answer:From the graph, let's locate the values of x for which f(x) = 2. Upon observation, f(x) = 2 when x = -2, 0, 4. Hence, f(f(x)) = 2 implies f(x) = -2, 0, 4. 1. For f(x) = -2, there is no x such that f(x) = -2 (as per graph). 2. For f(x) = 0, there is no x such that f(x) = 0 (as per graph). 3. For f(x) = 4, x = 2 such that f(x) = 4. Thus, there is only boxed{1} value of x which satisfies f(f(x)) = 2.

answer:Let's denote the number we are looking for as x. According to the problem, three times this number (3x) is the same as the difference of a certain multiple of the number and seven. Let's denote the certain multiple as k, so we have kx (where k is a constant). The equation based on the problem statement is: 3x = kx - 7 We are given that the number is -7.0, so we substitute x with -7.0: 3(-7.0) = k(-7.0) - 7 Now we solve for k: -21 = -7k - 7 Add 7 to both sides: -21 + 7 = -7k - 7 + 7 -14 = -7k Now divide both sides by -7 to solve for k: -14 / -7 = -7k / -7 2 = k So the certain multiple k is 2. The number we were looking for is indeed -7.0, and the equation is satisfied with k = boxed{2} .

answer:Let ( Y = (Y_1, Y_2, ldots) ) be an independent copy of the sequence ( X = (X_1, X_2, ldots) ). Then, the random sequences [ frac{X + Y}{sqrt{2}} quad text{and} quad frac{X - Y}{sqrt{2}} ] are independent and have the same distributions as ( X ). Let's denote ( |(x_1, x_2, ldots)| = sup_i |x_i| ). We need to show that: [ mathrm{E} sup_n |X_n| < infty ] Consider for all ( n geqslant 1 ): [ mathrm{P}(|X| < n) mathrm{P}(|X| > 2n) = mathrm{P}left( left| frac{X + Y}{sqrt{2}} right| < n, left| frac{X - Y}{sqrt{2}} right| > 2n right) ] Notice that: [ left{ left| frac{X + Y}{sqrt{2}} right| < n right} subseteq left{ left| frac{X - Y}{sqrt{2}} right| - sqrt{2}|X| land |Y| < n right} ] Thus, we have the following relationships: [ left{ left| frac{X + Y}{sqrt{2}} right| < n, left| frac{X - Y}{sqrt{2}} right| > 2n right} subseteq left{ 2n - sqrt{2}|X| wedge |Y| < n right} = left{ |X| > frac{n}{sqrt{2}}, |Y| > frac{n}{sqrt{2}} right} ] From this, we deduce that: [ mathrm{P}(|X| < n) mathrm{P}(|X| > 2n) leqslant mathrm{P}left( |X| > frac{n}{sqrt{2}} right)^2 ] Taking the limit as ( n to infty ): [ lim_{n to infty} frac{mathrm{P}(|X| > 2n)}{mathrm{P}(|X| > n / sqrt{2})} = 0 ] Because by assumption: [ mathrm{P}(|X| < n) to mathrm{P}(|X| < infty) = 1 quad text{and} quad mathrm{P}(|X| > n / sqrt{2}) to mathrm{P}(|X| = infty) = 0 ] Therefore ( mathrm{E}|X| ) is finite as derived from the observation in problem II.6.40. [ boxed{mathrm{E} sup_n |X_n| < infty} ]

answer:To solve this problem, we need to identify the angle whose cosine is -frac{1}{2}. We know from the unit circle and trigonometric values that: - cos left( frac{2pi}{3} right) = -frac{1}{2}, - cos left( frac{4pi}{3} right) = -frac{1}{2}. Since arccos function in its principal value range (from 0 to pi) returns the smallest positive angle corresponding to the cosine value, we consider only frac{2pi}{3}. Therefore, arccos left( -frac{1}{2} right) = boxed{frac{2pi}{3}}.