Write a method which, given a List of strings, returns what Strings appeared in the array and how many times each word appeared. You may assume that each String holds a single word. // INPUT -> OUTPUT // ["foo", "bar", "foo", "bar", "baz"] -> {"foo":2 , "bar":2, "baz":1} // ["a", "b", "a","c", "a", "d"] -> {"a":3, "b":1, "c":1,"d":1} public static Map wordCount(List words){

Exchanging which type of data uses the least bandwidth? music photographs video voice bandwidth- the amount of data that can be moved between two points in a set time period

Asoftware company hired ray, a college graduate to work in their development team. ray is assigned to work in the coding phase of a project. what happens during the coding phase of a software development project? a. the customer receives a working model of the software. b. developers convert the program design into code. c. developers gather requirements directly from the stakeholders. d. testing teams check the product for quality.

Fill in the following program so that it will correctly calculate the price of the orange juice the user is buying based on the buy one get one sale.#include //main functionint main() { int cartons; float price, total; //prompt user for input information printf("what is the cost of one container of oj in dollars? \n"); scanf(" [ select ] ["%d", "%c", "%f", "%lf"] ", & price); printf("how many containers are you buying? \n"); scanf(" [ select ] ["%d", "%c", "%f", "%lf"] ", & cartons); if ( [ select ] ["cartons / 2", "cartons % 1", "cartons % 2", "cartons % price", "cartons / price", "cartons / total"] [ select ] ["=", "==", "! =", "< =", "> =", "< "] 0) total = [ select ] ["price * cartons", "cartons * price / 2 + price", "(cartons / 2) * price", "cartons / (2.0 * price)", "(cartons / 2.0) * price + price", "((cartons / 2) * price) + price"] ; else total = ((cartons / 2) * price) + price; printf("the total cost is $%.2f.\n", total); return 0; }