Answer: Informal bench-marking
Explanation:
Informal bench-marking is defined as unconscious comparison of one's own behavior, skills, values etc with other and learning from them to improve. This leaning can be found in work-place, home, school etc.
- According to the question, Myles is using informal bench-marking through studying other stores complaint handling style and reduction technique so that he can learn from it.
- Other options are incorrect because designing analysis,outcome analysis, issue analysis and processing of complaining ta re not the comparison that unconsciously done by person .
- Thus, the correct option is informal bench- marking.
What are the options for this answer?
<span>In order to perform maintenance on a router and need to temporarily reroute traffic through another office the best way to perform this action would be to configure a static route on the router.
</span>By doing this, the router will use the manually-<span>configured routing entry to send the packets.</span>
Answer:
The solution code is written in Python 3:
- def modifyList(listNumber):
- posCount = 0
- negCount = 0
-
- for x in listNumber:
- if x > 0:
- posCount += 1
- else:
- negCount += 1
-
- if(posCount == len(listNumber)):
- listNumber.append(max(listNumber))
-
- if(negCount == len(listNumber)):
- listNumber.append(min(listNumber))
-
- print(listNumber)
-
- modifyList([-1,-99,-81])
- modifyList([1,99,8])
- modifyList([-1,99,-81])
Explanation:
The key step to solve this problem is to define two variables, posCount and negCount, to track the number of positive value and negative value from the input list (Line 2 - 3).
To track the posCount and negCount, we can traverse through the for-loop and create if else statement to check if the current number x is bigger than 0 then increment posCount by 1 otherwise increment negCount (Line 5- 9).
If all number in the list are positive, the posCount should be equal to the length of the input list and the same rule is applied to negCount. If one of them happens, the listNumber will append either the maximum number (Line 11 -12) or append the minimum number (Line 14-15).
If both posCount and negCount are not equal to the list length, the block of code Line 11 -15 will be skipped.
At last we can print the listNumber (Line 17).
If we test our function using the three sets of input list, we shall get the following results:
[-1, -99, -81, -99]
[1, 99, 8, 99]
[-1, 99, -81]
