Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
ssgayle
Group Title
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?
 11 months ago
 11 months ago
ssgayle Group Title
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?
 11 months ago
 11 months ago

This Question is Closed

rushal Group TitleBest 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!! :)
 11 months ago

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

rushal Group TitleBest ResponseYou've already chosen the best response.0
sure if i can figure it out
 11 months ago

ssgayle Group TitleBest 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.
 11 months ago

ssgayle Group TitleBest ResponseYou've already chosen the best response.0
if u can that would be great!
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
why is it \(\binom{9}{3}\) ?
 11 months ago

rushal Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 what bro ?
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
pretty sure that is not right
 11 months ago

rushal Group TitleBest ResponseYou've already chosen the best response.0
i guess but i m just trying to help _
 11 months ago

satellite73 Group TitleBest 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 \[\frac{9!}{2!3!4!}\]
 11 months ago

satellite73 Group TitleBest 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
 11 months ago

satellite73 Group TitleBest 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
 11 months ago

satellite73 Group TitleBest 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
 11 months ago

ssgayle Group TitleBest 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?
 11 months ago

rushal Group TitleBest ResponseYou've already chosen the best response.0
i got it too satellite73 is correct my bad _
 11 months ago

satellite73 Group TitleBest 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
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
it all comes from the counting principle dw:1378173059624:dw
 11 months ago

ssgayle Group TitleBest 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.
 11 months ago

satellite73 Group TitleBest 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
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
700 are done by john, 300 by meghan
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\(.02\times 700=14\) mistakes are made by john and \(.04\times 300=12\) by meghan
 11 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
so it is pretty damned close, but john makes more mistakes
 11 months ago

ssgayle Group TitleBest 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
 11 months ago

ssgayle Group TitleBest 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
 11 months ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.