What input and output devices does a mainframe computer have?

We as humans write math expression in infix notation, e.g. 5 + 2 (the operators are written in-between the operands). in a computer’s language, however, it is preferred to have the operators on the right side of the operands, i.e. 5 2 +. for more complex expressions that include parenthesis and multiple operators, a compiler has to convert the expression into postfix first and then evaluate the resulting postfix.write a program that takes an “infix” expression as input, uses stacks to convert it into postfix expression, and finally evaluates it. it must support the following operations: + - / * ^ % (example infix expression: (7 - 3) / (2 + 2)postfix expression: 7 3 - 2 2 + /result: 1guidelines: 1. you will need to use stacks in three placesa. one for the parenthesis check [char stack]b. one during infix to postfix [char stack]c. one during evaluation [int stack]for a and b above, you can use same array and same push, pop method as both ofthem are char. but for evaluation you have int stack and you might consider to createanother push pop method to handle it. maybe push_int, pop_int, etc. or find otherstrategy to utilize existing push pop method2. you can create a function for obtaining operator priority. that function should take anoperator as input and return its priority as an integer. this function will you a lot andreduce repeated code3. during evaluation you will need to convert char into integer. example for single digit: char c = '5'; int x = c - '0';

Which of the following factors would your hypothetical supervisor look at when deciding whether to test a biological material sample for dna? the amount of other evidence you have implicating a suspect in a crime the annual budget for the crime lab both of the above none of the above; you would almost always order a test

Write a function so that the main0 code below can be replaced by the simpler code that calls function mphandminutes tomiles0. original main0 int main) l double milesperhour-70.0; double minutestraveled = 100.0; double hourstraveled; double milestraveled; hourstraveled = minutestraveled / 60.0; milestraveled = hourstraveled * milesperhour; cout < "miles" 2 using namespace std; 4 /* your solution goes here/ 6 int maino 1 test passed 7 double milesperhour 70.0 all tests passed 8 double minutestraveled 100.0; 10 cout < < "miles: " < < mphandminutestomiles(milesper-hour, minutestraveled) < < endl; 12 return 0; 13