Elisa and josh need to access general . elisa will press the f1 key. josh will click on ? in the upper-right corner of the word document. who will access general ? a. only elisa b. only josh c. both elisa and josh d. neither elisa nor josh

Consider the following declarations (1, 2, 3, 5, 7)class bagtype{public: void set(string, double, double, double, double); void print() const; string getstyle() const; double getprice() const; void get(double, double, double, double); bagtype(); bagtype(string, double, double, double, double); private: string style: double l; double w; double h; double price; }; a.) write the definition of the number function set so that private members are set according to the parametersb.) write the definition of the member function print that prints the values of the data membersc.) write the definition of the default constructor of the class bagtype so that the private member variables are initialized to "", 0.0, 0.0, 0.0, 0.0, respectively d.) write a c++ statement that prints the value of the object newbag.e.) write a c++ statement that declares the object tempbag of type bagtype, and initialize the member variables of tempbag to "backpack", 15, 8, 20 and 49.99, respectively

Animations and transitions are added from the

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?