answer:To find the joint distribution of (B, C), we have: P(B = b, C = m) = P(A - B = m, B = b) = P(A = b + m, B = b) = P(A = b + m)P(B = b) since A and B are independent. To find the conditional distribution of B given C = m, we use Bayes' theorem: P(B = b | C = m) = frac{P(B = b, C = m)}{P(C = m)} where P(C = m) = sum_{b=0}^{infty} P(B = b, C = m) is the marginal distribution of C.

answer:To solve this limit without derivatives, we can use the following approach. First, let's make a substitution: x-1=t, which implies x=t+1. The limit then becomes lim_{trightarrow 0} frac{sin(pi (t+1)^a)}{sin(pi (t+1)^b)} Expanding (t+1)^a and (t+1)^b using the binomial theorem (assuming a and b are positive integers for simplicity), we get lim_{trightarrow 0} frac{sin(pi (t^a + at^{a-1} + dots + at))}{sin(pi (t^b + bt^{b-1} + dots + bt))} Now, apply the limit and the fact that lim_{xrightarrow 0} frac{sin(pi x)}{pi x} = 1: lim_{trightarrow 0} frac{pi (t^a + at^{a-1} + dots + at)}{pi (t^b + bt^{b-1} + dots + bt)} = frac{a}{b} This solution assumes a and b are positive integers. For general real numbers, the method needs to be adjusted. For instance, when a > 1, consider the left limit and let x = 1 - y as y rightarrow 0^+. We want to evaluate lim_{yrightarrow 0^+} frac{sin(pi (1 - y)^a)}{pi ay} Using inequalities (1 - y)^a geq 1 - ay and (1 - y)^a leq 1 - ay + frac{a(a-1)}{2}y^2 for small y > 0 and the decreasing property of sin(pi x) near pi, we can sandwich the limit: frac{sinleft(pileft(1-ay+frac{a(a-1)}{2}y^2right)right)}{pi ay} leq frac{sin(pi (1 - y)^a)}{pi ay} leq frac{sin(pi (1 - ay))}{pi ay} Since sin(pi (1 - ay)) = sin(pi ay) and sinleft(pileft(1-ay+frac{a(a-1)}{2}y^2right)right) = sinleft(pi ay-pifrac{a(a-1)}{2}y^2right), the limit converges to 1. Therefore, for the original limit, lim_{x rightarrow 1} frac{sin(pi x^a)}{sin(pi x^b)} = frac{a}{b} This solution does not use the exponential form of the sin function. However, if desired, the limit can be expressed using Euler's formula: lim_{x rightarrow 1} frac{sin(pi x^a)}{sin(pi x^b)} = lim_{x rightarrow 1} frac{frac{e^{-i pi x^a} - e^{i pi x^a}}{2i}}{frac{e^{-i pi x^b} - e^{i pi x^b}}{2i}} Simplifying, we obtain the same result: lim_{x rightarrow 1} frac{sin(pi x^a)}{sin(pi x^b)} = frac{a}{b}

answer:** The equilibrium constant {eq}rm (textit{K}_P) {/eq} for the reaction between carbon monoxide and molecular chlorine to form COCl{eq}_2 {/eq}(g) at 74° C is 0.758 atm{eq}^{-1}. **Explanation:** The equilibrium constant {eq}rm (textit{K}_P) {/eq} is calculated using the equilibrium concentrations of the reactants and products and the ideal gas constant {eq}rm (R) {/eq} and the temperature {eq}rm (T) {/eq}. The equilibrium reaction is given as: {eq}rm CO(g) + Cl_2(g) rightleftharpoons COCl_2(g) {/eq} The equilibrium constant expression is: {eq}rm textit{K}_P = dfrac{p_{COCl_2}}{p_{CO}p_{Cl_2}} {/eq} where {eq}rm p_{COCl_2}, p_{CO}, {/eq} and {eq}rm p_{Cl_2} {/eq} are the partial pressures of COCl{eq}_2 {/eq}, CO, and Cl{eq}_2 {/eq}, respectively. At equilibrium, the partial pressures are proportional to the molar concentrations: {eq}rm p_{COCl_2} = [COCl_2]RT {/eq} {eq}rm p_{CO} = [CO]RT {/eq} {eq}rm p_{Cl_2} = [Cl_2]RT {/eq} Substituting these expressions into the equilibrium constant expression, we get: {eq}rm textit{K}_P = dfrac{[COCl_2]RT}{[CO]RT[Cl_2]RT} {/eq} {eq}rm textit{K}_P = dfrac{[COCl_2]}{[CO][Cl_2]} {/eq} Substituting the given equilibrium concentrations, we get: {eq}rm textit{K}_P = dfrac{0.14 M}{0.012 M times 0.054 M} = 0.758 atm^{-1} {/eq} Therefore, the equilibrium constant {eq}rm (textit{K}_P) {/eq} for the reaction between carbon monoxide and molecular chlorine to form COCl{eq}_2 {/eq}(g) at 74° C is 0.758 atm{eq}^{-1}.

answer:To find the value of the infinite continued fraction, we can represent it as y = frac{1}{x+frac{1}{y}}. Solving for y gives us a quadratic equation in terms of y: [y^2 + xy - 1 = 0] Substitute x = frac{21683}{37672} into the equation: [left(y + frac{21683}{37672}right)^2 - 1 = 0] Now, solve for y: [y = frac{-21683 pm sqrt{(21683)^2 + 4 cdot 37672}}{2 cdot 37672}] We are only interested in the positive square root: [y = frac{-21683 + sqrt{245874833}}{75344}] Since the infinite continued fraction is the arithmetic mean of x and y, the final answer is: [frac{x + y}{2} = frac{frac{21683}{37672} + frac{-21683 + sqrt{245874833}}{75344}}{2}] Simplifying this expression: [frac{sqrt{245874833} - 21683}{75344}] So, the value of the infinite continued fraction is: [frac{sqrt{245874833} - 21683}{75344}]