How many comparisons are needed for a binary search in a set of 64 elements?

Let f(n) be the number of comparisons required to search for an element in a search sequence of size n. Then from the binary search algorithm we can have f(n) = f(n/2) + 2, where n is even.

The number of comparisons needed for a binary search in a set of 64 elements is f(64).

f(64) = f (32) + 2

= f(16) + 2 + 2

= f(8) + 2 + 2 + 2

= f(4) + 2 + 2 + 2 + 2

= f(2) + 2 + 2 + 2 + 2 + 2

= f(1) + 2 + 2 + 2 + 2 + 2 + 2

= 0 + 2 + 2 + 2 + 2 + 2 + 2+ 2

f(64) = 14

192/15 = 12.8.

12.8

step-by-step explanation:

192/15 is 12.8.

however, it says to round to the nearest foot, which would equal 13.

a.for lorena 14 hours makes $126   for 16 hours she makes 144

for julie 6 hours makes $72 for 21 hours its $252

b. julie makes more money than lorena julie makes $12 an hour and lorena makes $9  

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