Online note-taking tools can help students by providing a way to study for tests. showing them how to take notes. explaining how to do homework. providing the answers to tests. (in my logic I'd say it's "providing a way to study for tests, but I want to double check.)

In which job role will you be creating e-papers, newsletters and preiodicals

1. ask the user how many questions are in the quiz. 2. ask the user to enter the key (that is, the correct answers). there should be one answer for each question in the quiz, and each answer should be an integer. e. g., 34 7 13 100 81 3 9 10 321 12 might be the key for a 10-question quiz. you will need to store the key in an array. 3. ask the user to enter the answers for the quiz to be graded. there needs to be one answer for each question. note that these answers do not need to be stored; each answer can simply be compared to the key as it is entered. 4. when the user has entered all of the answers to be graded, print the number correct and the percent correct. 5. add a loop so that the user can grade any number of quizzes with a single key. after the results have been printed for each quiz, ask "grade another quiz? (y/n)." note: you only have one array (the key). you are not creating a new key for each set of quiz answers.

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?