A fleet of nine taxis is to be dispatched to three airports in such a way that three go to airport A, five go to airport B, and one goes to airport C. In how many distinct ways can this be accomplished?
2.44) Refer to Exercise 2.43. Assume that taxis are allocated to airports at random.
a) If exactly one of the taxis is in need of repair, what is the probability that it is dispatched to airport C?
b) If exactly three of the taxis are in need of repair, what is the probability that every airport receives one of the taxis requiring repairs?
So, my answer to 2.44a is 1/9. Hopefully this is correct at least :)
For 2.44b, my guess was
(3C1)(1/3)(2/3)2 * (5C1)(1/3)(2/3)4 * 1/3
The solutions manual on chegg (which seems to be riddled with errors) says something completely different. Is my calculation correct?
Answer: I think it’s f(x)=2x+2
Step-by-step explanation:
Fist add 203 and 220 and you get 423.
Then add all the dvds 11+3 and you get 14 then
Next add all the video games 11+14 and you get 25
next divide 25 from 423 then you get 16.92
then divide the dvds 423 divided by 14 and you get 30.21
Answer:
D
Step-by-step explanation:
If x is not equal to zero, then 1/x can be equal to x.
So which is k * 1/x if x =1?
D, because x is the same as 1x, which is equivalent to 1/1 (x).
I may be wrong though please look at other answers and if you have questions then comment
