answer:To find the probability that both smartphones are defective, we can use the formula for the probability of independent events, which is the product of the probabilities of each event occurring. First, we calculate the probability of the first phone being defective: Probability of first phone being defective = Number of defective phones / Total number of phones = 67 / 250 Next, if the first phone is defective, there are now 66 defective phones left out of a total of 249 phones (since one phone has already been taken out of the shipment). Probability of second phone being defective (given the first one is defective) = Number of remaining defective phones / Total remaining number of phones = 66 / 249 Now, we multiply the two probabilities together to get the probability that both phones are defective: Probability (both phones are defective) = (67 / 250) * (66 / 249) Let's calculate this: Probability (both phones are defective) = (67 / 250) * (66 / 249) ≈ 0.268 * 0.265 ≈ 0.071052 So, the approximate probability that both phones are defective is boxed{0.071052} or about 7.11%.

answer:1. We start with the given 32-digit integer: [ 64312311692944269609355712372657 ] We know that this number is the product of 6 consecutive prime numbers. 2. First, we will estimate the size of these primes. We approximate the product of these primes as: [ 64 times 10^{30} ] To find the average size of the primes, we take the sixth root of the product: [ (64 times 10^{30})^{1/6} = (2^6 times 10^{30})^{1/6} = 2^{6/6} times 10^{30/6} = 2 times 10^5 = 200000 ] Hence, each prime number is approximately 200000. Let's denote the primes by (200000 + x_i) for (i = 1, ldots, 6). 3. The product of these primes can be expressed as: [ prod_{i=1}^{6}(200000 + x_i) ] Expanding this product as a polynomial centered around 200000, we get: [ 200000^6 + text{(sum of lower order terms)} ] 4. Knowing that: [ prod_{i=1}^{6}(200000 + x_i) approx 64312 times 10^{25} implies 200000^5 (x_1 + cdots + x_6) + text{carry} ] By simplifying, we note: [ 200000^5 = (2 times 10^5)^5 = 2^5 times 10^{25} = 32 times 10^{25} ] Thus: [ 32 (x_1 + x_2 + x_3 + x_4 + x_5 + x_6) approx 64312 implies x_1 + x_2 + x_3 + x_4 + x_5 + x_6 approx frac{64312}{32} = 2009 ] 5. The actual calculation yields: [ 31231 = frac{64312}{32} ] So: [ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 leq 975 ] 6. Considering higher-order corrections, we estimate: [ 16(x_1 x_2 + x_1 x_3 + cdots + x_5 x_6) leq 16 times frac{5}{12}(x_1 + x_2 + cdots + x_6)^2 ] Simplifying: [ 16 times frac{5}{12} times 1000^2 approx 67 times 10^5 ] Therefore, the carry term from (200000^4 (x_1 x_2 + cdots + x_5 x_6)) is at most (67 times 10^{25}). 7. This clarifies that: [ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 approx 973 - 975 ] 8. Hence, the sum of the primes, which is: [ 6 times 200000 + (x_1 + x_2 + x_3 + x_4 + x_5 + x_6) ] Lies in: [ [1200973, 1200975] ] 9. Since all the primes are odd, their sum must be even. Thus, the only even number in the range is: [ 1200974 ] Conclusion: [ boxed{1200974} ]

answer:1. **Define the problem**: We are to minimize the total time spent fetching water with 10 different individuals. Each person i has a distinct time T_i to fill a bucket. We need to determine the optimal sequence for the 10 people to minimize the total time. 2. **Formulate the total time**: Suppose the 10 people draw water in the order i_1, i_2, ldots, i_{10}. The total time function can be expressed as: [ f(t) = T_{i_1} + (T_{i_1} + T_{i_2}) + (T_{i_1} + T_{i_2} + T_{i_3}) + cdots + (T_{i_1} + T_{i_2} + cdots + T_{i_{10}}) ] 3. **Simplify the function**: By breaking down the summation, we get: [ f(t) = (10)T_{i_1} + (9)T_{i_2} + (8)T_{i_3} + cdots + (1)T_{i_{10}} ] where the coefficients represent the weights of each person’s drawing time in the total time summation. 4. **Order the drawing times**: We assume T_1 < T_2 < ldots < T_{10}. According to the properties of weighted sums (arrange coefficients from largest to smallest with the smallest terms to minimize the weighted sum), this arrangement T_{i_1} = T_1, T_{i_2} = T_2, ldots, T_{i_{10}} = T_{10} is optimal. 5. **Proof and validation**: Using the principle of Rearrangement Inequality: - The Rearrangement Inequality states that for sequences (a_1 le a_2 le cdots le a_n) and (b_1 le b_2 le cdots le b_n), the sum ( sum_{i=1}^{n} a_i b_i ) is minimized when sequences are similarly sorted. - In our case, set a_i (time) and b_i (weights) both being in increasing order. Hence, to minimize our total time function f(t), the optimal ordering is T_1, T_2, ldots, T_{10}. 6. **Calculate the minimal total time**: [ f(t) = 10T_1 + 9T_2 + 8T_3 + cdots + 1T_{10} ] 7. **Conclusion**: The minimal total time when the sequence of drawing is optimally arranged ( (T_1 < T_2 < ldots < T_{10}) ): [ boxed{10T_1 + 9T_2 + 8T_3 + cdots + 1T_{10}} ]

answer:Let J be the number of junior girls, S the number of senior girls, and B the number of boys. Since every member is either a boy or a girl (junior or senior), we have: [ J + S + B = 32 ] From the attendance: [ frac{1}{3}J + frac{1}{2}S + B = 18 ] We need to isolate B from the equations. Subtract the attendance equation from the total membership equation: [ J - frac{1}{3}J + S - frac{1}{2}S = 32 - 18 ] [ frac{2}{3}J + frac{1}{2}S = 14 ] To eliminate fractions, multiply the whole equation by 6: [ 4J + 3S = 84 ] Now, expressing B from the total: [ B = 32 - J - S ] And from the attendance: [ B = 18 - frac{1}{3}J - frac{1}{2}S ] Equating both expressions for B: [ 32 - J - S = 18 - frac{1}{3}J - frac{1}{2}S ] [ 14 = frac{2}{3}J + frac{1}{2}S ] Substituting the value from earlier: [ 4J + 3S = 84 ] We solve this system of equations for J and S. From the second equation: [ S = frac{84 - 4J}{3} ] Substituting S back: [ 14 = frac{2}{3}J + frac{1}{2} left( frac{84 - 4J}{3} right) ] [ 14 = frac{2}{3}J + frac{42 - 2J}{3} ] [ 14 = frac{42 - J}{3} ] [ 42 - J = 42 ] [ J = 0 ] Then S = frac{84}{3} = 28. Thus: [ B = 32 - J - S = 32 - 0 - 28 = 4 ] Thus, there are boxed{4} boys on the chess team.