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Q. Suppose Jessica has 10 pairs of shorts and 5 pairs of jeans in her drawer. How many ways could she pick out something to wear for the day? What is the probability that she picks out shorts (assuming she picks out her clothing blindfolded)?
Solution:
Jessica could choose a pair of shorts in 10 ways or a pair of jeans in 5 ways, but she can't choose both types of clothing to wear today. Thus she can choose her clothes in 10 + 5 = 15 ways.
Since there are 10 pairs of shorts in the drawer, the probability of Jessica choosing one of them is
I need help with this question: Find the probability that two quarters and a nickel are chosen without replacement from a bag of 8 quarters and 12 nickles.
Find the 14th term in the arithmetic sequence. 60, 68, 76, 84, 92
200 + 406578
#question.2157\7625.
Let D(subscript12) = ({x,y : x^2 = e ; y^6 = e ; xy =(y^-1) x}) a) Which of the following subsets are subgroups of D(subscript12) ? Justify your answer. i) {x,y,xy,y^2,y^3,e}
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