[Numeracy 488] Re: Response & Essentials for productive learning

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Chip Burkitt chip.burkitt at orderingchaos.com
Fri Aug 13 11:31:38 EDT 2010

Sorry; you're right.

The width of the rectangular box is unknown, but we know that its length
is twice the width and its height is six times the width. We also know
that the surface area is 1000 cm^2. Let /x/ be the unknown width. The
surface area is the sum of the areas of the six sides of the box, so
2(/x/)(2/x/) + 2(/x/)(6/x/) + 2(2/x/)(6/x/) = 1000. This simplifies to
40/x/^2 = 1000, which yields /x/ = 5. We can now calculate the volume.
/V/ = 5(10)(30) = 1500. So the volume of the rectangular box is 1500 cm^3.

For the number of possible hands of seven cards containing four of a
kind, I reasoned as follows. There are 13 possible hands of four cards
containing four of a kind, one for each face value. If we add three
cards drawn at random to each of those four-card hands, we will find the
number of seven-card hands containing four of a kind. Since we already
have the four cards, the remainder of the deck consists of 52 - 4 = 48
cards. The number of combinations of 48 things taken 3 at a time is
given by (48!)/(45!3!) = (48·47·46)/(1·2·3) = 17,296. Therefore, the
number of seven-card hands containing four of a kind is 13·17,296 =
224,848. The total number of possible seven-card hands is (52!)/(45!7!)
= 133,784,560, so the probability of getting four of a kind in a
seven-card deal is 224,848/133,784,560 = 1/595. I think this is correct,
but I have to say that I find combinatorics and probability confusing.

Chip Burkitt
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