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ssgayle

Night engineers are going on an energy assessment in 3 cars that hold 2, 3, and 4 passengers, respectively. In how many ways is it possible to transport the 9 engineers to the manufacturing facility, using all cars?

  • 7 months ago
  • 7 months ago

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  1. ssgayle
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    what?

    • 7 months ago
  2. rushal
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    9C3 = . 9! ------- = 6! 3! 9 x 8 x 7 ------------- = 84 <==ANSWER 3 x 2 x 1 I hope that helps!! :-)

    • 7 months ago
  3. ssgayle
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    okay thats what i thought can you help me with one other problems?

    • 7 months ago
  4. rushal
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    sure if i can figure it out

    • 7 months ago
  5. ssgayle
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    A construction company employs two sales engineers, John and Meghan. John does the work of estimating cost for 70% of the jobs bid by the company. Meghan does the work for 30% of the jobs bid by the company. It is known that the error rate for John is such that 0.02 is the probability of an error when he does the work, whereas the probability of an error in the work of Meghan is 0.04. Suppose the bid arrives and a serious error occurs in estimating cost. Which engineer would you guess did the work? Explain and show work.

    • 7 months ago
  6. ssgayle
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    if u can that would be great!

    • 7 months ago
  7. satellite73
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    why is it \(\binom{9}{3}\) ?

    • 7 months ago
  8. rushal
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    @satellite73 what bro ?

    • 7 months ago
  9. satellite73
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    pretty sure that is not right

    • 7 months ago
  10. rushal
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    i guess but i m just trying to help -_-

    • 7 months ago
  11. satellite73
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    you have \(9\) people to order in groups of 2, 3, and 4 i am fairly sure that the number of ways to do this is \[\frac{9!}{2!3!4!}\]

    • 7 months ago
  12. satellite73
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    hey no problem, i just don't think "9 choose 3" is the right way to solve it

    • 7 months ago
  13. satellite73
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    you have \(9!\) ways to order the people then divide by the number of ways to permute the people in each car

    • 7 months ago
  14. satellite73
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    also notice that the "9 choose 3" approach ignores completely the number of people in each car

    • 7 months ago
  15. ssgayle
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    ooooo i see cause it has to be either 2 groups or 3 or 4... if it was just random groups then it could be 9 chose 3?

    • 7 months ago
  16. rushal
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    i got it too satellite73 is correct my bad -_-

    • 7 months ago
  17. satellite73
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    9 choose 3 has really nothing to do with this problem it is the way to choose 3 things from a set of 9

    • 7 months ago
  18. satellite73
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    it all comes from the counting principle |dw:1378173059624:dw|

    • 7 months ago
  19. ssgayle
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    would you be able to help me with another difficult problem satellite or rushal? A construction company employs two sales engineers, John and Meghan. John does the work of estimating cost for 70% of the jobs bid by the company. Meghan does the work for 30% of the jobs bid by the company. It is known that the error rate for John is such that 0.02 is the probability of an error when he does the work, whereas the probability of an error in the work of Meghan is 0.04. Suppose the bid arrives and a serious error occurs in estimating cost. Which engineer would you guess did the work? Explain and show work.

    • 7 months ago
  20. satellite73
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    this is probably easiest if you do it with numbers lets imagine there are 1000 jobs total

    • 7 months ago
  21. satellite73
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    700 are done by john, 300 by meghan

    • 7 months ago
  22. satellite73
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    \(.02\times 700=14\) mistakes are made by john and \(.04\times 300=12\) by meghan

    • 7 months ago
  23. satellite73
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    so it is pretty damned close, but john makes more mistakes

    • 7 months ago
  24. ssgayle
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    yea i wasnt quite sure what to do with the numbers but that makes sense since its a rate in in units work/ mistake

    • 7 months ago
  25. ssgayle
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    can we do one more proeblem... sorry i keep asking but these few have been bothering me

    • 7 months ago
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