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1000^(1/6) = 3 | 1000^(1/6) = 3 | ||
1000^(1/5) = 3 | 1000^(1/5) = 3 | ||
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How do you know it's the power of (1/6) and not (1/5)? | How do you know it's the power of (1/6) and not (1/5)? | ||
Revision as of 07:21, 5 September 2008
#12 in 7.5
It would be amazing if someone could explain the intersection between A and B on problem number 12. I have so far the squares being 31 and the cubes being 10. But I am not sure how to get the intersection. Any ideas? Also I do not understand 20 or 28 at all. If someone could please help me out that would ROCK! Thanks
Possible Solutions
I think that if you want numbers between 1 and 1,000 that are both cubes and squares you'd do (1,000)^1/6 but thats just my guess because when you want,
how many squares: (1,000)^1/2 = 31
how many cubics: (1,000)^1/3= 10
so how many squares and cubics: (1,000)^1/6 = 3
12. A: # of squares: 31 B: # of cubics: 10 A intersection B: 3
31+10-3=38
I also struggled with finding the intersection of the two, and although I got the same answer as you, I'm still lost. I need to know WHY you are choosing to raise 1000 to the power of (1/6).
Also:
1000^(1/6) = 3 1000^(1/5) = 3
How do you know it's the power of (1/6) and not (1/5)?
#20 in 7.5
If anyone is still having trouble with 20, it might help to look at the answer to 19 in the back of the book. It gives an explicit formula for the number of elements in the union of 5 sets. It should help make the Principle of Inclusion-Exclusion a little more clear.
#28 in 7.5
Can someone rephrase the question or shed some light on what this question is asking? I looked at the solution for #29 which seems to be quite similar, but it was a notation we haven't learned in class.
I looked at the solution for #27, which seems to be a similar problem to #28. I think the inclusion/exclusion property for sets also holds for probabilities, and I think the point of the "no two events can occur at the same time" stipulation is to tell you that the probability of any intersection of events, whether 2, 3, or n, will be zero. That said, I think you just need to add the probabilities of the independent events.
Related: 4.1 Homework_MA375Fall2008walther
