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tresset_1 [31]
1 year ago
15

Identify three errors in this HTML code snippet:

Computers and Technology
2 answers:
Kazeer [188]1 year ago
5 0

I have annotated the errors in the picture.

<em />

<em>Hope this helps :)</em>

Zarrin [17]1 year ago
3 0
<h3>The three errors in the HTML code are as follows:</h3>
  1. The opening tag is heading 1 while the closing tag is heading 2.
  2. The closing tag should include “</li>”.
  3. The opening tag is an unordered list, while the closing tag is an ordered list.

<em>It is a pleasure to work with you :)</em>

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What are the uses of computer in educational setting?
Arada [10]

Answer:

Quick Communication & Correspondence

Explanation:

Another main advantage of using computers in the education field is the improvement in the quality of teaching-learning process and communication between students & teachers. For this, they use Microsoft PowerPoint to prepare electronic presentations about their lectures.

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Construct a SQL query that displays a list of colleges, their sity/state, the accrediting agency, and whether or not the school
Alex787 [66]

Answer:

SELECT college, city_state, accre_agency, distance LIMIT 10

Explanation:

Given

Table name: College

See attachment for table

Required

Retrieve top 10 college, state, agency and school distance from the table

To retrieve from a table, we make use of the SELECT query

The select statement is then followed by the columns to be selected (separated by comma (,))

So, we have:

SELECT college, city_state, accre_agency, distance

From the question, we are to select only first 10 records.

This is achieved using the LIMIT clause

i.e. LIMIT 10 for first 10

So, the complete query is:

SELECT college, city_state, accre_agency, distance LIMIT 10

6 0
3 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
An experimenter discovers that the time for people to detect a string of letters on a computer screen is 400 milliseconds, and t
nadya68 [22]

Answer:

The correct answer to the following question will be Option D (Subtractive method ).

Explanation:

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  • This method will be much more valuable when deciding the period for phonemic awareness, despite text-string identification.

The other three options are not related to the given scenario. So that Option D is the right answer.

3 0
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