anonymous
  • anonymous
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?
Probability
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what?
anonymous
  • anonymous
9C3 = . 9! ------- = 6! 3! 9 x 8 x 7 ------------- = 84 <==ANSWER 3 x 2 x 1 I hope that helps!! :-)
anonymous
  • anonymous
okay thats what i thought can you help me with one other problems?

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anonymous
  • anonymous
sure if i can figure it out
anonymous
  • anonymous
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.
anonymous
  • anonymous
if u can that would be great!
anonymous
  • anonymous
why is it \(\binom{9}{3}\) ?
anonymous
  • anonymous
@satellite73 what bro ?
anonymous
  • anonymous
pretty sure that is not right
anonymous
  • anonymous
i guess but i m just trying to help -_-
anonymous
  • anonymous
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 \[\frac{9!}{2!3!4!}\]
anonymous
  • anonymous
hey no problem, i just don't think "9 choose 3" is the right way to solve it
anonymous
  • anonymous
you have \(9!\) ways to order the people then divide by the number of ways to permute the people in each car
anonymous
  • anonymous
also notice that the "9 choose 3" approach ignores completely the number of people in each car
anonymous
  • anonymous
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?
anonymous
  • anonymous
i got it too satellite73 is correct my bad -_-
anonymous
  • anonymous
9 choose 3 has really nothing to do with this problem it is the way to choose 3 things from a set of 9
anonymous
  • anonymous
it all comes from the counting principle |dw:1378173059624:dw|
anonymous
  • anonymous
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.
anonymous
  • anonymous
this is probably easiest if you do it with numbers lets imagine there are 1000 jobs total
anonymous
  • anonymous
700 are done by john, 300 by meghan
anonymous
  • anonymous
\(.02\times 700=14\) mistakes are made by john and \(.04\times 300=12\) by meghan
anonymous
  • anonymous
so it is pretty damned close, but john makes more mistakes
anonymous
  • anonymous
yea i wasnt quite sure what to do with the numbers but that makes sense since its a rate in in units work/ mistake
anonymous
  • anonymous
can we do one more proeblem... sorry i keep asking but these few have been bothering me

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