Answered: Run time Arrival Time .Job No 12 1 4 18…

Run time Arrival Time .Job No 12 1 4 18 4 What is average turnaround time if you use ?OS with multi-programming and no 1/O overlapped :Select one T=32.256 T=24.99 T=25.625 T=26.025 Clear my choice

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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4. Create and run a program that will compute the total cost charges of a telephone call rendered from a particular network using the given table. The user will input call destination code CDA (1 to 5), time code TC (1 = off-peak, 2 = peak), and the duration of call. Save your program as Exp5_ex4.cpp and write the code below. Sample Process: Call Destination Code: 1 Time Code: 2 Duration: 4 Computation: 3 * 4 = 12 pesos harge Time Code (2 ) Peak (1 )Off- peak Rate Rate Sample Output CDA Description Good day! Same P8 P3 1 Network Other Enter Call Destination Code: , 2 Network 3 P8 P4 Enter Time Code: P4 Enter Call Duration (minutes): Landline National P8 P11 P7 4 Direct International The total cost charge is pesos. P20 P20 Direct
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QUESTION 51 Suppose you have defined x = [1,2,3] and y = [2,4,6] in the Command Window. Determine the result of the following MATLAB operation: >>x +y O ans = 369 O ans = 123246 O ans = 6 12 O ans = 18
zy Section 1.3 - IT 140: Introduct zy Section 1.3 - IT 140: Introduct X O (158) SNHU IT140 Module Or x b Answered: Run the program G Many useful programs allow + Ô https://learn.zybooks.com/zybook/SNHUIT140V3/chapter/1/section/3?content_resource_id=40688398 = zyBooks My library > IT 140: Introduction to Scripting v3 home > 1.3: Input/Output E zyBooks catalog ? Help/FAQ 8 Jose Roque CHALLENGE 1.3.4: Read user input numbers and perform a calculation. АCTIVITY Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type code where the questions marks are to finish the code. num1 = int(input ()) num2 = ? print(num1 + ?) Note: Our system automatically runs your program several times, trying different input values each time to ensure your program works for any values. Notes on how to solve this. See How to Use zyBooks for info on how our automated program grader works. 247772.2002516.qx3zqy7 1 2 ' Your solution goes here '' 1…
Which of the following state changes is a process able to perform by itself? (do not select state changes that the OS performs) ready -> waiting ready -> running running -> waiting running -> ready running ->terminated new -> ready
Study the code listing below: #include <xc.h> // Required for all MPLAB XC8 source files void main(void) { INTCON = 0xA0; // SET GIE, TMR0IE OPTION_REGbits.T0CS = 0; // Set clock source to internal (timer mode) OPTION_REGbits.PSA = 0; // Set prescaler to Timer 0 OPTION_REGbits.PS = 7; // Set prescaler bits while(1); // Do nothing } void interrupt myISR(void) { if(TMR0IF) // Timer0 overflow interrupt specific code { INTCONbits.TMR0IF = 0; // Reset flag } } Calculate the total time that elapses between Timer0 overflows. Assume the oscillator frequency is 3.2768 MHz. Enter you answer in milliseconds.
Consider the following IF statement: if ( ( (se == s1) || ( s2 == s3) ) && (se < s2) ) s1 = s3 ; what would be the missing statement 1 and 2 in the following equivalent MIPS implementation: ---# statement 1 beq $se,$s1, process --# statement 2 process : add $s1, $s3, $zero exit Select one: a.statement 1 is ( bgt $s0, $s2, exit ) and statement 2 is (bne $s2, $3, exit) b.None of the given answers c.statement 1 is ( bge $s0, $s2, exit ) and statement 2 is ( bne $s2, $s3, exit) d.statement 1 is ( bit $s0, $s2, process ) and statement 2 is ( bne $s2, $s3, exit ) e.statement 1 is ( bge $s0, $s2, exit ) statement 2 is ( beq $s2, $3, process)
I want a minimal Ubuntu of less than 300 MB Ubuntu with only terminal no GUI If any export has a solution on how to do it feel free to answer, Thank you , God bless you. :(
/* GCC *//* gcc -Wa,-adhln -g -masm=intel -m32 "Project 2.c" > "Project 2-g.asm" *//* gcc -Wa,-adhln -O -masm=intel -m32 "Project 2.c" > "Project 2-o.asm" */ #include <stdio.h> #define NOINLINE __attribute__ ((noinline)) static NOINLINE int function1(int x, int y){ int i;int sum;int values[10];sum = 0;for (i = 0; i < 10; i++) {values[i] = 10 * i + x * y;sum += values[i];} return (sum);} static int NOINLINE function2(int *values, int valuesLen){ int i;int sum; sum = 0;for (i = 0; i < valuesLen; i++) {sum += values[i];} return (sum);} static NOINLINE int function3(int x){ int y; y = x / 10;return (y);} static NOINLINE int function4(int a, int b, int c, int d){ int r; if (a > b)r = a;else if (a > c)r = 2 * a;else if (a > d)r = 3 * a;elser = -1;return (r);} int main(int argc, char **argv){ int i;int j;int k;int values[10]; i = 1;j = 2;k = function1(i, j);printf("function1: i = %d, j = %d, k = %d\n", i, j, k); for (i = 0; i < 10; i++) {values[i] = i;}k =…
can you modify it to use more than two cpus, i compiled and ran and the time intervals are still showing 0.0 but everything else is done as desired thank you. i think it shows the 0.0 time becuase of my computer, i see on yours your cpu count is 8
Which of the following state changes is a process able to perform by itself? (do not select state changes that the OS performs) ready -> waiting ready -> running running -> waiting running -> ready running - ->terminated new -> ready
i got this error when i run could you help me please thank you /unix_hw/output/unix_hw$ gcc helloworld.c -o output/helloworld/usr/bin/ld: /usr/lib/gcc/x86_64-linux-gnu/9/../../../x86_64-linux-gnu/Scrt1.o: in function `_start':(.text+0x24): undefined reference to `main'collect2: error: ld returned 1 exit status