answer:Given the series sum: [frac{c/d}{1 - 1/d} = 3.] Thus, frac{c}{d - 1} = 3, leading to c = 3(d - 1). Substituting into the new series: begin{align*} frac{c}{c + 2d} + frac{c}{(c + 2d)^2} + frac{c}{(c + 2d)^3} + dotsb &= frac{c/(c + 2d)}{1 - 1/(c + 2d)} &= frac{c}{c + 2d - 1} &= frac{3(d - 1)}{3(d - 1) + 2d - 1} &= frac{3(d - 1)}{5d - 4}. end{align*} Upon simplifying: begin{align*} frac{3(d - 1)}{5d - 4} &= frac{3d - 3}{5d - 4}. end{align*} Since the final expression depends on the value of d, and no contradiction arises, the answer is: [boxed{frac{3d - 3}{5d - 4}}.]

answer:From left(frac{1}{pi}right)^{-x+1} > left(frac{1}{pi}right)^{x^2-x}, we get -x+1 < x^2-x, which simplifies to x^2 > 1. Solving this, we find: x < -1 or x > 1. Therefore, the solution set for the original inequality is: boxed{{x|x > 1 text{ or } x < -1}}. This solution directly uses the properties of exponential functions to convert the exponential inequality into a quadratic inequality for solving. This question tests the method of solving exponential inequalities and examines the properties of exponential functions, making it a basic question.

answer:To find the probability that both smartphones are defective, we can use the formula for the probability of independent events, which is: P(A and B) = P(A) * P(B given A) First, we calculate the probability of selecting one defective smartphone (event A): P(A) = Number of defective smartphones / Total number of smartphones P(A) = 84 / 250 Next, we calculate the probability of selecting another defective smartphone after the first one has been selected (event B given A). Since one defective smartphone has already been taken, there are now 83 defective smartphones left out of a total of 249 smartphones: P(B given A) = Number of remaining defective smartphones / Total remaining smartphones P(B given A) = 83 / 249 Now we can calculate the probability of both events happening: P(A and B) = P(A) * P(B given A) P(A and B) = (84 / 250) * (83 / 249) Let's calculate the values: P(A) = 84 / 250 = 0.336 P(B given A) = 83 / 249 ≈ 0.3333 P(A and B) ≈ 0.336 * 0.3333 ≈ 0.112 So the approximate probability that both phones are defective is boxed{0.112} or 11.2%.

answer:Let (0,0), (4,0), (0,4), and (4,4) be the vertices of the square. Consider endpoints of a segment on sides adjacent to (0,0), with coordinates (x,0) and (0,y). The segment's length condition, x^2 + y^2 = 3^2 = 9, still holds. The midpoint formula gives left(frac{x}{2}, frac{y}{2}right) for these endpoints. The distance from (0,0) to left(frac{x}{2},frac{y}{2}right) is calculated by d = sqrt{left(frac{x}{2}right)^2 + left(frac{y}{2}right)^2} = sqrt{frac{1}{4}(x^2 + y^2)} = sqrt{frac{1}{4} times 9} = frac{3}{2}. Thus, the midpoints for segments on sides adjacent to (0,0) form a quarter-circle with radius frac{3}{2}. Repeating this for all four corners, the midpoints' enclosed region is comprised of four quarter-circles of radius frac{3}{2} subtracted from the square: Total area of four quarter-circles: [ 4 times left(frac{pi left(frac{3}{2}right)^2}{4}right) = pi left(frac{9}{4}right) = frac{9pi}{4} ] Area of the square: [ 4^2 = 16 ] Area enclosed by midpoints: [ 16 - frac{9pi}{4} ] [ = 16 - 2.25pi ] [ approx 16 - 7.07 ] [ approx 8.93 ] Thus, 100k = 100 times 8.93 = boxed{893}.