What term defines internet applications that people connect with each other through a virtual environment by providing them with a collection of online tools?
a. blogs
b. forums
c. podcasts
d. social networks

Selective incapacitation is a strategy to reduce prison population

Write a function so that the main0 code below can be replaced by the simpler code that calls function mphandminutes tomiles0. original main0 int main) l double milesperhour-70.0; double minutestraveled = 100.0; double hourstraveled; double milestraveled; hourstraveled = minutestraveled / 60.0; milestraveled = hourstraveled * milesperhour; cout < "miles" 2 using namespace std; 4 /* your solution goes here/ 6 int maino 1 test passed 7 double milesperhour 70.0 all tests passed 8 double minutestraveled 100.0; 10 cout < < "miles: " < < mphandminutestomiles(milesper-hour, minutestraveled) < < endl; 12 return 0; 13

The idea that, for each pair of devices v and w, there’s a strict dichotomy between being “in range” or “out of range” is a simplified abstraction. more accurately, there’s a power decay function f (·) that specifies, for a pair of devices at distance δ, the signal strength f(δ) that they’ll be able to achieve on their wireless connection. (we’ll assume that f (δ) decreases with increasing δ.) we might want to build this into our notion of back-up sets as follows: among the k devices in the back-up set of v, there should be at least one that can be reached with very high signal strength, at least one other that can be reached with moderately high signal strength, and so forth. more concretely, we have values p1 ≥ p2 ≥ . . ≥ pk, so that if the back-up set for v consists of devices at distances d1≤d2≤≤dk,thenweshouldhavef(dj)≥pj foreachj. give an algorithm that determines whether it is possible to choose a back-up set for each device subject to this more detailed condition, still requiring that no device should appear in the back-up set of more than b other devices. again, the algorithm should output the back-up sets themselves, provided they can be found.\