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2 years ago

In the code snippet below, pick which instructions will have pipeline bubbles between them due to hazards.

Computers and Technology

1 answer:

denpristay [2]2 years ago

7 0

Answer:

b. lw $t4, 4($t0)

c. add $t3, $t5, $t4

Explanation:

Pipeline hazard prevents other instruction from execution while one instruction is already in process. There is pipeline bubbles through which there is break in the structural hazard which preclude data. It helps to stop fetching any new instruction during clock cycle.

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Which branch of science helps avoid or minimize stress related injuries at workplace? _____ is a branch of science that aims to

QveST [7]

Answer:

Ergonomics

Explanation:

We can get stress injuries in the workplace, and the study of the stress injuries at workplaces is a very important branch of scientific study currently, and all companies from all the fields are working on it, and so are the academic and research institutions. And this branch has gained heights in the past 10 years as companies want to increase worker productivity as well as bring down the downtime as well as various injury claims related to the job. And we know this branch as "ergonomics".

5 0

3 years ago

Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi

Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

5 0

3 years ago

In GamePoints' constructor, assign teamWhales with 500 and teamLions with 500. #include using namespace std; class GamePoints {

xxMikexx [17]

Answer:

Here is the GamePoints constructor:

GamePoints::GamePoints() :

/* Your code goes here */

{

teamLions = 500;

teamDolphins = 500;

}

Explanation:

Here is the complete program:

#include //to use input output functions

using namespace std; //to identify objects like cin cout

class GamePoints { //class GamePoints

public: GamePoints(); // constructor of GamePoints class

void Start() const; //method of GamePoints class

private: //declare data members of GamePoints class

int teamDolphins; // integer type private member variable of GamePoints

int teamLions; }; // integer type private member variable of GamePoints

GamePoints::GamePoints() //constructor

{ teamLions = (500), teamDolphins= (500); } //assigns 500 to the data members of teamLions and teamDolphins of GamePoints class

void GamePoints::Start() const { //method Start of classs GamePoints

cout << "Game started: Dolphins " << teamDolphins << " - " << teamLions << " Lions" << endl; } //displays the values of teamDolphins and teamLions i.e. 500 assigned by the constructor

int main() //start of main() function

{ GamePoints myGame; // creates an object of GamePoints class

myGame.Start(); //calls Start method of GamePoints class using the object

return 0; }

The output of the program is:

Game started: Dolphins 500 - 500 Lions

8 0

3 years ago

Which of the following is a hand-drawn animation that takes a large amount of time to complete?

elixir [45]

Answer:

Where are the choices?

Thanks for adding the choices at the comment secion

2D animation
3D animation
Stop Motion
Celluloid animation

A- Celluloid Animation

4 0

3 years ago

What is the definition of overflow in binary?

asambeis [7]

Overflow occurs when the magnitude of a number exceeds the range allowed by the size of the bit field. The sum of two identically-signed numbers may very well exceed the range of the bit field of those two numbers, and so in this case overflow is a possibility.

4 0

3 years ago