Answer:
2720
Step-by-step explanation:
because I used a calculator
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:
Ones place: 456.3000
tenths place: 456.3000
hundredths place: 456.2990
thousandths place: 456.2991
She needs more then 472 and has 296 so she still needs 176 because 472-296=176
Each bracelets is 4 so then 176/4 is 44
1. She needs to sell 44 more bracelets,45 if she needs more than 472
2 the inequality would be x>=44 bracelets
Answer: it’s half as much of the radius so the relationship is 4
Step-by-step explanation: