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icang [17]
4 years ago
6

Mention three types of pipeline hazards​

Computers and Technology
2 answers:
kaheart [24]4 years ago
7 0

Answer:

Structural Hazards.

Data Hazards.  

Control Hazards.

Explanation:

boyakko [2]4 years ago
3 0

Answer:

Structural Hazards, Data Hazards, Control Hazards

Explanation:

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What are the basic problem that my pc can have?
Lena [83]
A virus, malware, unplugged, used too much storage, etc.
5 0
4 years ago
You are given a list of n positive integers a1, a2, . . . an and a positive integer t. Use dynamic programming to design an algo
Anna11 [10]

Answer:

See explaination for the program code

Explanation:

The code below

Pseudo-code:

//each item ai is used at most once

isSubsetSum(A[],n,t)//takes array of items of size n, and sum t

{

boolean subset[n+1][t+1];//creating a boolean mtraix

for i=1 to n+1

subset[i][1] = true; //initially setting all first column values as true

for i = 2 to t+1

subset[1][i] = false; //initialy setting all first row values as false

for i=2 to n

{

for j=2 to t

{

if(j<A[i-1])

subset[i][j] = subset[i-1][j];

if (j >= A[i-1])

subset[i][j] = subset[i-1][j] ||

subset[i - 1][j-set[i-1]];

}

}

//returns true if there is a subset with given sum t

//other wise returns false

return subset[n][t];

}

Recurrence relation:

T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times

T(1)=1

solving recurrence:

T(n)=T(n-1)+t

T(n)=T(n-2)+t+t

T(n)=T(n-2)+2t

T(n)=T(n-3)+3t

,,

,

T(n)=T(n-n-1)+(n-1)t

T(n)=T(1)+(n-1)t

T(n)=1+(n-1)t = O(nt)

//so complexity is :O(nt)//where n is number of element, t is given sum

6 0
3 years ago
If you would like to give another user permissions on your mailbox or to particular folders within your mailbox, which
jenyasd209 [6]
Manager there SUbMiT
7 0
3 years ago
Read 2 more answers
In Python please:
g100num [7]

Answer:

def compute_pay(number_of_hours, rate_of_pay):

   if number_of_hours > 40:

       pay_for_week = (40*rate_of_pay)+((number_of_hours-40)*\

                                                (rate_of_pay+rate_of_pay*0.5))

   else:

       pay_for_week = number_of_hours*rate_of_pay

   if pay_for_week >= 375:

       print("Paying %d by direct deposit" % pay_for_week)

   else:

       print("Paying %d by mailed check" % pay_for_week)

Explanation:

  1. We define the computer pay function that receives the number of hours worked in a week and the rate of pay
  2. From the test cases we deduce that if a worker works more than 40 hours a week an extra payment is given, you can calculated it as follow: (40 * rate_of_pay) + ((number_of_hours - 40) * (rate_of_pay + rate_of_pay * 0.5))
  3. If a worker works less than 40 hours the payment is calculated as follow: pay_for_week = number_of_hours * rate_of_pay
  4. If the pay for week is equal or greater than 375 we print a payment by direct deposit otherwise we print payment by mailed check

3 0
3 years ago
Write a program that contains three methods:
Reil [10]

Answer:

#include <stdio.h>//defining header file

int max (int x, int y, int z) //defining a method max that hold three parameters

{

   if(x>=y && x>=z)//defining if block that checks x is greater then y and x is greater then z

   {

       return x;//return the value x

   }

   else if(y>z)//defining else if block it check y is greater then z

   {

       return y;//return the value y

   }

   else//else block

   {

       return z;//return the value z

   }

}

int min (int x, int y, int z) //defining a method max that holds three parameters

{

if(x<=y && x<=z) //defining if block that check x value is less then equal to y and less then equal to z

{

return x;//return the value of x

}

if(y<=x && y<=z) //defining if block that check y value is less then equal to x and less then equal to z

{

return y;//return the value of y

}

if(z<=x && z<=x)//defining if block that check z value is less then equal to x  

{

return z;//return the value of z

}

return x;//return the value of z  

}

int average (int x, int y, int z) //defining average method that take three parameters

{

int avg= (x+y+z)/3;//defining integer variable avg that calculate the average

return avg;//return avg value

}

int main()//defining main method

{

   int x,y,z;//defining integer variable

   printf("Enter first value: ");//print message

   scanf("%d",&x);//input value from the user end

   printf("Enter Second value: ");//print message

   scanf("%d",&y);//input value from the user end

   printf("Enter third value: ");//print message

   scanf("%d",&z);//input value from the user end

   printf("The maximum value is: %d\n", max(x,y,z));//calling the method max

   printf("The minimum value is: %d\n", min(x,y,z));//calling the method min

   printf("The average value is: %d\n", average(x,y,z));//calling the method average

   return 0;

}

Output:

Enter first value: 45

Enter Second value: 35

Enter third value: 10

The maximum value is: 45

The minimum value is: 10

The average value is: 30

Explanation:

In the above-given code, three methods "max, min, and average" is declared that holds three integer variable "x,y, and z" as a parameter and all the method work as their respective name.

  • In the max method, it uses a conditional statement to find the highest number among them.
  • In the min method, it also uses the conditional statement to find the minimum value from them.
  • In the average method, it defined an integer variable "avg" that holds the average value of the given parameter variable.
  • In the main method, three variable is defined that inputs the value from the user end and passes to the method and print its value.      
7 0
3 years ago
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