Answered: Find the error in each of the following…

Find the error in each of the following code segments and explain how to correct it. a) x=%3; while (x<=10); X++; b) for (y=.1; y!=1.0;y+=.1) cout<

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

See similar textbooks

Similar questions

7) What is the value of r after the following code executes? mySet (2, 3, 5, 7, 11) yourSet = (5, 7, 2, 11, 3) r = yourSet == mySet a. (5,7,2, 11, 3) b. [2, 3, 5, 7, 11) c. True d. error 13) How can the following algorithm be described? position = 0 while position < len(lyst): if target == lyst[position]: return position position += 1 return -1 a. sequential search b. bubble sort c. binary search d. selection sort
Find the errors in this code: double gpa[] = {2.2, 3.8, 4.0, 3.6, 1.7, 2.8}; int countSuper = 0; // number of students whose GPA >= 3.5 for (int count = 0; count < 6; count++){ if(gpa[count] >= 3.5){ System.out.print(”\n You are super!”); } countSuper++; } System.out.print(”\n Number of super students is: ”+ countSuper);
Module/Week 2 ASSIGNMENT (INPUT/OUTPUT)The number of permutations of a set of n items taken r at a time is given by the following formulan !⁄r !(n-r)!: where n! is the factorial of n, r! is the factorial of r, and (n-r)! is the factorial of the result of n-r. The factorial of a number n can be solved using the following formula: 〖n!=e〗^(-n) n^n√2πn. If there are 18 people in your class and you want to divide the class into programming teams of 3 members, you can compute the number of different teams that can be arranged using this formula (n !⁄r !(n-r)!). Write a C++ program that determines the number of potential team arrangements. You will need to use the double type for this computation. Use the Lab Template you set-up last week, proper formatting, and appropriate comments in your code. The output must be labeled clearly and formatted neatly. Submit C++ Programming Assignment 2 by 11:59 p.m. (ET) on Monday of Module/Week 2.
Write a program that prompts the user to enter two positive integers. Implement a function to validate the user input. The program will use a loop to re-prompt a user for an input until a user inserts a valid input. Implement a summation function.The program will sum those two integers and show a result on a terminal screen. The program must have a lease two functions. #1 Examples output: input1 - 7 input2 = 6 7+6 - 13 # 2 Examples output: input1 = 0 try again = -5 try again = 10 input2 - 6 10 + 6 - 16 Please fill in the source code below: #include using namespace std; int input); (int a, int b); int int input1= 0; int input2 = 0; cout<<"Input1 = "; input1 = (input1); cout<<"Input2 = "; input2 = getPositivelnput( átv S MacBook Air DII DD 80 888 F7 F8 F9 F4 F5 F6 F3 * & #3 2$ 7 CO
Whats wrong with this code? # This program is to find roots of quadratic equationimport math # function for finding the rootsdef equationroots(a, b, c): # calculating discriminant using formuladis = b * b - 4 * a * csqrt_val = math.sqrt(abs(dis)) # checking condition for discriminantelif dis > 0:print("\nreal and different roots ")print("First solution =",(-b + sqrt_val)/(2 * a))print("second solution =",(-b - sqrt_val/(2 * a)) elif dis == 0:print("\nReal and same roots")print(-b/(2*a)) # when discriminant is less than 0else:print("\nComplex Roots")print("First solution =",-b/(2 * a), "+i", sqrt_val)print("Second solution =",-b/(2 * a), "+i", sqrt_val) # Driver Program a = int(input("Enter a: "))b = int(input("Enter b: "))c = int(input("Enter c: ")) # if a is 0, then incorrect equationif a == 0:print("Input correct quadratic equation") else:equationroots(a, b, c)
C language Q4. Write a program that asks the user to enter an integer n, then prints the first n multiple of n. The program should keep asking to enter a number until -1 is entered. (0.5 point) Q5: a. What does the following program print? char c='Z'; for(i=1;i<= 4; i++) { for(j= 1; j<= i; j++) { printf("%c", C--); } printf("\n"); } b. Re-write the above code using while loop. Output: Sample Output: Enter a number: 4 48 12 16 Enter a number: 6 6 12 18 24 30 36 Enter a number: 1 1 Enter a number: -1 End Using while loop: (2x0.5 =1 point)
in C++ use this function to calcul the range: function range step Specified value of y y = 0.05*x3+6sin(3x)+4 From -2.5 to 2.5 0.5 -2 The program will calculate average value of y for the given range. To find the average, usestd::accumulate using algorithm header and divide by the number of elements. The output must be write with the result:Average value of function y for the given range is : ;
function avg and pass x and y Tin printf ("the avg of x and y is %d\n', avgl) give_sqrt (avgl); return 0, float avg (float m, float n) I/ Return the average of n and m void give sqrt (float x) printf ('the sqrt of is Mn'x :/ Print the sqrt value of x. return;
Problem Statement The barcode used by the U.S. Postal System to route mail is defined as follows: Each decimal digit in the ZIP code is encoded using a sequence of three half-height and two full-height bars. The barcode starts and ends with a full-height bar (the guard rail) and includes a checksum digit (after the five-digit ZIP code or ZIP + 4), computed by summing up the original digits modulo 10. Define the following functions: Draw a half-height or full-height bar on stddraw. Given a digit, draw its sequence of bars. Compute the checksum digit. Also define global code that read in a five- (or nine-) digit ZIP code as the command-line argument and draws the corresponding postal barcode.
y=[w(1:6),0,0,3] Enter ans= 1600 3<< Correct Incorrect
The program segment has errors. Find as many as you can. // This code should display the sum of two numbers. int choice, num1, num2; do {cout << "Enter a number: "; cin >> num1;cout << "Enter another number: "; cin >> num2;cout << "Their sum is " << (num1 + num2) << endl;cout << "Do you want to do this again?\n";cout << "1 = yes, 0 = no\n"; cin >> choice;} while (choice = 1)
8. Select the value of x, y so that the following code outputs: containedCircle c = { {10, 15}, 3};Point2D p = {x, y};if(c.contains(p)){std::cout << "contained" << std::endl;}A. x=7, y=18B. x=10, y=20C. x=11, y=14D. None of the above.