answer:To find out how many students from the class are going to the regional chess tournament, we can follow these steps: 1. First, we determine how many students are in the after-school chess program. Since one-third of the 24 students are in the program, we calculate this as: [ frac{24}{3} = 8 ] 2. Next, we know that half of those in the chess program will be absent from school on Friday for the regional chess tournament. So, to find out how many students that is, we divide the number of students in the chess program by 2: [ frac{8}{2} = 4 ] Therefore, the number of students from this class going to the tournament is boxed{4}.

answer:(1) When k = frac{1}{2}, we can solve the system of equations: begin{cases} y = frac{1}{2}x + frac{p}{4} y^2 = 2px end{cases} Eliminating y, we get 4x^2 + 28px + p^2 = 0. Thus, begin{align} x_1 + x_2 &= 7p, x_1x_2 &= frac{p^2}{4}. end{align} Hence, (x_1 - x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2 = 49p^2 - p^2 = 48p^2. Since the chord MN has a length of 4sqrt{15}, we have: |MN|^2 = left(1 + frac{1}{4}right)(x_1 - x_2)^2 = frac{5}{4} times 48p^2 = 16 times 15, which implies p^2 = 4, and so p = 2. Therefore, the standard equation of the parabola C is y^2 = 4x. (2) From part (1), we know that the equation of the line l is y = k(x + 1). Substituting this into the equation of the parabola, we get ky^2 - 4y + 4k = 0. Let M(x_1, y_1), N(x_2, y_2), and Q(x_3, y_3) be the coordinates of points M, N, and Q, respectively. Since M and N lie on the parabola, we have y_1y_2 = 4. The slope of line MQ is given by: k_{MQ} = frac{y_1 - y_3}{x_1 - x_3} = frac{y_1 - y_3}{frac{y_1^2}{4} - frac{y_3^2}{4}} = frac{4}{y_1 + y_3}. The equation of line MB is y + 1 = frac{4}{y_1 + y_3}(x - 1). Thus, y_1 + 1 = frac{4}{y_1 + y_3}(x_1 - 1), and we can solve for y_1 to get: y_1 = -frac{4 + y_3}{1 + y_3}. Since y_1y_2 = 4, we have: frac{4}{y_2} = -frac{4 + y_3}{1 + y_3}, which simplifies to y_2y_3 + 4(y_2 + y_3) + 4 = 0. The equation of line QN is y - y_2 = frac{4}{y_2 + y_3}(x - x_2). Solving for x, we find that x = 1, and substituting this in the equation, we find that y = -4. Thus, the line QN passes through the fixed point boxed{(1, -4)}.

answer:Since f(1) < 2, and since f(x) is an odd function defined on mathbb{R}, therefore f(-1) = -f(1), thus f(-1) > -2. Since the smallest positive period of f(x) is 3, therefore f(2) = f(3-1) = f(-1), thus f(2) > -2. Hence, the answer is boxed{(-2, +infty)}.

answer:To find the total pounds of snacks Kelly bought, we need to add the weights of the peanuts, raisins, and almonds together: Peanuts: 0.1 pounds Raisins: 0.4 pounds Almonds: 0.3 pounds Total weight = 0.1 + 0.4 + 0.3 Total weight = 0.8 pounds Kelly bought a total of boxed{0.8} pounds of snacks.