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In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30. As 120 ÷ 30 = 4, and 90 ÷ 30 = 3, one gets
The picture to the right illustrates 3 / 4 of a cake. Fractions can be used to represent ratios and division. [1] Thus the fraction 3 / 4 can be used to represent the ratio 3:4 (the ratio of the part to the whole), and the division 3 ÷ 4 (three divided by four).
) Usually the resulting fraction should be simplified: the result of the division of 52 by 22 is also . This simplification may be done by factoring out the greatest common divisor . Give the answer as an integer quotient and a remainder , so 26 11 = 2 remainder 4. {\displaystyle {\tfrac {26}{11}}=2{\mbox{ remainder }}4.}
Fair random assignment (also called probabilistic one-sided matching) is a kind of a fair division problem. In an assignment problem (also called house-allocation problem or one-sided matching ), there are m objects and they have to be allocated among n agents, such that each agent receives at most one object.
Informally, the expectation of a random variable with a ... Suppose the random variable X takes values 1, −2,3, −4, ... as well as in the simplified form obtained ...
What is the largest fraction r such that there always exists an allocation giving each agent a utility of at least r times his 1-of-maximin share? Known cases: For two agents: = by divide-and-choose. For three agents, even with additive valuations: <. By a carefully crafted example. [13]
Instead, the division is reduced to small steps. Starting from the left, enough digits are selected to form a number (called the partial dividend) that is at least 4×1 but smaller than 4×10 (4 being the divisor in this problem). Here, the partial dividend is 9. The first number to be divided by the divisor (4) is the partial dividend (9).
Long division is the standard algorithm used for pen-and-paper division of multi-digit numbers expressed in decimal notation. It shifts gradually from the left to the right end of the dividend, subtracting the largest possible multiple of the divisor (at the digit level) at each stage; the multiples then become the digits of the quotient, and the final difference is then the remainder.