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algol13
3 years ago
6

Is majority intent determined by how many times the same type of result is shown on the search engine result page?

Computers and Technology
1 answer:
laila [671]3 years ago
8 0

According to the search engine algorithm, it is <u>True</u> that the majority intent is determined by how many times the same result is shown on the search engine result page.

<h3>What is Search Intent?</h3>

Search Intent is a term used to describe a user's reason when typing a question or words into a search engine.

Generally, if a user found that no search results match his wants, he would likely not click on any link before performing a similar query search. This would make search engines return with more links that have higher clicks.

<h3>Different types of Search Intent</h3>
  • Informational
  • Commercial
  • Navigation
  • Transactional

Hence, in this case, it is concluded that the correct answer is True.

Learn more about Search Engine here: brainly.com/question/13709771

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Write the definition of a function named max that receives an int parameter and returns the largest value that it has been calle
kkurt [141]

Answer:

Following are the code in the C language .

int max(int x) // function definition

{

static int largest=0; // variable declaration

if (x>largest) // checking the condition

largest=x; // assign the value of input in the largest variable

return largest; // return the largest number

}

Explanation:

Following are the description of code .

  • Decalred a function max of int type which hold a parameter x of int datatype.
  • In this function compared the number x with the largest .If this condition is true then largest number hold the value of x variable
  • Finally return the largest number .
8 0
3 years ago
the first day anna read a quarter of the book. on the second day she read a third of the remainder. noticed that after two days
Misha Larkins [42]

Answer:

160 pages

Explanation:

Day\ 1 = \frac{1}{4}

Day\ 2 = \frac{1}{3}Remainder

Left = 80

Required

The number of pages

Let the number of pages be x.

So, on day 1; we have:

Day\ 1 = \frac{1}{4}x

After day 1, there are:\frac{3}{4}x left ----------------- i.e x - 1/4x

On day 2, we have:

Day\ 2 = \frac{1}{3} * \frac{3}{4}x

Day\ 2 = \frac{1}{4}x

At this point, we have:

Day\ 1 = \frac{1}{4}x

Day\ 2 = \frac{1}{4}x

Left = 80 ---- pages left

The summation of all must equal x, the book pages

Day\ 1 + Day\ 2 + Left = Total\\

\frac{1}{4}x + \frac{1}{4}x+ 80= x

Simplify the left-hand side

\frac{1}{2}x+ 80= x

Collect like terms

x - \frac{1}{2}x= 80

Simplify

\frac{2-1}{2}x= 80

\frac{1}{2}x= 80

Multiply by 2

2 * \frac{1}{2}x= 80*2

x = 160

4 0
3 years ago
Program MATH_SCORES: Your math instructor gives three tests worth 50 points each. You can drop one of the test scores. The final
Simora [160]

Answer:

import java.util.Scanner;

public class num5 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       //Prompt and receive the three Scores

       int score1;

       int score2;

       int score3;

       do {

           System.out.println("Enter first Score Enter score between 1 -50");

           score1 = in.nextInt();

       } while(score1>50 || score1<0);

       do {

           System.out.println("Enter second Score.The second score must be between 1 -50");

           score2 = in.nextInt();

       } while(score2>50 || score2<0);

       do {

           System.out.println("Enter Third Score Third score must between 1 -50");

           score3 = in.nextInt();

       } while(score3>50 || score3<0);

       //Find the minimum of the three to drop

       int min, min2, max;

       if(score1<score2 && score1<score3){

           min = score1;

           min2 = score2;

           max = score3;

       }

       else if(score2 < score1 && score2<score3){

           min = score2;

           min2 = score1;

           max = score3;

       }

       else{

           min = score3;

           min2 = score1;

           max = score2;

       }

       System.out.println("your entered "+max+", "+min2+" and "+min+" the min is");

       int total = max+min2;

       System.out.println("Total of the two highest is "+total);

       //Finding the grade based on the cut-off points given

       if(total>=90){

           System.out.println("Grade is A");

       }

       else if(total>=80){

           System.out.println("Grade is B");

       }

       else if(total>=70){

           System.out.println("Grade is C");

       }

       else if(total>=60){

           System.out.println("Grade is D");

       }

       else{

           System.out.println("Grade is F");

       }

   }

}

Explanation:

  • Implemented with Java
  • Use the scanner class to receive user input
  • Use a do.....while loop to validate user input for each of the variables. A valid score must be between 0 and 50 while(score>50 || score<0);  
  • Use if and else to find the minimum of the three values and drop
  • Add the two highest numbers
  • use if/else if /else statements to print the corresponding grade
8 0
3 years ago
This motherboard already has 1GB of RAM installed in the DIMM1 slot. The customerwould like to upgrade to 4GB total memory, use
Luden [163]

Answer:

The two phases to the context of this discussion are listed follows.

Explanation:

  • <u>Solution 1</u>: Delete 1 GB of current RAM as well as install another DIMM0 Chan A slot through one 2 GB of double-channel RAM. (thinkable unless the 2 GB RAM is provided by the motherboard in what seems like a DIMM0 Chan A slot)  
  • <u>Solution 2</u>: whether there's an unused or blank slot, perhaps one 1 GB dual-channel Ram could be mounted in some other slot at around the same speed or frequency.  

It's quite safer to mount memory with appropriate frequencies across both situations.

6 0
3 years ago
Abby is creating a professional development plan to advance in her current career. She has identified what job she wants and wha
Andre45 [30]

Evaluate her current skills and identify areas of improvement

5 0
3 years ago
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