Suppose that an element is known to be among the first four elements in an ordered list of 32. Would a linear search or a binary search locate this element more rapidly? Linear Search Now suppose it is know to be among the 5th through 8th elements in a list of 32 elements. On average, which search would be faster (in terms of number of comparisons)? Binary Search Suppose that an element is known to be among the first four elements in an ordered list of 32. What is the worst case number of comparisons that will be done to locate the element with a linear search? Neglect any comparisons that are done to check whether or not you are still in the for loop. Choose... Suppose that an element is known to be among the first four elements in an ordered list of 32. What is the worst case number of comparisons that will be done to locate the element with a binary search? Neglect any comparisons that are done to check whether or not you are still in the for loop Choose Check

step-by-step explanation:

i need a picture

step-by-step explanation:

tip $2.90

total $17.40

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages