

  }

  // Function for calculating median

  public static double findMedian(double a[], int n)

  {

      // First we sort the array

      Arrays.sort(a);

      // check for even case

      if (n % 2 != 0)

      return (double)a[n / 2];

 

      return (double)(a[(n - 1) / 2] + a[n / 2]) / 2.0;

  }

  public static double findMode(double a[], int n)

{

// The output array b[] will

// have sorted array

//int []b = new int[n];

 

// variable to store max of

// input array which will

// to have size of count array

double max = Arrays.stream(a).max().getAsDouble();

 

// auxiliary(count) array to

// store count. Initialize

// count array as 0. Size

// of count array will be

// equal to (max + 1).

double t = max + 1;

double[] count = new double[(int)t];

for (int i = 0; i < t; i++)

{

count[i] = 0;

}

 

// Store count of each element

// of input array

for (int i = 0; i < n; i++)

{

count[(int)(10*a[i])]++;

}

 

// mode is the index with maximum count

double mode = 0;

double k = count[0];

for (int i = 1; i < t; i++)

{

if (count[i] > k)

{

k = count[i];

mode = i;

}

}

return mode;

}

public static double findSmallest(double [] A, int total){

Arrays.sort(A);

return A[0];

}

 

public static void printAboveAvg(double arr[], int n)

{

 

// Find average

double avg = 0;

for (int i = 0; i < n; i++)

avg += arr[i];

avg = avg / n;

 

// Print elements greater than average

for (int i = 0; i < n; i++)

if (arr[i] > avg)

System.out.print(arr[i] + " ");

System.out.println();

}

 

public static void printrand(double [] A, int n){

Arrays.sort(A);

for(int i=0;i<n;i++){

System.out.print(A[0]+"/t");

}

System.out.println();

}

 

public static void printHist(double [] arr, int n) {

 

for (double i = 1.0; i >= 0; i-=0.1) {

System.out.print(i+" | ");

for (int j = 0; j < n; j++) {

 

// if array of element is greater

// then array it print x

if (arr[j] >= i)

System.out.print("x");

 

// else print blank spaces

else

System.out.print(" ");

}

System.out.println();

}

// print last line denoted by ----

for(int l = 0; l < (n + 3); l++){    

System.out.print("---");

}

 

System.out.println();

System.out.print(" ");

 

for (int k = 0; k < n; k++) {

System.out.print(arr[k]+" ");

}

}

  // Driver program

  public static void main(String args[]) throws IOException

  {

      //Enter data using BufferReader

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

double [] A = new double[100];

int i=0;

System.out.println("Enter the numbers(0.0-1.0) /n Enter 9 if u have entered the numbers. /n");

do

{

A[i++]=Double.parseDouble(br.readLine());

}while(A[i-1]==9);

      i--;

      System.out.println("Average = " + findMean(A,i) );

      System.out.println("Median = " + findMedian(A,i));

      System.out.println("Element that occured most frequently = " + findMode(A,i));

      System.out.println("number closest to 0.0 =" + findSmallest(A,i));

      System.out.println("Numbers that are greater than the average are follows:");

      printAboveAvg(A,i);

      System.out.println("Numbers in random order are as follows:");

      printrand(A,i);

      System.out.println("Histogram is bellow:");

      printHist(A,i);

  }

}

Explanation:

3 0
3 years ago
Your corporation hosts a Web site at the static public IP address 92.110.30.123.
ivann1987 [24]

Answer:

Check the explanation

Explanation:

In line with the question, we can now derive that:

The router's outside interface IP address will be 92.110.30.65.

The router's inside interface IP address will be 192.168.11.254.

The Web site's IP public IP address will be 92.110.30.123.

The private IP address of the backup Web server will be 192.168.11.110.

and when we say IP address, it stands for Internet Protocol, it is a set of usual predefined rules which are utilized to administrate the manner to which data packets are sent over the internet. An IP address, which is typically just identified as an IP, is a sequence of figures used to uniquely recognize a computer/device on a particular network or on the internet space.

7 0
3 years ago
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