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Murrr4er [49]
3 years ago
7

Help PLEASESSS due in 20

Mathematics
1 answer:
kicyunya [14]3 years ago
7 0
  1. 4 - 4 + 4 ÷ 4
  2. 4 ÷ 4 + 4 ÷ 4
  3. (4 + 4 + 4) ÷ 4
  4. √4 + √4 + 4 - 4
  5. √4 + 4 + 4 ÷ 4
  6. √4 + 4 + 4 - 4
  7. 4 + 4 - 4 ÷ 4
  8. 4 + 4 + 4 - 4
  9. 4 + 4 + 4 ÷ 4
  10. √4 + √4 + √4 + 4
  11. 44/(√4 + √4)
  12. √4 + √4 + 4 + 4
  13. 44/4 + 4
  14. 4 + 4 + 4 + √4
  15. 44/4 + 4
  16. 4 * 4 * 4 ÷ 4
  17. 4 * 4 + 4 ÷ 4
  18. 4 * 4 - √4 + 4
  19. 4! - 4 - 4 ÷ 4
  20. 4 * (4 + 4 ÷ 4)
  21. 4! - 4 + 4 ÷ 4
  22. 4 * 4 + 4 + √4
  23. 4! - √4 + 4/4
  24. 4 * (√4 + √4 + √4)
  25. 4! + √2 - 4 ÷ 4
  26. 4! + √4 + 4 - 4
  27. 4! + √4 + 4 ÷ 4
  28. 4! + 4 + 4 - 4
  29. 4! + 4 + 4 ÷ 4
  30. 4! + √4 + √4 + √4

Lol, that took a while, hope it helps!

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g A law firm has six senior and seven junior partners. A committee of three partners is selected at random to represent the firm
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Answer:

133/143

Step-by-step explanation:

Let S be the sample space

Let E be the event of selecting three committee partners with at least one junior partner.

Partners in the law firm include:

Senior partners = 6

Junior partners = 7

Total partners = 13

n(S) = number of ways of selecting 3 partners from 13 = 13C3

n(S) = 13C3 = 13!/(10!3!) = (13x12x11)/(3x2x1) = 286

To get n(E) i.e least 1 junior partner in the selected committee, we may have:

(2 senior and 1 junior) or ( 1 senior and 2 junior) or (3 junior).

Therefore, the required number of way is given below:

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= 105 + 126 + 35

n(E) = 266

Therefore, the probability P(E) that at least one of the junior partners is on the​ committee is given below:

P(E) = n(E) /n(S)

P(E) = 266/286

P(E) = 133/143

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