Assuming that a user enters 68 as the score, what is the output of the following code snippet? int score = 68; if (score < 50) { System. out. println("F"); } else if (score >= 50 || score < 55) { System. out. println("D"); } else if (score >= 55 || score < 65) { System. out. println("C"); } else if (score >= 65 || score < 75) { System. out. println("B"); } else if (score >= 75 || score < 80) { System. out. println("B+"); } else { System. out. println("A"); } D C B A

Acase study allows a more detailed look at the life of a single subject than any other study.

The camera still is bad even with the new iphone xr and especially in low light it is even worst because you can see the pixels more if its in low light. if all you apple customers want apple to fix this then lets fill there feedback with complaints about the can apple do to fix this issue?

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?