Let Ω be the set of all permutations of the numbers 1,2, ..., n. Let an almostfixed point be defined as follows: If we put the numbersi∈1,2, ..., naround acircle in clockwise order (such that 1 andnare next to each other) and thenassign another numberω(i)∈1,2, ..., nto it, if the numberω(i) is next toi(oris equal toi), we will say thatiis almost a fixed point. So, for the permutationω(1) = 5, ω(2) = 3, ω(3) = 1, ω(4) = 4, ω(5) = 2, we have that 1, 2, and 4 arealmost fixed points. Now, letX(ω) denote the number of almost fixed points inω∈Ω. FindE[X]and var(X). You may assume thatn≥5

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the equation of the line that passes through the points and has o form

if line passes through the points (-1, -3) and (2, 1), then

the equation in the standard form is

  b:   ∠x = 28.6°, ∠y = 31.2°, ∠z = 120.2°

step-by-step explanation:

this is one of those questions where "test taking skill" is all you need. the only answer that has angles that sum to 180° is the one shown above. the remaining choices do not describe the angles of a triangle.

if you want to do more work than simply adding the offered numbers, you can also check them against the law of sines. it should be true that

  x/sin(x) = y/sin(y) = z/sin(z)

and all of these ratios must be greater than 15, the longest side length.

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if you want to solve the triangle from scratch, i'd suggest solving for the largest angle using the law of cosines.

  z² = x² +y² -2xy·cos(z)

  z = arccos((x² +y² -z²)/(2xy)) = arccos(-75.11/149.4) = 120.182°

now you can find y from the law of sines

  y = arcsin(y/z·sin(z)) = 31.242°