Answer:
990 ways to choose 6 starters out of 14 with exactly two of the three triplets.
Step-by-step explanation:
Ways to choose 2 of the triplets
= C(3,2) = 3! / (2!1!) = 3
Ways to choose the remaining 4 starters out of 11 players left
= C(11,4) = 11! / (4!7!) = 330
Total number of ways to choose 6 starters
= 3*330 = 990
Answer:
p > 220
Explanation:
This inequality says, "p (the weight of the pig) is less than 220." This describes the situation shown. So, the answer is p > 220.
Good luck ^^
Real part of z = 9
Imaginary Part = 3i
Conjugate of z is 9 - 3i
Answer:
Step-by-step explanation:
