answer:First, calculate the side lengths based on the isosceles triangle's and the rectangle's dimensions: - **Isosceles Right Triangle**: Let each leg be (x) inches. The perimeter is then (2x + sqrt{2}x = 48). Solving for (x), [ x(2 + sqrt{2}) = 48 ] [ x = frac{48}{2 + sqrt{2}} = frac{48}{2 + sqrt{2}} cdot frac{2 - sqrt{2}}{2 - sqrt{2}} = frac{48(2 - sqrt{2})}{4 - 2} = 24(2 - sqrt{2}) ] [ x = 24(2 - sqrt{2}) ] - **Rectangle**: With sides (y) and (2y), its perimeter is (2(y + 2y) = 48). Solving for (y), [ 6y = 48 Rightarrow y = 8 ] Now, compute the hypotenuse of the triangle: - **Hypotenuse**: ( text{hypotenuse} = x sqrt{2} = 24(2 - sqrt{2})sqrt{2} = 24(2sqrt{2} - 2) ) Calculate the desired ratio (hypotenuse( / small side of the rectangle)): - **Ratio**: [ frac{text{hypotenuse}}{y} = frac{24(2sqrt{2}-2)}{8} = 3(2sqrt{2}-2) ] [ boxed{3(2sqrt{2}-2)} ] Conclusion: The ratio of the length of the hypotenuse of the right isosceles triangle to the shorter side of the rectangle is ( boxed{3(2sqrt{2}-2)} ).

answer:Since the universal set U={0,1,2,3}, and complement_U A={2}, Therefore, A={0,1,3}, Hence, the correct choice is: boxed{text{B}}. By determining set A from the universal set U and the complement of A, we can solve this problem. This question tests the understanding of complements and their operations. Mastering the definition of complements is key to solving this problem.

answer:1. Find the least common multiple (LCM) of the primes 2, 3, 5, 7, and 11. The LCM is the product of these primes (since they are all distinct and prime): [ text{lcm}[2, 3, 5, 7, 11] = 2 cdot 3 cdot 5 cdot 7 cdot 11 = 2310 ] 2. Determine the smallest n such that 2310n is a five-digit number. The inequality is: [ 2310n geq 10000 ] Dividing both sides by 2310 gives: [ n geq frac{10000}{2310} approx 4.329004329 ] Since n must be an integer, we take n = 5. 3. Calculate 2310 cdot 5 to find the smallest five-digit number satisfying the condition: [ 2310 cdot 5 = 11550 ] Conclusion: [ boxed{11550} ]

answer:To solve this problem, we start by noting that we are dealing with a binomial distribution, where ( n = 500 ) (the total number of items) and ( p = 0.02 ) (the probability of a single item being defective). 1. Calculation for exactly 10 defective items: 1. **Identify Parameters:** Given: [ n = 500, quad p = 0.02 ] Hence, [ q = 1 - p = 0.98 ] Calculate: [ np = 500 cdot 0.02 = 10 ] and [ npq = 500 cdot 0.02 cdot 0.98 = 9.8 ] 2. **Use the Normal Approximation:** According to the Normal approximation to the Binomial Distribution, we can approximate the probability as: [ P(X = 10) approx frac{1}{sqrt{2pi npq}} e^{-frac{(10 - np)^2}{2npq}} ] Given: [ np = 10, quad npq = 9.8 ] 3. **Calculate the Approximation:** The formula can be written as: [ P_{500}(10) = frac{1}{sqrt{9.8}} varphileft(frac{10 - 10}{sqrt{9.8}}right) ] Simplifying further: [ P_{500}(10) = frac{1}{sqrt{9.8}} varphi(0) ] The value of the standard normal density function at 0 is: [ varphi(0) = frac{1}{sqrt{2pi}} approx 0.399 ] Therefore: [ P_{500}(10) = frac{1}{sqrt{9.8}} cdot 0.399 ] Since: [ frac{1}{sqrt{9.8}} approx 0.319 ] We get: [ P_{500}(10) approx 0.319 cdot 0.399 = 0.127 ] 2. Calculation for the number of defective items between 10 and 20: 1. **Use Phi Function:** To find the probability that the number of defective items is between 10 and 20, we use the cumulative distribution function: [ Phi(10 leq k leq 20) = Phileft(frac{20 - 10}{sqrt{9.8}}right) - Phileft(frac{10 - 10}{sqrt{9.8}}right) ] 2. **Calculate Individual Terms:** [ Phileft(frac{20 - 10}{sqrt{9.8}}right) = Phileft(frac{10}{sqrt{9.8}}right) ] First, calculate: [ frac{10}{sqrt{9.8}} approx frac{10}{3.13} approx 3.2 ] 3. **Find Phi Values:** Using standard normal tables or a calculator: [ Phi(3.2) approx 0.9993 ] and [ Phi(0) = 0.5 ] 4. **Subtract to Find the Desired Probability:** [ Phi(10 leq k leq 20) = Phi(3.2) - Phi(0) = 0.9993 - 0.5 = 0.4993 ] 5. **Final Result:** [ Phi(10 leq k leq 20) = 0.499 ] # Conclusion: 1. Probability of exactly 10 defective items: [ boxed{0.127} ] 2. Probability of number of defective items being between 10 and 20: [ boxed{0.499} ]