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sukhopar [10]
3 years ago
14

A(n) _____________ is a simple tool that can help identify computers/devices or communication circuits that have higher-than-ave

rage error or usage rates.
Computers and Technology
1 answer:
jok3333 [9.3K]3 years ago
3 0
Fdddfdashddxzvcsssgtd
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Universal Containers wants line items to be organized into pre-defined groups on the output document. The name of these pre-defi
devlian [24]

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Set the Group field on the Line Items type of Template Content record.

Set the New Quote Group field on each Product record with the associated group name.

Explanation:

The two array that can be implemented are:

  • Set the Group field on the Line Items type of Template Content record.
  • Set the New Quote Group field on each Product record with the associated group name.
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You can use the ____ to change the order of the wat windows are stacked.
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A

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3 years ago
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pogonyaev

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Write an application that uses String method region-Matches to compare two stringsinput by the user. The application should inpu
SOVA2 [1]

Answer:

Explanation:

The following program creates a function called region_Matches that takes in two strings as arguments as well as an int for starting point and an int for amount of characters to compare. Then it compares those characters in each of the words. If they match (ignoring case) then the function outputs True, else it ouputs False. The test cases compare the words "moving" and "loving", the first test case compares the words starting at point 0 which outputs false because m and l are different. Test case 2 ouputs True since it starts at point 1 which is o and o.

class Brainly {

   public static void main(String[] args) {

       String word1 = "moving";

       String word2 = "loving";

       boolean result = region_Matches(word1, word2, 0, 4);

       boolean result2 = region_Matches(word1, word2, 1, 4);

       System.out.println(result);

       System.out.println(result2);

   }

   public static boolean region_Matches(String word1, String word2, int start, int numberOfChars) {

       boolean same = true;

       for (int x = 0; x < numberOfChars; x++) {

           if (Character.toLowerCase(word1.charAt(start + x)) == Character.toLowerCase(word2.charAt(start + x))) {

               continue;

           } else {

               same = false;

               break;

           }

       }

       return same;

   }

}

3 0
2 years ago
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