Suppose we modify the deterministic linear-time selection algorithm by grouping the elements into groups of 7, rather than groups of 5. (use the "median-of-medians" as the pivot, as before.) does the algorithm still run in time? what if we use groups of 3?

What is the missing step in this proof ?

Aschool allots ? 1500 to spend on a trip to the theatre. theatre tickets have a regular cost of ? 55 each and are on offer for 1 5 off. a train ticket for the day will cost ? 12 each. if 2 teachers and the maximum number of students attend, how much money will the school have left over?

Carla was trying to decide what to do saturday evening. she has $100 available to spend for the evening. she has narrowed her choices down to these events: braves game, a concert of an artist she likes, or the georgia aquarium. admission to the braves game is $25, hot dogs cost $10, sodas cost $5. the concert costs $25 for admission, $20 for a souvernir t-shirt, and $10 for a soda. the georgia aquarium has the show of a lifetime with an all you can eat and drink that costs $100. carla does not want to take any driving chances, so in addition to the cost of the aquarium, she will also need to pay for her uber/lyft, which is $15 each way. which option should carla choose if she wants to spend the least? based on that choice, what then is carla's oppotunity cost? fully explain your answer in at least one paragraph.