🧠:<think>Alright, so I need to calculate the definite integral of the exponential function from 0 to 1, which is written as ∫₀¹ e⁻ˣ dx. Hmm, okay. I remember that integrating exponential functions often involves finding an antiderivative and then applying the Fundamental Theorem of Calculus. But let me think through this step by step to make sure I get it right.First, I recall that the integral of eˣ is eˣ plus a constant, right? But in this case, we have e⁻ˣ instead of eˣ. So, I wonder if the integral of e⁻ˣ is similar or if there's a difference because of the negative exponent. Maybe I need to adjust for that negative sign somehow.Let me try to find the antiderivative of e⁻ˣ. If I let F(x) be the antiderivative of e⁻ˣ, then F'(x) should equal e⁻ˣ. I know that the derivative of e⁻ˣ with respect to x is -e⁻ˣ because of the chain rule. So, if I have F(x) = -e⁻ˣ, then F'(x) = e⁻ˣ. That seems to work! So, the antiderivative of e⁻ˣ is -e⁻ˣ.Now that I have the antiderivative, I can use the Fundamental Theorem of Calculus to evaluate the definite integral from 0 to 1. The theorem states that ∫ₐᵇ f(x) dx = F(b) - F(a), where F is the antiderivative of f. Applying this to our problem:∫₀¹ e⁻ˣ dx = F(1) - F(0)Substituting F(x) = -e⁻ˣ into this expression:= (-e⁻¹) - (-e⁰)Simplifying each term:- e⁻¹ is just -1/e, and e⁰ is 1, so:= (-1/e) - (-1)= (-1/e) + 1= 1 - 1/eSo, the value of the integral ∫₀¹ e⁻ˣ dx is 1 minus 1 over e.Wait, let me double-check my steps to make sure I didn't make any mistakes. I found the antiderivative correctly by considering the chain rule and adjusting for the negative exponent. Then, I applied the Fundamental Theorem of Calculus by plugging in the upper and lower limits. The algebra seems straightforward, and I correctly simplified the expression to 1 - 1/e.Is there another way to approach this problem to confirm my answer? Maybe by using substitution or another method of integration? Let's see.If I let u = -x, then du/dx = -1, which means du = -dx. So, dx = -du. When x = 0, u = 0, and when x = 1, u = -1. Substituting into the integral:∫₀¹ e⁻ˣ dx = ∫₀⁻¹ eᵘ (-du) = ∫₋₁⁰ eᵘ duBut integrating from -1 to 0 is the same as integrating from 0 to -1 and then taking the negative, so:= -∫₋₁⁰ eᵘ du = ∫₀⁻¹ eᵘ duWait, that seems a bit confusing. Maybe I should have kept the limits as they are and adjusted the integral accordingly. Let me try that again.With u = -x, du = -dx, so dx = -du. When x = 0, u = 0; when x = 1, u = -1. So, the integral becomes:∫₀¹ e⁻ˣ dx = ∫₀⁻¹ eᵘ (-du) = ∫₋₁⁰ eᵘ duNow, integrating eᵘ from -1 to 0:= [eᵘ]₋₁⁰ = e⁰ - e⁻¹ = 1 - 1/eOkay, that matches my previous result. So, both methods give me the same answer, which is reassuring.Another way to verify could be to approximate the integral numerically and see if it's close to 1 - 1/e. Let's calculate 1 - 1/e numerically. Since e is approximately 2.71828, 1/e is about 0.36788. So, 1 - 0.36788 ≈ 0.63212.If I approximate the integral ∫₀¹ e⁻ˣ dx numerically, say using the trapezoidal rule or Simpson's rule, I should get a value around 0.63212. Let's try a simple approximation with, say, two intervals using the trapezoidal rule.Divide the interval [0,1] into two equal parts: x₀ = 0, x₁ = 0.5, x₂ = 1. The function values are:f(x₀) = e⁰ = 1f(x₁) = e⁻⁰·⁵ ≈ 0.6065f(x₂) = e⁻¹ ≈ 0.3679Using the trapezoidal rule formula:∫₀¹ e⁻ˣ dx ≈ (Δx/2) [f(x₀) + 2f(x₁) + f(x₂)]= (0.5/2) [1 + 2(0.6065) + 0.3679]= 0.25 [1 + 1.213 + 0.3679]= 0.25 [2.5809]= 0.645225Hmm, that's a bit higher than 0.63212. Maybe with more intervals, the approximation would be better. Let's try with four intervals.x₀ = 0, x₁ = 0.25, x₂ = 0.5, x₃ = 0.75, x₄ = 1f(x₀) = 1f(x₁) = e⁻⁰·²⁵ ≈ 0.7788f(x₂) = e⁻⁰·⁵ ≈ 0.6065f(x₃) = e⁻⁰·⁷⁵ ≈ 0.4724f(x₄) = e⁻¹ ≈ 0.3679Trapezoidal rule with four intervals:∫₀¹ e⁻ˣ dx ≈ (Δx/2) [f(x₀) + 2(f(x₁) + f(x₂) + f(x₃)) + f(x₄)]= (0.25/2) [1 + 2(0.7788 + 0.6065 + 0.4724) + 0.3679]= 0.125 [1 + 2(1.8577) + 0.3679]= 0.125 [1 + 3.7154 + 0.3679]= 0.125 [5.0833]= 0.6354125That's closer to 0.63212. If I increase the number of intervals further, the approximation should get even better, converging to the exact value of 1 - 1/e ≈ 0.63212.So, through both symbolic integration and numerical approximation, I've confirmed that the value of the integral ∫₀¹ e⁻ˣ dx is indeed 1 - 1/e.</think>

🧠:<think>Okay, so I have this problem about Celeste and her o-Pod. Let me try to understand it step by step. She has 10 songs, each 40 seconds longer than the previous one, starting from 40 seconds. Her favorite song is 4 minutes long, which is 240 seconds. The o-Pod plays all the songs in a random order without repeating any until all have been played. I need to find the probability that in the first 5 minutes (which is 300 seconds) of continuous play, she does not hear her favorite song in its entirety. First, let me list out the lengths of all her songs. Since each subsequent song is 40 seconds longer, starting from 40 seconds, the lengths are:1. 40 seconds2. 80 seconds3. 120 seconds4. 160 seconds5. 200 seconds6. 240 seconds (favorite song)7. 280 seconds8. 320 seconds9. 360 seconds10. 400 secondsSo, her favorite song is the 6th song in this list. Now, the o-Pod plays these songs in a random order. So, the total number of possible orders is 10 factorial, which is 10! = 3,628,800. That's the total number of permutations.I need to find the number of permutations where her favorite song (240 seconds) is not played in its entirety within the first 300 seconds. That means either her favorite song is played partially or not at all within the first 300 seconds. But since the o-Pod plays songs without pause, if her favorite song starts playing before the 300-second mark, it will finish playing only if the total time up to and including her favorite song is less than or equal to 300 seconds.Wait, actually, if her favorite song is played, it will take up 240 seconds. So, if the favorite song is played as the first song, it will finish at 240 seconds, which is within 300 seconds. If it's played as the second song, the first song must be short enough so that the total time doesn't exceed 300 seconds when adding the favorite song.Let me think about this more carefully.To hear her favorite song in its entirety, it must start playing before 300 - 240 = 60 seconds. So, if the favorite song is played first, it will finish at 240 seconds. If it's played second, the first song must be less than or equal to 60 seconds. But looking at the song lengths, the shortest song is 40 seconds. So, if the favorite song is second, the first song is 40 seconds, then the favorite song starts at 40 seconds and ends at 280 seconds, which is still within 300 seconds. Wait, 40 + 240 = 280, which is less than 300. So, actually, if the favorite song is played second, it will still finish within 300 seconds.Wait, hold on. If the favorite song is played second, the first song is 40 seconds, then the favorite song starts at 40 seconds and ends at 280 seconds. So, yes, it's still within 300 seconds. So, the favorite song can be played second and still finish in time.Similarly, if the favorite song is played third, we need to check if the total time of the first two songs plus 240 seconds is less than or equal to 300 seconds. So, the sum of the first two songs must be less than or equal to 60 seconds. But the shortest two songs are 40 and 80 seconds, which add up to 120 seconds. 120 + 240 = 360, which is more than 300. So, if the favorite song is played third, it will not finish within 300 seconds. Therefore, the favorite song can only be played first or second to finish within 300 seconds.Wait, let me verify that. If the favorite song is played third, the first two songs must be short enough so that their total length plus 240 seconds is less than or equal to 300. So, the first two songs must sum to <= 60 seconds. But the shortest two songs are 40 and 80, which sum to 120. 120 + 240 = 360 > 300. So, indeed, if the favorite song is played third or later, it will not finish within the first 5 minutes.Therefore, the favorite song can only be played first or second to be heard in its entirety within the first 5 minutes. So, the number of favorable permutations where the favorite song is played first or second is the number of ways where the favorite song is in the first or second position.So, the number of permutations where the favorite song is first: 1 (for the favorite song) * 9! (for the remaining songs) = 9!.Similarly, the number of permutations where the favorite song is second: 9 (choices for the first song) * 1 (favorite song) * 8! (for the remaining songs) = 9 * 8!.Wait, but hold on. If the favorite song is second, the first song must be such that the total time of the first song plus 240 seconds is <= 300. So, the first song must be <= 60 seconds. But the shortest song is 40 seconds. So, the first song can only be 40 seconds. Therefore, if the favorite song is second, the first song must be 40 seconds.Therefore, the number of permutations where the favorite song is second is 1 (for the 40-second song) * 1 (for the favorite song) * 8! (for the remaining songs) = 8!.So, total number of favorable permutations where the favorite song is heard in its entirety is 9! + 8!.Therefore, the number of permutations where the favorite song is not heard in its entirety is total permutations minus favorable permutations: 10! - (9! + 8!).But wait, let me think again. The favorite song can be played first or second, but only if the first song is 40 seconds when it's played second. So, actually, the number of favorable permutations is:- Favorite song first: 9! permutations.- Favorite song second: 1 (40 seconds) * 1 (favorite song) * 8! = 8!.So, total favorable permutations: 9! + 8!.Therefore, the probability that the favorite song is heard in its entirety is (9! + 8!)/10!.Simplify that:(9! + 8!)/10! = (9*8! + 8!)/10! = (8!(9 + 1))/10! = (8! * 10)/10! = (10 * 8!)/(10 * 9 * 8!) = 1/9.Wait, that can't be right because 10! = 10 * 9 * 8!, so (8! * 10)/10! = 10/(10*9) = 1/9.But wait, that would mean the probability of hearing the favorite song in its entirety is 1/9, so the probability of not hearing it is 1 - 1/9 = 8/9.But that contradicts my earlier thought where I considered that the favorite song could be first or second, but only if the first song is 40 seconds when it's second.Wait, let me check the calculation again.Total permutations: 10!.Favorable permutations:- Favorite song first: 9!.- Favorite song second: 1 (40 seconds) * 1 (favorite song) * 8! = 8!.So, total favorable: 9! + 8!.So, probability: (9! + 8!)/10! = (9*8! + 8!)/10! = (8!(9 + 1))/10! = (8! * 10)/10! = (10 * 8!)/(10 * 9 * 8!) = 1/9.Yes, that's correct. So, the probability of hearing the favorite song in its entirety is 1/9, so the probability of not hearing it is 8/9.Wait, but earlier I thought that if the favorite song is played second, the first song must be 40 seconds. So, the number of favorable permutations is 9! (for favorite first) plus 8! (for favorite second with first song 40 seconds). So, total favorable is 9! + 8!.Therefore, the probability is (9! + 8!)/10! = 1/9, so the probability of not hearing it is 8/9.But wait, let me think about this again. If the favorite song is played second, the first song must be 40 seconds. So, the number of ways is 1 (for 40 seconds) * 1 (favorite) * 8! = 8!.Similarly, if the favorite song is played first, it's 9!.So, total favorable is 9! + 8!.Thus, the probability is (9! + 8!)/10! = (9! + 8!)/(10*9!) = (1 + 1/9)/10 = (10/9)/10 = 1/9.Yes, that's correct.Therefore, the probability that she does not hear her favorite song in its entirety is 1 - 1/9 = 8/9.Wait, but in the initial problem, it's stated that the o-Pod plays all the songs in a random order before repeating any song. So, the first 5 minutes could include multiple songs, but the favorite song must be played in its entirety within those 5 minutes.Wait, but 5 minutes is 300 seconds. The favorite song is 240 seconds. So, if the favorite song is played first, it will finish at 240 seconds, leaving 60 seconds. If it's played second, the first song must be 40 seconds, so the favorite song starts at 40 and ends at 280, leaving 20 seconds. If it's played third, the first two songs must be 40 and 80, totaling 120, so the favorite song would end at 360, which is over 300. So, indeed, only if the favorite song is first or second (with first song 40 seconds) will it finish within 300 seconds.Therefore, the number of favorable permutations is 9! + 8!.Thus, the probability is (9! + 8!)/10! = 1/9, so the probability of not hearing it is 8/9.Wait, but let me think about another angle. The total number of ways the favorite song can be played in the first two positions is 2 * 9! (since for each position, the favorite song can be there and the rest arranged). But no, that's not correct because if the favorite song is second, the first song must be 40 seconds, so it's not just any song.So, actually, the number of favorable permutations is:- Favorite first: 9!.- Favorite second: 1 (40 seconds) * 1 (favorite) * 8! = 8!.So, total favorable: 9! + 8!.Therefore, the probability is (9! + 8!)/10! = 1/9.Thus, the probability of not hearing it is 8/9.Wait, but I'm getting confused because earlier I thought it was 9/10, but now it's 8/9. Let me double-check.Total permutations: 10!.Favorable permutations where favorite is heard in entirety: 9! (favorite first) + 8! (favorite second with first song 40 seconds).So, total favorable: 9! + 8!.Probability: (9! + 8!)/10! = (9*8! + 8!)/10! = (8!(9 + 1))/10! = (8! * 10)/10! = (10 * 8!)/(10 * 9 * 8!) = 1/9.So, probability of hearing it is 1/9, so not hearing it is 8/9.But wait, in the initial problem, the favorite song is 4 minutes, which is 240 seconds. The total time is 300 seconds.If the favorite song is played first, it takes 240 seconds, leaving 60 seconds. If it's played second, the first song must be 40 seconds, so total time is 40 + 240 = 280 seconds, leaving 20 seconds. If it's played third, the first two songs must be 40 and 80, totaling 120, so the favorite song would end at 360, which is over 300. So, only first and second positions with the first song being 40 seconds allow the favorite song to finish within 300 seconds.Therefore, the number of favorable permutations is 9! (favorite first) + 8! (favorite second with first song 40 seconds).Thus, the probability is (9! + 8!)/10! = 1/9.Therefore, the probability of not hearing it is 8/9.Wait, but I'm getting conflicting results. Earlier, I thought it was 9/10, but now it's 8/9. Let me see where I went wrong.In the initial problem, the favorite song is 4 minutes, which is 240 seconds. The total time is 5 minutes, which is 300 seconds.If the favorite song is played first, it will finish at 240 seconds. If it's played second, the first song must be 40 seconds, so the favorite song starts at 40 and ends at 280. If it's played third, the first two songs must be 40 and 80, totaling 120, so the favorite song would end at 360, which is over 300.Therefore, only if the favorite song is first or second (with first song 40 seconds) will it finish within 300 seconds.So, the number of favorable permutations is:- Favorite first: 9!.- Favorite second: 1 (40 seconds) * 1 (favorite) * 8! = 8!.Total favorable: 9! + 8!.Probability: (9! + 8!)/10! = 1/9.Therefore, the probability of not hearing it is 1 - 1/9 = 8/9.Wait, but in the initial problem, the user's answer was 9/10, but my calculation shows 8/9. Let me see why.In the initial problem, the user considered that the favorite song could only be played first to finish within 300 seconds, but that's incorrect because if it's played second with the first song being 40 seconds, it still finishes within 300 seconds.Therefore, the correct number of favorable permutations is 9! + 8!, leading to a probability of 1/9 for hearing it, and 8/9 for not hearing it.Wait, but let me think again. The total number of permutations where the favorite song is in the first position is 9!.The number of permutations where the favorite song is in the second position and the first song is 40 seconds is 8!.So, total favorable permutations: 9! + 8!.Total permutations: 10!.Therefore, probability of hearing it: (9! + 8!)/10! = (9*8! + 8!)/10! = (8!(9 + 1))/10! = (8! * 10)/10! = (10 * 8!)/(10 * 9 * 8!) = 1/9.So, the probability of not hearing it is 1 - 1/9 = 8/9.Therefore, the correct answer should be 8/9.Wait, but in the initial problem, the user's answer was 9/10, which is incorrect because they only considered the favorite song being first, but not second with the first song being 40 seconds.Therefore, the correct probability is 8/9.</think>

🧠:<think>Alright, so I'm trying to solve this domino placement problem on an 8x8 chessboard. The dominoes are 1x2 in size, and they can extend beyond the borders of the board, but the center of each domino must lie strictly within the board. That means the center can't be on the border itself. First, I need to understand the constraints better. The chessboard is 8x8, so there are 64 squares in total. Each domino covers two squares, so theoretically, if I could place dominoes without any overlap and entirely within the board, I could fit up to 32 dominoes. But the problem allows dominoes to extend beyond the border, which means I can potentially fit more dominoes because parts of them can be outside the board. However, the center of each domino must still be strictly inside the board.For part (a), I need to place at least 40 dominoes. That seems like a lot more than 32, so I need to figure out how to arrange them so that they can extend beyond the borders while keeping their centers inside. Maybe if I arrange them in a way that they overlap the borders but still maintain their centers within the board.I think about how the center of a domino is determined. For a 1x2 domino, the center would be at the midpoint of its length. So, if the domino is placed horizontally, the center would be at the midpoint of its two squares. Similarly, if placed vertically, the center would be at the midpoint of its two squares. If I want the center to be strictly inside the board, that means the domino can't be placed too close to the edge. For example, if I place a domino horizontally near the top edge, part of it would extend beyond the top border, but its center must still be within the board. So, the domino can't be placed too close to the edge; otherwise, the center would be on the border or outside.Maybe I can create a grid where dominoes are slightly offset so that their centers are within the board, even if parts of them extend beyond the borders. If I divide the board into smaller sections, maybe 4x4 quadrants, and place dominoes in each quadrant in a way that they extend beyond the quadrant's borders but still keep their centers within the overall board.I also think about rotating dominoes. If I rotate them at an angle, maybe I can fit more dominoes without overlapping. However, rotating dominoes might complicate the placement because the centers need to stay strictly inside the board.Another idea is to use a checkerboard pattern, alternating the placement of dominoes horizontally and vertically to maximize coverage. But I'm not sure if that would help in fitting more dominoes beyond the standard 32.Wait, the problem says the dominoes can extend beyond the border, but the center must lie strictly within the board. So, maybe if I place dominoes near the edges, extending beyond, but ensuring that their centers are still within the board. That might allow me to fit more dominoes because parts of them can be outside, but their centers are still inside.Let me try to visualize this. If I place a domino horizontally near the top edge, part of it extends beyond the top, but its center is still within the board. Similarly, I can do this for the bottom, left, and right edges. By doing this, I can potentially fit more dominoes because they're using space outside the board, but their centers are still inside.But how many dominoes can I fit this way? If I can place dominoes near each edge, extending beyond, I might be able to fit more than 32. Maybe 40 is achievable by carefully arranging them in such a way.For part (b), I need to place at least 41 dominoes. That's just one more than part (a). So, if I can fit 40, I should be able to fit 41 by adjusting the placement slightly.For part (c), I need to place more than 41 dominoes. That means I need to find a configuration that allows me to fit at least 42 dominoes. This might be more challenging, but perhaps by optimizing the placement and ensuring that as many dominoes as possible extend beyond the borders while keeping their centers inside, I can achieve this.I think the key is to maximize the number of dominoes by utilizing the space outside the board while keeping their centers within the board. Maybe by creating a grid that slightly overlaps the borders, but the centers remain inside.I also need to ensure that dominoes don't overlap. So, I have to be careful with their placement to avoid any overlaps, even if parts of them are outside the board.Maybe I can use a mathematical approach to determine the maximum number of dominoes that can be placed under these conditions. If I consider the area covered by each domino and the constraints on their placement, I might be able to calculate the maximum number.Each domino covers an area of 2 square units, but since they can extend beyond the board, the effective area they cover within the board is less than 2. So, the total number of dominoes would be more than 32, but I need to figure out exactly how many.I think I need to consider the positions where the centers of the dominoes can be placed. Since the center must lie strictly within the board, the domino can extend beyond the borders, but the center can't be on the border. So, the center must be at least 0.5 units away from the border in all directions.This means that the effective area where the centers can be placed is a smaller square inside the board, with borders reduced by 0.5 units on each side. So, the effective area is 7x7, which is 49 square units. Since each domino's center occupies a point, but the domino itself covers an area, I need to ensure that the dominoes don't overlap.But this is getting a bit complicated. Maybe I should look for patterns or existing solutions to similar problems. I recall that in some tiling problems, people use offset grids or staggered placements to fit more tiles.Perhaps I can divide the board into smaller regions and place dominoes in each region in a way that maximizes the number while keeping their centers inside. For example, dividing the board into 2x2 blocks and placing dominoes in each block, but allowing them to extend beyond the block's borders.Alternatively, I could think about the problem in terms of coordinates. If I assign coordinates to the centers of the dominoes, ensuring they are within the board, and then check if the dominoes placed at those centers overlap or extend beyond the board in a way that's allowed.This is a bit abstract, but I think the key takeaway is that by allowing dominoes to extend beyond the borders, I can fit more than the standard 32 dominoes. The exact number depends on how efficiently I can place them without overlapping and while keeping their centers inside.So, for part (a), placing at least 40 dominoes seems feasible by carefully arranging them near the borders, extending beyond, but keeping their centers within. For part (b), adding one more domino to make it 41 should be possible with a slight adjustment. For part (c), fitting more than 41, like 42, would require an optimized placement to ensure all dominoes meet the center condition without overlapping.I think I need to sketch this out or use a grid to visualize the placements better. Maybe by creating a grid with coordinates and marking where the centers of the dominoes can be, I can determine the maximum number.Overall, the strategy involves:1. Understanding the constraints on domino placement.2. Visualizing the board and possible placements.3. Experimenting with different configurations to maximize the number of dominoes.4. Ensuring no overlaps and centers are strictly within the board.By following these steps, I should be able to determine the maximum number of dominoes that can be placed under the given conditions.</think>

🧠:<think>Okay, so I have this problem about an examination where different percentages of students failed in Hindi, English, and Mathematics. I need to find out the percentage of students who passed in all three subjects. Hmm, let me try to understand what's given and how I can approach this.First, let's list out all the information provided:- 35% of students failed in Hindi.- 45% failed in English.- 25% failed in Mathematics.- 18% failed in all three subjects.- 32% failed in both Hindi and English.- 22% failed in both English and Mathematics.- 24% failed in both Hindi and Mathematics.The examination has a minimum passing threshold of 40% in each subject. So, to pass, a student needs to score at least 40% in each subject. But I'm not sure if the passing threshold affects the calculation directly, or if it's just additional information. Maybe it's just there to set the context that failing means scoring below 40%.I need to find the percentage of students who passed in all three subjects. That means they didn't fail in any of the subjects. So, if I can find the percentage of students who didn't fail in Hindi, English, or Mathematics, that should give me the answer.To do this, I think I can use the principle of inclusion-exclusion. This principle helps in calculating the probability of the union of multiple events by considering their intersections. In this case, the events are failing in Hindi, English, or Mathematics.Let me denote the following:- H = Percentage of students who failed in Hindi = 35%- E = Percentage of students who failed in English = 45%- M = Percentage of students who failed in Mathematics = 25%- H ∩ E = Percentage of students who failed in both Hindi and English = 32%- E ∩ M = Percentage of students who failed in both English and Mathematics = 22%- H ∩ M = Percentage of students who failed in both Hindi and Mathematics = 24%- H ∩ E ∩ M = Percentage of students who failed in all three subjects = 18%According to the inclusion-exclusion principle, the percentage of students who failed in at least one subject (F) can be calculated as:F = H + E + M - (H ∩ E) - (E ∩ M) - (H ∩ M) + (H ∩ E ∩ M)Plugging in the values:F = 35% + 45% + 25% - 32% - 22% - 24% + 18%Let me compute this step by step:First, add up the individual failure percentages:35% + 45% + 25% = 105%Now, subtract the pairwise overlaps:105% - 32% - 22% - 24% = 105% - 78% = 27%Finally, add back the percentage of students who failed in all three subjects:27% + 18% = 45%So, 45% of the students failed in at least one subject. Therefore, the percentage of students who passed in all three subjects would be the complement of this, which is:100% - 45% = 55%Wait, that seems straightforward, but let me double-check to make sure I didn't make any mistakes.Let me verify the inclusion-exclusion formula again. It states that the union of three sets is equal to the sum of the individual sets minus the sum of all pairwise intersections plus the intersection of all three sets. So, in formula terms:|H ∪ E ∪ M| = |H| + |E| + |M| - |H ∩ E| - |E ∩ M| - |H ∩ M| + |H ∩ E ∩ M|Yes, that's correct. So, plugging in the numbers:|H ∪ E ∪ M| = 35% + 45% + 25% - 32% - 22% - 24% + 18% = 45%Therefore, 45% failed in at least one subject, so 55% passed in all three.But just to ensure I haven't overlooked anything, let me consider if there's another way to approach this problem or if there are any potential pitfalls.One thing to consider is whether the overlaps are correctly accounted for. For instance, the 18% who failed in all three subjects are included in each of the pairwise overlaps (H ∩ E, E ∩ M, H ∩ M). So, when we subtract the pairwise overlaps, we're removing the triple overlap three times, which is why we need to add it back once at the end.Let me break it down:- Start with the total failures in each subject: 35% + 45% + 25% = 105%- This counts students who failed in two subjects twice and those who failed in all three subjects three times.- Subtract the pairwise overlaps: 32% (H ∩ E) + 22% (E ∩ M) + 24% (H ∩ M) = 78%- Now, we've subtracted the triple overlap three times, so we need to add it back once: +18%- So, total failure percentage: 105% - 78% + 18% = 45%Yes, that makes sense. So, 45% failed in at least one subject, and thus 55% passed in all three.Another way to think about it is to consider the Venn diagram representation. Each circle represents the percentage of students failing in a subject. The overlaps represent students failing in multiple subjects. By using inclusion-exclusion, we ensure that we're not double-counting or missing any overlaps.I also wonder if the passing threshold of 40% affects this calculation. Since failing is defined as scoring below 40%, and passing is scoring 40% or above, the threshold doesn't directly affect the calculation of the percentages who passed or failed. It's more about defining what constitutes a pass or fail.So, in conclusion, using the inclusion-exclusion principle, I've calculated that 45% of students failed in at least one subject, which means 55% passed in all three subjects.</think>