Induction Proof a. Consider the following induction "proof" that all sheep in a flock are the same color: Base case: One sheep. It is clearly the same color as itself. Induction step: A flock of n sheep. Take a sheep, a, out of the flock. The remaining n - 1 are all the same color by induction. Now put sheep a back in the flock, and take out a different sheep, b. By induction, the n - 1 sheep (now with a in their group) are all the same color. Therefore, a is the same color as all the other sheep; hence, all the sheep in the flock are the same color. What is wrong with this "proof"?

So im doing this school challenge and the teachers said whats the average text a student gets a day so i need to get about 20 in a day but dont know how can you guys 2163371293

What are ways to switch windows in excel? check all that apply. on the status bar, click the windows button, and then click the file name. on the task bar, click to display the excel jump list, and then click the file name. on the view tab, in the window group, click switch windows, and then click the file name. on the review tab, in the viewing group, click files, and then click the file name.

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?