On 4 Jul 2003 at 9:13, Henry Baker wrote:
Interspersed comments, below:
Indeed! Where they belong!! :o)
At 11:11 AM 7/4/03 -0400, William Thurston wrote:
20. How many different five-member teams can be made from a group of eight students, if each student has an equal chance of being chosen? (1) 40 (2) 56 (3) 336 (4) 6720
I think the correct answer is not among these choices: 1 It's very poorly phrased, being neither colloquial English or typical mathematical/combinatorial English.
I agree that there is a reading that would give the choice "1", but since that choice is not available, you have to assume that that interpretation is not intended.
Could you elaborate on how to get the choice '1'? Hmm...well... I guess you could say that you can only make *ONE* five-man team, but you could do it in 56 different ways... Is that what you had in minde?
21. The student scores on Mrs. Frederick's mathematics test are shown on the stem and leaf plot below: 4 | 3 6 | 0 5 5 7 9 7 | 2 5 6 8 9 9 9 9 | 0 1 2 5 9
Find the median of these scores.
--this is a new one for me. I'm clueless, even seeing the key, which I'm not printing.
Ditto. It would be interesting to see how many Fields medal winners would be able to understand this one.
I must be looking at this wrong, because I didn't have any trouble with it. I've never heard of a 'stem and leaf plot', but it seemed obvious enough to me just looking at it and guessing [later confirmed with a quick google search] what the blasted thing is.... [and I assume that the students taking the test HAD already heard of/used such a thing, so that shouldn't be a mystery to them]. And i get the answer '77' [assuming that for the median of an even set you average the two 'middle' elements]. Is there some subtlety/difficulty that I'm not seeing?? /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--