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ssgayle

  • one year ago

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?

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  1. ssgayle
    • one year ago
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    what?

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

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

  4. rushal
    • one year ago
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    sure if i can figure it out

  5. ssgayle
    • one year ago
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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.

  6. ssgayle
    • one year ago
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    if u can that would be great!

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

  8. rushal
    • one year ago
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    @satellite73 what bro ?

  9. satellite73
    • one year ago
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    pretty sure that is not right

  10. rushal
    • one year ago
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    i guess but i m just trying to help -_-

  11. satellite73
    • one year ago
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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 \[\frac{9!}{2!3!4!}\]

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

  13. satellite73
    • one year ago
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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

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

  15. ssgayle
    • one year ago
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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?

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

  17. satellite73
    • one year ago
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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

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

  19. ssgayle
    • one year ago
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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.

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

  21. satellite73
    • one year ago
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    700 are done by john, 300 by meghan

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

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

  24. ssgayle
    • one year ago
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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

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

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