Answer: What is the complete question? Can you comment it?
Step-by-step explanation: I can’t help unless I know what the full question is. Sorry.
Answer:
Step-by-step explanation:
Let us assume that if possible in the group of 101, each had a different number of friends. Then the no of friends 101 persons have 0,1,2,....100 only since number of friends are integers and non negative.
Say P has 100 friends then P has all other persons as friends. In this case, there cannot be any one who has 0 friend. So a contradiction. Hence proved
Part ii: EVen if instead of 101, say n people are there same proof follows as
if different number of friends then they would be 0,1,2...n-1
If one person has n-1 friends then there cannot be any one who does not have any friend.
Thus same proof follows.
The correct question is
<span>Michael sold 2/3 of his CDs and purchased 16 more. If Michael has 25 CDs now, how many CDs did he have to begin with?
Let
x--------------> numbers of CDs Michael have to begin
if Michael sold (2/3)x-----------> </span><span>are pending to sell (1/3)x
we know that
(1/3)x+16=25--------> </span>(1/3)x=25-16-------> (1/3)x=9---------> x=27
the answer is 27 CDs
206 X .4 = 82.4
That's the answer.
