9514 1404 393
Answer:
a) 3003
b) 1419
Step-by-step explanation:
a) With no restrictions, the problem amounts to counting the number of combinations of 14 people taken 6 at a time. That is ...
14!/(6!(14-6)!) = 3003 . . . teams with no restrictions
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b) With the restrictions, there are two cases:
(i) all 6 people are taken from the 12 that exclude the couple. That number of combinations is ...
12!/(6!(12-6)!) = 924
(ii) 4 people are taken from the 12 that exclude the couple, and the couple fills the last two team slots. That number is ...
12!/(4!(12-4)!) = 495
The sum of the counts of these two cases is ...
924 +495 = 1419 . . . teams with inseparable couple
f-1 (1/2) =2, f-1(8) = 6
i did it on edge
C) p(x) has a zero between x=2 and x=4
For a polynomial p(x), means that, when p(x) is divided by , the remainder is .
This is called the remainder theorem
Therefore , means that when p(x) is divided and is 1 and respectively.
Also the Intermediate value theorem says that, if p(a) and p(b) have opposite signs then there is a certain on where i.e there is a zero between and
Therefore the correct answer is C
10/12 or 5/6
1/2 + 1/2 = 1
5/12 + 5/12 = 10/12 or 5/6