answer:Let's denote the number of boxes as ( x ). Each bag has 7 cookies, so 9 bags have ( 9 times 7 = 63 ) cookies. Each box has 12 cookies, so ( x ) boxes have ( 12x ) cookies. According to the problem, there are 33 more cookies in the boxes than in the 9 bags. So we can write the equation: ( 12x = 63 + 33 ) Solving for ( x ): ( 12x = 96 ) ( x = frac{96}{12} ) ( x = 8 ) Therefore, there are boxed{8} boxes.

answer:Given that the vectors left{ begin{pmatrix} 2 3 end{pmatrix}, begin{pmatrix} 4 k end{pmatrix} right} are linearly dependent, there exist non-zero constants a and b such that: [a begin{pmatrix} 2 3 end{pmatrix} + b begin{pmatrix} 4 k end{pmatrix} = begin{pmatrix} 0 0 end{pmatrix}.] This yields the system: [2a + 4b = 0,] [3a + kb = 0.] From the first equation, a = -2b. Substituting this into the second equation gives: [3(-2b) + kb = 0,] [-6b + kb = 0,] [(k - 6)b = 0.] Since b neq 0 for a non-trivial solution (as the vectors are linearly dependent), we must have k - 6 = 0. Therefore, k = boxed{6}.

answer:This problem primarily tests geometric probability. The key to solving it is to transform the original problem into a geometric probability problem and then apply the geometric probability calculation formula. Let the "3m long segment AB" correspond to the interval [0,3]. The event "A" denotes "the distance from both endpoints A and B of the segment is greater than 1m". Therefore, the interval satisfying A is [1,2]. Applying the geometric probability calculation formula, we get P(A)=frac{3-2}{3-0}=frac{1}{3}. So the answer is boxed{B}.

answer:# Step-by-Step Solution Part 1: Cost of Each Basketball and Soccer Ball Let's denote: - x as the cost of each basketball in yuan, - y as the cost of each soccer ball in yuan. From the given information, we can set up the following system of equations: 1. For 2 basketballs and 3 soccer balls costing 310 yuan, we have: [2x + 3y = 310] 2. For 5 basketballs and 2 soccer balls costing 500 yuan, we have: [5x + 2y = 500] Solving this system of equations, we aim to find the values of x and y. - Multiplying the first equation by 2 and the second by 3, we get: [4x + 6y = 620] [15x + 6y = 1500] - Subtracting the first modified equation from the second gives us: [11x = 880] [x = 80] - Substituting x = 80 into the first original equation: [2(80) + 3y = 310] [160 + 3y = 310] [3y = 150] [y = 50] Therefore, the cost of each basketball is 80 yuan, and the cost of each soccer ball is 50 yuan. Final answer for part (1): [ boxed{text{Each basketball costs 80 yuan, and each soccer ball costs 50 yuan.}} ] Part 2: Maximum Number of Basketballs Let m represent the number of basketballs to be purchased. Then, the number of soccer balls to be purchased is (60 - m). Given the cost constraints, the total cost for purchasing m basketballs and (60 - m) soccer balls should not exceed 4000 yuan. Thus, we have: [80m + 50(60 - m) leqslant 4000] Expanding and simplifying the inequality: [80m + 3000 - 50m leqslant 4000] [30m leqslant 1000] [m leqslant frac{1000}{30}] [m leqslant frac{100}{3}] Since m must be an integer (as you cannot purchase a fraction of a basketball), the maximum integer value of m that satisfies this inequality is 33. Final answer for part (2): [ boxed{text{At most, they can purchase 33 basketballs.}} ]