The simplest fraction for is . Write the upper bound as a fraction with the same denominator:
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Hence the range for would be:
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If the denominator of is also , then the range for its numerator (call it ) would be . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than .
To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)
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At this point, the difference between the numerators is now . That allows a number ( in this case) to fit between the bounds. However, can't be written as finite decimals.
Try multiplying the numerator and the denominator by a different number.
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It is important to note that some expressions for can be simplified. For example, because of the common factor .
From the figure, it can be seen that the mage A'B'C' is obtained from the pre-image ABC by translating/shifting the vertices of the image by 7 units to the right and 6 units down.
Therefore, the rule represents the translation from the pre-image, ΔABC, to the image, ΔA'B'C' is (x, y) → (x + 7, y – 6).
(A) These events are pairwise disjoint. This is false. Pairwise disjoint are also known as mutually exclusive events. Here we can see that both events are occurring at same time.
(B) These events are independent events. This is also false.
(C) These events are both independent and pairwise disjoint. False
(D) A worker is either married or a college graduate always. False
Here Probability(A or B) shall be 1
= Pr(A) + Pr(B) - Pr( A and B) = 0.74 + 0.42 - 0.56 * 0.42 = 0.9248