Complete The test as quickly as possible
The answer is the team. Teams are self-establishing which no one not even the scrum master tells the team how to turn product backlog into augmentations of stoppable functionality. Each team member put on his or her knowledge to all of the complications. The interaction that outcomes progresses the entire scrum team’s general competence and usefulness. The optimal size for a scrum team is seven people, plus or minus two. When there are fewer than five team members, there is less interface and as a result less productivity improvement. The product owner and scrum master roles are not comprised in this count. Team arrangement may variation at the end of a sprint. Every time Team membership is altered, the productivity increased from self-organization is reduced.
user_str1 = str ( input ("Please enter a phrase: "))
user_str2 = str ( input("Please enter a second phrase: "))
def strcmp (word):
user_in1 = int (len(user_str1))
user_in2 = int (len(user_str2))
if user_in1 > user_in2:
return "Your first phrase is longer"
elif user_in1 < user_in2:
return "Your second phrase is longer"
else:
return "Your phrases are of equal length"
Answer: D
Local news channels
Explanation:
Since Jack is trying to improve local service the local news would be the best fit to reach the maximum amount of people in his area.
Answer:
Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list
Explanation:
Assuming that both lists have firs_t and last_ pointers.
For a singly-linked list ; when locating a kth element, you have iterate through a number of k-1 elements which means that locating an element will be done only in one ( 1 ) direction
For a Doubly-linked list : To locate the Kth element can be done from two ( directions ) i.e. if the Kth element can found either by traversing the number of elements before it or after it . This makes finding the Kth element faster because the shortest route can be taken.
<em>Finding kth element is more efficient in a doubly-linked list when compared to a singly-linked list </em>
