"In a(n) file, each alphabetic, numeric, or special character is represented with a 7-bit binary number."

1. the program must provide following functions to extract some statistics. note that the data_list parameter specified in these functions may be the same for all functions or different for different functions—that is your choice. a skeleton file is provided on mirmir. a) open_file()prompts the user to enter a year number for the data file. the program will check whether the year is between 1990 and 2015 (both inclusive). if year number is valid, the program will try to open data file with file name ‘year.txt’, where is the year. appropriate error message should be shown if the data file cannot be opened or if the year number is invalid. this function will loop until it receives proper input and successfully opens the file. it returns a file pointer and year. i. hint: use string concatenation to construct the file name b) read_file(fp)has one parameter, a file pointer read. this function returns a list of your choosing containing data you need for other parts of this project. c) find_average(data_list) takes a list of data (of some organization of your choosing) and returns the average salary. the function does not print anything. hints: i. this is not the average of the last column of data. it is not mathematically valid to find an average by finding the average of averages—for example, in this case there are many more in the lowest category than in the highest category. ii. how many wage earners are considered in finding the average (denominator)

What is a rush associated with alcohol?

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }