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anonymous
 3 years 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?
anonymous
 3 years 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.09C3 = . 9!  = 6! 3! 9 x 8 x 7  = 84 <==ANSWER 3 x 2 x 1 I hope that helps!! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0okay thats what i thought can you help me with one other problems?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sure if i can figure it out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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
 3 years ago
Best ResponseYou've already chosen the best response.0if u can that would be great!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why is it \(\binom{9}{3}\) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 what bro ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0pretty sure that is not right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i guess but i m just trying to help _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you 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
 3 years ago
Best ResponseYou've already chosen the best response.0hey no problem, i just don't think "9 choose 3" is the right way to solve it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you have \(9!\) ways to order the people then divide by the number of ways to permute the people in each car

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0also notice that the "9 choose 3" approach ignores completely the number of people in each car

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ooooo 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
 3 years ago
Best ResponseYou've already chosen the best response.0i got it too satellite73 is correct my bad _

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.09 choose 3 has really nothing to do with this problem it is the way to choose 3 things from a set of 9

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it all comes from the counting principle dw:1378173059624:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would 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
 3 years ago
Best ResponseYou've already chosen the best response.0this is probably easiest if you do it with numbers lets imagine there are 1000 jobs total

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0700 are done by john, 300 by meghan

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(.02\times 700=14\) mistakes are made by john and \(.04\times 300=12\) by meghan

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so it is pretty damned close, but john makes more mistakes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea i wasnt quite sure what to do with the numbers but that makes sense since its a rate in in units work/ mistake

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can we do one more proeblem... sorry i keep asking but these few have been bothering me
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