answer：First, I need to determine the production rates of Machine A and Machine B. Let’s denote the number of sprockets Machine A produces per hour as ( A ) and Machine B as ( B ).According to the problem, Machine B produces 10% more sprockets per hour than Machine A. This can be expressed as:[ B = A + 0.10A ][ B = 1.10A ]Next, I know that Machine A takes 10 hours longer than Machine B to produce 220 sprockets. Let’s denote the time it takes for Machine B to produce 220 sprockets as ( T ) hours. Therefore, Machine A takes ( T + 10 ) hours.The total number of sprockets produced by each machine is 220, so I can set up the following equations based on their production rates:For Machine A:[ 220 = A times (T + 10) ]For Machine B:[ 220 = B times T ]Substituting ( B = 1.10A ) into the equation for Machine B:[ 220 = 1.10A times T ]Now I have two equations:1) ( 220 = A times (T + 10) )2) ( 220 = 1.10A times T )I'll solve the second equation for ( T ):[ T = frac{220}{1.10A} ]Substituting this value of ( T ) into the first equation:[ 220 = A times left(frac{220}{1.10A} + 10right) ]Simplifying the equation:[ 220 = frac{220A}{1.10A} + 10A ][ 220 = 200 + 10A ]Solving for ( A ):[ 10A = 220 - 200 ][ 10A = 20 ][ A = 2 ]Therefore, Machine A produces 2 sprockets per hour.

answer：Alright, so I have this probability problem here, and I need to figure out how many independent trials are required so that the probability of the inequality |m/n - p| < 0.2 exceeds 0.96. The probability of success in a single trial is given as p = 0.7. Hmm, okay, let me try to unpack this step by step.First, let me make sure I understand the problem correctly. We're dealing with a series of independent trials, each with a probability of success p = 0.7. After n trials, we observe m successes. We want to find the smallest n such that the probability that the sample proportion m/n is within 0.2 of the true probability p is greater than 0.96. So, in other words, we want P(|m/n - 0.7| < 0.2) > 0.96.I think this is related to the Law of Large Numbers, which tells us that as n increases, the sample proportion m/n should get closer to the true probability p. But since we're dealing with probabilities, maybe I need to use some sort of probability inequality, like Chebyshev's inequality or the Central Limit Theorem.Let me recall Chebyshev's inequality. It states that for any random variable with mean μ and variance σ², the probability that it deviates from the mean by more than k standard deviations is at most 1/k². In formula terms: P(|X - μ| ≥ kσ) ≤ 1/k². So, maybe I can apply this here.In our case, the random variable is m/n, the sample proportion. The mean of m/n is p, which is 0.7, and the variance of m/n is pq/n, where q = 1 - p = 0.3. So, variance = (0.7)(0.3)/n = 0.21/n. Therefore, the standard deviation is sqrt(0.21/n).We want P(|m/n - 0.7| < 0.2) > 0.96. Let's rewrite this as P(|m/n - 0.7| ≥ 0.2) ≤ 0.04. So, using Chebyshev's inequality, we can say that P(|m/n - 0.7| ≥ k * sqrt(0.21/n)) ≤ 1/k². We need this probability to be less than or equal to 0.04.So, setting 1/k² ≤ 0.04, we get k² ≥ 25, so k ≥ 5. Therefore, we need 0.2 ≥ k * sqrt(0.21/n). But wait, k is 5, so 0.2 ≥ 5 * sqrt(0.21/n). Let me solve for n.First, divide both sides by 5: 0.04 ≥ sqrt(0.21/n). Then, square both sides: 0.0016 ≥ 0.21/n. So, n ≥ 0.21 / 0.0016. Calculating that, 0.21 divided by 0.0016 is 131.25. Since n must be an integer, we round up to 132.Wait, but is Chebyshev's inequality the best approach here? I remember that Chebyshev's inequality is quite conservative and might give a larger n than necessary. Maybe using the Central Limit Theorem would give a better estimate since it provides a more precise approximation for large n.The Central Limit Theorem tells us that the distribution of m/n is approximately normal with mean p and variance pq/n. So, we can standardize the variable and use the standard normal distribution to find the required n.Let me try that approach. We want P(|m/n - 0.7| < 0.2) > 0.96. This can be rewritten as P(-0.2 < m/n - 0.7 < 0.2) > 0.96. Dividing by the standard deviation, we get:P((-0.2)/sqrt(0.21/n) < (m/n - 0.7)/sqrt(0.21/n) < 0.2/sqrt(0.21/n)) > 0.96.The term (m/n - 0.7)/sqrt(0.21/n) follows a standard normal distribution approximately. So, we need the probability that a standard normal variable Z lies between -0.2/sqrt(0.21/n) and 0.2/sqrt(0.21/n) to be greater than 0.96.Looking at standard normal tables, the z-score corresponding to 0.96 probability in the middle is approximately 2.05. That is, P(-2.05 < Z < 2.05) ≈ 0.96. So, we set 0.2/sqrt(0.21/n) = 2.05.Solving for n: 0.2 = 2.05 * sqrt(0.21/n). Squaring both sides: 0.04 = (2.05)² * (0.21/n). Calculating (2.05)²: 2.05 * 2.05 = 4.2025. So, 0.04 = 4.2025 * 0.21 / n. Multiply 4.2025 by 0.21: 4.2025 * 0.21 ≈ 0.882525. So, 0.04 = 0.882525 / n. Solving for n: n = 0.882525 / 0.04 ≈ 22.0631. Since n must be an integer, we round up to 23.Wait, that's a big difference from the 132 we got earlier. Which one is correct? I think the Central Limit Theorem gives a better approximation for large n, but since n=23 is relatively small, maybe the normal approximation isn't very accurate here. On the other hand, Chebyshev's inequality is more conservative and gives a larger n, ensuring that the probability is definitely above 0.96.But the problem doesn't specify which inequality or theorem to use, so perhaps both approaches are valid, but they give different answers. Maybe I need to check which one is more appropriate here.Alternatively, perhaps I can use the exact binomial distribution to calculate the required n. But that might be more complicated since it involves summing binomial probabilities, which isn't straightforward.Wait, another thought: maybe the problem is expecting the use of the inequality that relates to the Law of Large Numbers, which is similar to Chebyshev's inequality. The problem mentions "the probability of satisfying the inequality |m/n - p| < 0.2 exceeds 0.96." So, perhaps it's referring to a specific inequality, maybe the one given in the problem statement or a standard one.Looking back, the user provided the problem statement twice, so maybe it's expecting a specific method. Let me see if there's a standard formula for this.I recall that for the inequality |m/n - p| < ε, the probability can be bounded using the inequality:P(|m/n - p| < ε) ≥ 1 - (pq)/(n ε²)This is similar to Chebyshev's inequality. So, setting 1 - (pq)/(n ε²) > 0.96, we get (pq)/(n ε²) < 0.04. Therefore, n > pq/(0.04 ε²).Given p = 0.7, q = 0.3, ε = 0.2, let's compute pq = 0.7 * 0.3 = 0.21. ε² = 0.04. So, n > 0.21 / (0.04 * 0.04) = 0.21 / 0.0016 = 131.25. So, n must be at least 132.This matches the earlier result from Chebyshev's inequality. So, perhaps the problem expects this method. The Central Limit Theorem gives a lower n, but it's an approximation and might not guarantee the probability exceeds 0.96.Therefore, to be safe and ensure that the probability is definitely above 0.96, we should use the Chebyshev-based inequality, which gives n = 132.But just to double-check, let's see what happens if we use n = 132. The variance of m/n is pq/n = 0.21/132 ≈ 0.0016. The standard deviation is sqrt(0.0016) = 0.04. So, the interval |m/n - 0.7| < 0.2 is 5 standard deviations away from the mean. According to Chebyshev's inequality, the probability of being within 5 standard deviations is at least 1 - 1/25 = 0.96. So, exactly 0.96. But the problem says "exceed 0.96," so maybe we need n > 131.25, which would be 132.Alternatively, if we use n = 131, then pq/(n ε²) = 0.21 / (131 * 0.04) ≈ 0.21 / 5.24 ≈ 0.04007, which is just above 0.04, so the probability would be less than 0.96. Therefore, n must be at least 132.So, after considering both approaches and verifying the calculations, I'm confident that the required number of trials is 132.

answer：Alright, so I have this problem about a sequence where the sum of the first n terms is S_n. They give me that a_1 is 1/2, and there's this equation: 2n(S_{n+1} - S_n) = (n + 1)a_n. I need to find the general formula for a_n and then figure out the range of λ such that b_n = n(2 - S_n) is less than or equal to λ for all natural numbers n.Okay, let's start with part (1). I know that S_{n+1} - S_n is just a_{n+1}, right? Because S_{n+1} is the sum up to n+1 terms, so subtracting S_n gives the (n+1)th term. So, substituting that into the equation, I get 2n * a_{n+1} = (n + 1)a_n. That simplifies to 2n a_{n+1} = (n + 1) a_n. So, if I solve for a_{n+1}, I get a_{n+1} = [(n + 1)/(2n)] a_n.Hmm, that looks like a recursive formula. Maybe I can express a_n in terms of a_{n-1}, and so on, back to a_1. Let me try writing out the first few terms to see if I can spot a pattern.Given a_1 = 1/2.Then, a_2 = [(2)/(2*1)] a_1 = (2/2)(1/2) = 1/2.Wait, that's interesting. a_2 is also 1/2. Let's do a_3.a_3 = [(3)/(2*2)] a_2 = (3/4)(1/2) = 3/8.a_4 = [(4)/(2*3)] a_3 = (4/6)(3/8) = (2/3)(3/8) = 1/4.a_5 = [(5)/(2*4)] a_4 = (5/8)(1/4) = 5/32.Hmm, so the terms are: 1/2, 1/2, 3/8, 1/4, 5/32,...I wonder if there's a pattern here. Let me see:n: 1, 2, 3, 4, 5,...a_n: 1/2, 1/2, 3/8, 1/4, 5/32,...Looking at the numerators: 1, 1, 3, 1, 5,... It seems like for odd n, the numerator is n, and for even n, it's 1? Wait, no, that doesn't hold because a_2 is 1/2, which is 1/2, but a_4 is 1/4, which is 1/4. Hmm, maybe not.Alternatively, maybe the numerators are n for all n, but the denominators are 2^n. Let's check:For n=1: 1/2^1 = 1/2. Correct.n=2: 2/2^2 = 2/4 = 1/2. Correct.n=3: 3/2^3 = 3/8. Correct.n=4: 4/2^4 = 4/16 = 1/4. Correct.n=5: 5/2^5 = 5/32. Correct.Oh! So it seems like a_n = n / 2^n. That fits all the terms I calculated. So maybe that's the general formula.But let me verify this with the recursive formula to make sure.Given a_{n+1} = [(n + 1)/(2n)] a_n.If a_n = n / 2^n, then a_{n+1} should be (n + 1)/2^{n + 1}.Let's plug into the recursive formula:[(n + 1)/(2n)] * [n / 2^n] = [(n + 1)/2n] * [n / 2^n] = (n + 1) / 2^{n + 1}.Which is exactly a_{n+1}. So yes, this formula satisfies the recursion. And since a_1 = 1/2, which is 1 / 2^1, it holds for all n. So part (1) is solved: a_n = n / 2^n.Now, moving on to part (2). They define b_n = n(2 - S_n). I need to find the range of λ such that b_n ≤ λ for all n in natural numbers.First, I need to find S_n, the sum of the first n terms of a_n. Since a_n = n / 2^n, S_n = sum_{k=1}^n (k / 2^k).I remember that the sum of k / 2^k from k=1 to infinity is 2. So S_n is approaching 2 as n increases. But since we're dealing with finite n, S_n will be less than 2.Let me recall the formula for the finite sum S_n = sum_{k=1}^n (k / 2^k). I think it's S_n = 2 - (n + 2)/2^n. Let me verify this.Yes, I remember that for the infinite sum, it's 2, and the finite sum is 2 - (n + 2)/2^n. So S_n = 2 - (n + 2)/2^n.Therefore, 2 - S_n = (n + 2)/2^n.So b_n = n(2 - S_n) = n * (n + 2)/2^n = [n(n + 2)] / 2^n.So now, I need to find the maximum value of b_n over all natural numbers n, because λ needs to be greater than or equal to all b_n. So the smallest such λ is the maximum of b_n.Therefore, I need to find the maximum of the sequence b_n = [n(n + 2)] / 2^n.To find the maximum, I can consider the ratio of consecutive terms b_{n+1} / b_n and see when it's greater than 1 or less than 1.Compute b_{n+1} / b_n:= [ (n + 1)(n + 3) / 2^{n + 1} ] / [ n(n + 2) / 2^n ]= [ (n + 1)(n + 3) / 2^{n + 1} ] * [ 2^n / (n(n + 2)) ]= [ (n + 1)(n + 3) / 2 ] * [ 1 / (n(n + 2)) ]= [ (n + 1)(n + 3) ] / [ 2n(n + 2) ]Simplify numerator: (n + 1)(n + 3) = n^2 + 4n + 3Denominator: 2n(n + 2) = 2n^2 + 4nSo the ratio is (n^2 + 4n + 3) / (2n^2 + 4n) = [n^2 + 4n + 3] / [2n^2 + 4n]Let me write it as:= [n^2 + 4n + 3] / [2n^2 + 4n] = [n^2 + 4n + 3] / [2n(n + 2)]I can factor numerator: n^2 + 4n + 3 = (n + 1)(n + 3)So,= (n + 1)(n + 3) / [2n(n + 2)]Hmm, not sure if that helps. Maybe I can compare the ratio to 1.Compute when b_{n+1} / b_n > 1:(n + 1)(n + 3) / [2n(n + 2)] > 1Multiply both sides by 2n(n + 2):(n + 1)(n + 3) > 2n(n + 2)Expand both sides:Left: n^2 + 4n + 3Right: 2n^2 + 4nSubtract right from left:n^2 + 4n + 3 - 2n^2 - 4n = -n^2 + 3So, -n^2 + 3 > 0 => n^2 < 3 => n < sqrt(3) ≈ 1.732Since n is a natural number, n=1.So for n=1, b_{2}/b_1 > 1, meaning b_2 > b_1.For n ≥ 2, b_{n+1}/b_n < 1, meaning the sequence b_n is increasing up to n=2 and then decreasing.Therefore, the maximum of b_n occurs at n=2.Compute b_2:b_2 = [2(2 + 2)] / 2^2 = (2*4)/4 = 8/4 = 2.So the maximum value of b_n is 2, achieved at n=2.Therefore, for all n, b_n ≤ 2. So λ must be at least 2.Hence, the range of λ is [2, ∞).Wait, let me double-check. For n=1, b_1 = [1(1 + 2)] / 2^1 = 3/2 = 1.5.n=2: 2.n=3: [3*5]/8 = 15/8 = 1.875.n=4: [4*6]/16 = 24/16 = 1.5.n=5: [5*7]/32 = 35/32 ≈ 1.09375.So yes, it peaks at n=2 with 2, then decreases.Therefore, the maximum is indeed 2, so λ must be ≥ 2.**Final Answer**(1) The general term is boxed{dfrac{n}{2^n}}.(2) The range of λ is boxed{[2, infty)}.

answer：First, I recognize that 4% of the units produced are defective. This means that out of every 100 units produced, 4 units are defective.Next, I understand that 4% of these defective units are shipped for sale. So, out of the 4 defective units, 4% of them are sent to be sold.To find the percentage of the total units produced that are both defective and shipped for sale, I need to calculate 4% of 4%. This involves multiplying the two percentages together.Finally, by performing the calculation, I determine the final percentage of units that are both defective and shipped for sale.

answer：Okay, so I have this triangle ABC where AB is 3 units, AC is 2 units, and the angle at A is 60 degrees. I need to find the area of the circumcircle of this triangle. Hmm, let me think about how to approach this.First, I remember that the circumcircle of a triangle is the circle that passes through all three vertices of the triangle. The center of this circle is called the circumcenter, and the radius is called the circumradius. To find the area of the circumcircle, I need to find the radius first and then use the formula for the area of a circle, which is π times the radius squared.But how do I find the circumradius? I think there's a formula that relates the sides of the triangle and the angles to the circumradius. Let me recall... Oh, right! The formula is:[ 2R = frac{a}{sin A} ]Where ( R ) is the circumradius, ( a ) is the length of the side opposite angle A, and ( sin A ) is the sine of angle A. So, in this case, angle A is 60 degrees, and the side opposite to it is BC. But wait, I don't know the length of BC yet. I only know AB and AC. So, I need to find BC first.To find BC, I can use the Law of Cosines. The Law of Cosines states that:[ c^2 = a^2 + b^2 - 2ab cos C ]In this case, if I consider angle A, which is 60 degrees, the sides adjacent to angle A are AB and AC, which are 3 and 2 respectively. So, applying the Law of Cosines:[ BC^2 = AB^2 + AC^2 - 2 cdot AB cdot AC cdot cos A ]Plugging in the values:[ BC^2 = 3^2 + 2^2 - 2 cdot 3 cdot 2 cdot cos 60^circ ]Calculating each term:- ( 3^2 = 9 )- ( 2^2 = 4 )- ( 2 cdot 3 cdot 2 = 12 )- ( cos 60^circ = 0.5 )So,[ BC^2 = 9 + 4 - 12 cdot 0.5 ][ BC^2 = 13 - 6 ][ BC^2 = 7 ]Therefore, BC is the square root of 7, so:[ BC = sqrt{7} ]Alright, now that I have BC, I can use the formula for the circumradius. Let me write that down again:[ 2R = frac{BC}{sin A} ]We know BC is ( sqrt{7} ) and angle A is 60 degrees, so ( sin 60^circ ) is ( frac{sqrt{3}}{2} ). Plugging these values in:[ 2R = frac{sqrt{7}}{frac{sqrt{3}}{2}} ]Dividing by a fraction is the same as multiplying by its reciprocal, so:[ 2R = sqrt{7} cdot frac{2}{sqrt{3}} ][ 2R = frac{2sqrt{7}}{sqrt{3}} ]To rationalize the denominator, I can multiply numerator and denominator by ( sqrt{3} ):[ 2R = frac{2sqrt{7} cdot sqrt{3}}{3} ][ 2R = frac{2sqrt{21}}{3} ]Therefore, solving for R:[ R = frac{sqrt{21}}{3} ]Now that I have the radius, I can find the area of the circumcircle. The area formula is:[ text{Area} = pi R^2 ]Substituting R:[ text{Area} = pi left( frac{sqrt{21}}{3} right)^2 ]Calculating the square:[ left( frac{sqrt{21}}{3} right)^2 = frac{21}{9} = frac{7}{3} ]So, the area is:[ text{Area} = pi cdot frac{7}{3} = frac{7pi}{3} ]Wait, let me double-check my steps to make sure I didn't make a mistake. I used the Law of Cosines correctly to find BC, got ( sqrt{7} ). Then applied the formula for the circumradius, which I think is correct. Calculated ( sin 60^circ ) correctly as ( frac{sqrt{3}}{2} ). Then, when solving for R, I correctly isolated R by dividing both sides by 2. Then, squared R correctly for the area. Seems solid.Just to be thorough, maybe I can use another method to find the circumradius and see if I get the same result. I remember another formula for the circumradius:[ R = frac{abc}{4K} ]Where a, b, c are the sides of the triangle, and K is the area of the triangle. Let's try that.First, I need to find the area K of triangle ABC. Since I have two sides and the included angle, I can use the formula:[ K = frac{1}{2}ab sin C ]In this case, sides AB and AC are 3 and 2, and the included angle is 60 degrees. So,[ K = frac{1}{2} cdot 3 cdot 2 cdot sin 60^circ ][ K = 3 cdot frac{sqrt{3}}{2} ][ K = frac{3sqrt{3}}{2} ]Now, using the formula for R:[ R = frac{abc}{4K} ]Here, a, b, c are the sides of the triangle. Let me assign:- a = BC = ( sqrt{7} )- b = AC = 2- c = AB = 3So,[ R = frac{sqrt{7} cdot 2 cdot 3}{4 cdot frac{3sqrt{3}}{2}} ]Simplify numerator and denominator:Numerator: ( sqrt{7} cdot 2 cdot 3 = 6sqrt{7} )Denominator: ( 4 cdot frac{3sqrt{3}}{2} = 2 cdot 3sqrt{3} = 6sqrt{3} )So,[ R = frac{6sqrt{7}}{6sqrt{3}} ][ R = frac{sqrt{7}}{sqrt{3}} ][ R = frac{sqrt{21}}{3} ]Same result! Okay, so that confirms that R is indeed ( frac{sqrt{21}}{3} ). Therefore, the area of the circumcircle is ( frac{7pi}{3} ).Just to recap, the steps were:1. Use the Law of Cosines to find the unknown side BC.2. Use the formula relating side, sine of the opposite angle, and the circumradius to find R.3. Alternatively, use another formula for R involving the sides and the area of the triangle to confirm the result.4. Calculate the area of the circumcircle using the radius found.Everything checks out, so I'm confident the area is ( frac{7pi}{3} ).**Final Answer**The area of the circumcircle of triangle ( triangle ABC ) is boxed{dfrac{7pi}{3}}.

answer：First, I need to understand the problem. We are given two sequences: a geometric progression starting with 5, x, y, and an arithmetic progression starting with x, y, 15. The goal is to find the values of x and y and then calculate their sum.For the geometric progression, the ratio between consecutive terms is constant. Let’s denote this common ratio as r. Therefore, x can be expressed as 5 multiplied by r, and y can be expressed as 5 multiplied by r squared.For the arithmetic progression, the difference between consecutive terms is constant. Let’s denote this common difference as d. Therefore, y can be expressed as x plus d, and 15 can be expressed as y plus d.Next, I'll set up the equations based on these relationships. From the geometric progression, I have:x = 5ry = 5r²From the arithmetic progression, I have:y = x + d15 = y + dI'll substitute the expression for y from the geometric progression into the arithmetic progression equations. This gives me:5r² = 5r + d15 = 5r² + dNow, I'll solve these equations simultaneously. From the first equation, I can express d in terms of r:d = 5r² - 5rSubstituting this into the second equation:15 = 5r² + (5r² - 5r)15 = 10r² - 5rThis simplifies to a quadratic equation:10r² - 5r - 15 = 0Dividing the entire equation by 5 to simplify:2r² - r - 3 = 0Solving this quadratic equation, I find the roots r = -3/2 and r = 1. Since a negative ratio would result in negative terms in the geometric progression, which doesn't make sense in this context, I'll choose r = 1.Substituting r = 1 back into the expressions for x and y:x = 5 * 1 = 5y = 5 * (1)² = 5Finally, calculating the sum of x and y:x + y = 5 + 5 = 10Therefore, the sum of the two numbers is 10.