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?

the answer would be the last one (constat) she stays at the same rat every two mintues it goes up by 0.4

step-by-step explanation:

step-by-step explanation:

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: