"Six is a number perfect in itself, and not because God created the world in six days; rather the contrary is true. God created the world in six days because this number is perfect, and it would remain perfect, even if the work of the six days did not exist." - (St. Augustine, 354-430)   Ancient Egyptian Mathematics: Papyrus is a form of paper made from a reed that grows along the banks of the Nile River. Four very ancient documents provide us information about the amazing mathematics of the Egyptians. (Two will be mentioned here; the other two were referenced in Assignment #74.) The Rollin Papyrus (ca 1350 B.C.): This document, now housed in the Louvre in Paris, supplies examples of Egyptian bread accounts. The Harris Papyrus (ca 1167 B.C.): A document prepared by Rameses IV to document the accomplishments of his father, Rameses III. These documents somehow managed to escape the ravages of time and human conquests. Other sources of knowledge about Egyptian mathematics comes from inscriptions on stone tombs and monuments.

When Herkimer was lazy, why did he want to work in a candle making factory? Answer: He was told you only have to work on wick ends. Herky wants to know: If a dating agency for chickens goes bankrupt, it is because it couldn't make hens meet? Is it fair if the guy who stole calendars only gets 12 months?

ASSIGNMENT #75 Reading: Read Section 12.2, pages 708-712. Written: Page 712/18-25 (can write answers in textbook).

Mathematical word analysis: CARDIOID : From the Greek kardia (heart). In mathematics, a cardioid is a path of a point on a circle as it rolls around another circle. This path is somewhat heart shaped.

Algebra II lass lists are provided here to be used for probability problems. Females are listed in bold type, and are listed first.

Here is a sample probability problem:

If I randomly pick four students from one of the algebra II sections, what is the probability that the group of four students would contain 2 girls and 2 boys?

SECTION 2: The answer is (₄C₂)(₅C₂)/(₉C₄) = (6)(10)/126 = 60/126 = 47.62%.

SECTION 3: The answer is (₅C₂)(₆C₂)/(₁₁C₄) = (10)(15)/330 = 150/330 = 45.45%.

SECTION 4: The answer is (₈C₂)(₅C₂)/(₁₃C₄) = (28)(10)/715 = 280/715 = 39.16%.

Suppose I combine the three classes, and then randomly pick four students from the combined grouping. Since the combined group contains 17 girls and 16 boys, the probability that I would get two girls and two boys would be

(₁₇C₂)(₁₆C₂)/(₃₃C₄) = (136)(120)/(40920) = 16320/340920 = 46.74%.

Problem: From a standard deck of 52 cards, I randomly select five cards. What is the probability that exactly two of the five chosen cards are aces? Solution (with communication): The two aces must be chosen from the four aces in the deck, and the remaining three cards must come from the forty-eight non-aces. The requested probability is (4C2)(48C3)/(52C5) = 0.0399, or approximately 4%.

Problem: What is the probability that a randomly selected bridge hand of 13 cards contains no hearts? Solution: A bridge hand containing no hearts must be chosen from the 39 cards that are not hearts. The requested probability is (39C13)/(52C13) = 0.01279, or about 1.3%.