1-Compute the weakest precondition for the following statements A. x = 2* (y + 2*x); y=3*x-6 {y > 9} B. if (x >y) y=6*x+ 2 else y= 3*x+9; {y > 2}
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- Compute the weakest preconditions of the following statements for the given postconditions. Assume all variables are integers1)a = 2*b + 1;b = a*a + 1;{ a > 10 } 2)if (b > 0)a = 4*b - 10;elsea = b + 172;{ a > 0 }T/F 2. In Java, only if and if-else expressions are used in selection statements.Compute the weakest precondition for the sequence of statements: (Statement 1) b = 2b + 1; (Statement 2) b = b-3{b < 0}
- Explain these c++ statements void bar(int panam) {panam +=1;}(Numerical) a. Write a C++ program that accepts an integer argument and determines whether the passed integer is even or odd. (Hint: Use the % operator.) b. Enter, compile, and run the program written for Exercise 8a.Compute the weakest precondition for each of the following assignment statements and postconditions. x = 2 * y -5; y = x + 2; {y < 2}
- Write assignment statements for the following: Assign a value of 1 to between if n is in the range -k through +k, inclusive; otherwise assign a value of 0 Assign a value of 1 to uppercase if ch is an uppercase letter; otherwise, assign a value of 0. Assign a value of 1 to divisor if m is a divisor of n; otherwise, assign a value of 0.Can I please have the answer in C++ language 4.7: Time Machine Your time machine is capable of going forward in time up to 24 hours. The machine is configured to jump ahead in minutes. To enter the proper number of minutes into your machine, you would like a program that can take a start time (in hours, minutes, and a Boolean indicating AM or PM) and a future time (in hours, minutes, and a Boolean indicating AM or PM) and calculate the difference in minutes between the start and future time. A time is specified in your program with three variables: int hours, minutes; bool isAM; // You can also use a char, i.e. A or P for example, to represent 11:50 PM, you would store: hours = 11,minutes = 50, isAM = false or if using a char, hours = 11,minutes = 50, isAM = 'A' This means that you need six variables to store a start and future time. Write a program that allows the user to enter a start time and a future time. Include a function named computeDifference that takes the six variables as…Mark the following statements as true or false. i. An identifier can be any sequence of digits and letters. ii. In C++, there is no difference between a reserved word and a predefined identifier. iii. A C++ identifier can start with a digit. iv. The operands of the modulus operator must be integers. v. If a = 4; and b = 3; then after the statement a = b; the value of b is still 3. vi. The result of a logical expression cannot be assigned to an int variable. vii. Every if statement must have a corresponding else. viii. The expression in the if statement: if (score = 30) grade = 'A'; always evaluates to true. ix. The expression: (ch >= 'A' && ch = 'Z'. x. The expression !(x > 0) is true only if x is a negative number.
- State the order of evaluation of the operators in each of the following Java statements, and show the value of x after each statement is performed: a) int x =7+3 *6/2-1; b) int x =2% 2+2*2-2/2;(De Morgan’s Laws) In this chapter, we discussed the logical operators &&, ||, and !. DeMorgan’s Laws can sometimes make it more convenient for us to express a logical expression. Theselaws state that the expression !(condition1 && condition2) is logically equivalent to the expression(!condition1 || !condition2). Also, the expression !(condition1 || condition2) is logically equivalentto the expression (!condition1 && !condition2). Use De Morgan’s Laws to write equivalent expressions for each of the following, and then write a program to show that both the original expressionand the new expression in each case are equivalent.a) !(x < 5) && !(y >= 7)b) !(a == b) || !(g != 5)c) !((x <= 8) && (y > 4))d) !((i > 4) || (j <= 6))Solve the following questions: a)Solve the following postfix expressions i)2 3 + 1 - ii)3 4 - 10 + 7 2 3 * - 9 * /a) Convert the following infix to postfix expressions i)2 * 3 - 5 / 2 + 4 ii)1 * 2 * 3 * 4
