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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
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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rushalBest ResponseYou've already chosen the best response.0
9C3 = . 9!  = 6! 3! 9 x 8 x 7  = 84 <==ANSWER 3 x 2 x 1 I hope that helps!! :)
 7 months ago

ssgayleBest ResponseYou've already chosen the best response.0
okay thats what i thought can you help me with one other problems?
 7 months ago

rushalBest ResponseYou've already chosen the best response.0
sure if i can figure it out
 7 months ago

ssgayleBest ResponseYou've already chosen the best response.0
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

ssgayleBest ResponseYou've already chosen the best response.0
if u can that would be great!
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
why is it \(\binom{9}{3}\) ?
 7 months ago

rushalBest ResponseYou've already chosen the best response.0
@satellite73 what bro ?
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
pretty sure that is not right
 7 months ago

rushalBest ResponseYou've already chosen the best response.0
i guess but i m just trying to help _
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
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

satellite73Best ResponseYou've already chosen the best response.0
hey no problem, i just don't think "9 choose 3" is the right way to solve it
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
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

satellite73Best ResponseYou've already chosen the best response.0
also notice that the "9 choose 3" approach ignores completely the number of people in each car
 7 months ago

ssgayleBest ResponseYou've already chosen the best response.0
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

rushalBest ResponseYou've already chosen the best response.0
i got it too satellite73 is correct my bad _
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
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

satellite73Best ResponseYou've already chosen the best response.0
it all comes from the counting principle dw:1378173059624:dw
 7 months ago

ssgayleBest ResponseYou've already chosen the best response.0
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

satellite73Best ResponseYou've already chosen the best response.0
this is probably easiest if you do it with numbers lets imagine there are 1000 jobs total
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
700 are done by john, 300 by meghan
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
\(.02\times 700=14\) mistakes are made by john and \(.04\times 300=12\) by meghan
 7 months ago

satellite73Best ResponseYou've already chosen the best response.0
so it is pretty damned close, but john makes more mistakes
 7 months ago

ssgayleBest ResponseYou've already chosen the best response.0
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

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