What type of device is a projector?

Input
Memory
Output
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Draw the hierarchy chart and design the logic for a program that calculates service charges for hazel's housecleaning service. the program contains housekeeping, detail loop, and end-of-job modules. the main program declares any needed global variables and constants and calls the other modules. the housekeeping module displays a prompt for and accepts a customer's last name. while the user does not enter for the name, the detail loop accepts the number of bathrooms and the number of other rooms to be cleaned. the service charge is computed as $40 plus $15 for each bathroom and $10 for each of the other rooms. the detail loop also displays the service charge and then prompts the user for the next customer's name. the end-of-job module, which executes after the user enters the sentinel value for the name, displays a message that indicates the program is complete.

Writing a modular program in visual c++. i am new to this and not sure what i am missing. i am getting the following error: baddate.cpp: in function ‘int main()’: baddate.cpp: 50: 3: error: ‘else’ without a previous ‘if’elsehere are the instructions and code: writing a modular program in c++in this lab, you add the input and output statements to a partially completed c++ program. when completed, the user should be able to enter a year, a month, and a day. the program then determines if the date is valid. valid years are those that are greater than 0, valid months include the values 1 through 12, and valid days include the values 1 through 31.notice that variables have been declared for you.write the simulated housekeeping() function that contains the prompts and input statements to retrieve a year, a month, and a day from the user.include the output statements in the simulated endofjob() function. the format of the output is as follows: month/day/year is a valid date.ormonth/day/year is an invalid date.execute the program entering the following date: month = 5, day = 32, year = 2014. record the output of this program.execute the program entering the following date: month = 9, day = 21, year = 2002. record the output of this /* program name: baddate.cppfunction: this program determines if a date entered by the user is valid.input: interactiveoutput: valid date is printed or user is alerted that an invalid date was entered*/#include bool validatedate(int, int, int); using namespace std; int main(){// declare variablesint year; int month; int day; const int min_year = 0, min_month = 1, max_month = 12, min_day = 1, max_day = 31; bool validdate = true; // this is the work of the housekeeping() method// get the year, then the month, then the daycout< < "enter the year"< > year; cout< < "enter the month"< > month; cout< < "enter the day"< > day; // this is the work of the detailloop() method// check to be sure date is validif(year < = min_year) // invalid yearvaliddate = false; else if (month < min_month || month > max_month) // invalid monthvaliddate = false; else if (day < min_day || day > max_day) // invalid dayvaliddate = false; // this is the work of the endofjob() method// test to see if date is valid and output date and whether it is valid or notif(validdate == true); {// output statementcout<

When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2ă—, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5ă— slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?