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3 years ago

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A research team is studying parallel computing. They want to run parallel processes without having to use multiple processors. H

ow can this be done?
A. by using cache
B. by using cores
C. by using threads
D. by using locks

Computers and Technology

2 answers:

kicyunya [14]3 years ago

8 0

B. by using cores is the correct answer

valentina_108 [34]3 years ago

6 0

D by using locks okkkkkkkk

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Write the method makeNames that creates and returns a String array of new names based on the method’s two parameter arrays, arra

Gwar [14]

Answer:

import java.util.Arrays;
public class Main {
public static void main(String[] args) {
String [] first = {"David", "Mike", "Katie", "Lucy"};
String [] middle = {"A", "B", "C", "D"};
String [] names = makeNames(first, middle);
System.out.println(Arrays.toString(names));
}
public static String [] makeNames(String [] array1, String [] array2){
if(array1.length == 0){
return array1;
}
if(array2.length == 0){
return array2;
}
String [] newNames = new String[array1.length];
for(int i=0; i < array1.length; i++){
newNames[i] = array1[i] + " " + array2[i];
}
return newNames;
}
}

Explanation:

The solution code is written in Java.

Firstly, create the makeNames method by following the method signature as required by the question (Line 12). Check if any one the input string array is with size 0, return the another string array (Line 14 - 20). Else, create a string array, newNames (Line 22). Use a for loop to repeatedly concatenate the string from array1 with a single space " " and followed with the string from array2 and set it as item of the newNames array (Line 24-26). Lastly, return the newNames array (Line 28).

In the main program create two string array, first and middle, and pass the two arrays to the makeNames methods as arguments (Line 5-6). The returned array is assigned to names array (Line 7). Display the names array to terminal (Line 9) and we shall get the sample output: [David A, Mike B, Katie C, Lucy D]

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Answer:

four heads are better than one

Explanation:

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2 years ago

Disadvantages of Batch<br>operation system

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Answer:

Disadvantages of Batch Operating System:

The computer operators should be well known with batch systems.
Batch systems are hard to debug.
It is sometime costly.
The other jobs will have to wait for an unknown time if any job fails.

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3 years ago

You are a visitor at a political convention with delegates; each delegate is a member of exactly one political party. It is impo

Fed [463]

Answer:

The algorithm is as follows:

Step 1: Start

Step 2: Parties = [All delegates in the party]

Step 3: Lent = Count(Parties)

Step 4: Individual = 0

Step 5: Index = 1

Step 6: For I in Lent:

Step 6.1: If Parties[Individual] == Parties[I]:

Step 6.1.1: Index = Index + 1

Step 6.2: Else:

Step 6.2.1 If Index == 0:

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

Step 7: Else

Step 7.1: Index = Index - 1

Step 8: Print(Party[Individual])

Step 9: Stop

Explanation:

The algorithm begins here

Step 1: Start

This gets the political parties as a list

Step 2: Parties = [All delegates in the party]

This counts the number of delegates i.e. the length of the list

Step 3: Lent = Count(Parties)

This initializes the first individual you come in contact with, to delegate 0 [list index begins from 0]

Step 4: Individual = 0

The next person on the list is set to index 1

Step 5: Index = 1

This begins an iteration

Step 6: For I in Lent:

If Parties[Individual] greets, shakes or smile to Party[i]

Step 6.1: If Parties[Individual] == Parties[I]:

Then they belong to the same party. Increment count by 1

Step 6.1.1: Index = Index + 1

If otherwise

Step 6.2: Else:

This checks if the first person is still in check

Step 6.2.1 If Index == 0:

If yes, the iteration is shifted up

Step 6.2.2: Individual = I

Step 6.2.3: Index = 1

If the first person is not being checked

Step 7: Else

The index is reduced by 1

Step 7.1: Index = Index - 1

This prints the highest occurrence party

Step 8: Print(Party[Individual])

This ends the algorithm

Step 9: Stop

The algorithm, implemented in Python is added as an attachment

<em>Because there is an iteration which performs repetitive operation, the algorithm running time is: O(n) </em>

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2 years ago