Identify the best big-O estimate for the number of comparisons used by the algorithm that determines the number of 1s in a bit string by examining each bit of the string to determine whether it is a 1 bit.

0(n)

Step-by-step explanation:

Result previous exercise:  

procedure count(a1a2...an : string with n > 1)

i:=0

for k:=1 to n

if ak =1 then i:=i + I

return i  

Note: If you use a different algorithm, then you could possible get different results.  

SOLUTION  

There is only one part of the code that contains an operation (comparison), namely if a_k =1  

This comparison is executed in every iteration of the for-loop  

k can take on the values from 1 to n (for k:=I to n), thus k can take on n values.  

Thus in total there are then n comparisons, while n is 0(n).

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.  

c 1 hour

step-by-step explanation:

m = 2e

m = 2 so 2 = 2e

1 = e