💡Okay, so I have this problem here with four different operations involving exponents, and I need to figure out which one is correct. Let me take it step by step because I'm still getting the hang of exponent rules.First, let's look at option A: ( a cdot a^{2} = a^{2} ). Hmm, I remember that when you multiply two exponents with the same base, you add the exponents. So, ( a ) is the same as ( a^{1} ), right? So, ( a^{1} cdot a^{2} ) should be ( a^{1+2} = a^{3} ). But the answer here is ( a^{2} ), which doesn't match. So, option A doesn't seem correct.Next, option B: ( (ab)^{3} = ab^{3} ). I think when you raise a product to a power, you raise each factor to that power. So, ( (ab)^{3} ) should be ( a^{3}b^{3} ). But the answer given is ( ab^{3} ), which only raises ( b ) to the power of 3 and leaves ( a ) as ( a^{1} ). That doesn't follow the exponent rule correctly, so option B is also incorrect.Moving on to option C: ( (a^{2})^{3} = a^{6} ). This one seems familiar. When you raise a power to another power, you multiply the exponents. So, ( (a^{2})^{3} ) should be ( a^{2 cdot 3} = a^{6} ). That matches the given answer, so option C looks correct.Finally, option D: ( a^{10} div a^{2} = a^{5} ). I recall that when dividing exponents with the same base, you subtract the exponents. So, ( a^{10} div a^{2} ) should be ( a^{10-2} = a^{8} ). But the answer here is ( a^{5} ), which is not right. So, option D is incorrect.After going through each option, it seems like only option C is correct. I think I understand the exponent rules better now, especially how to handle multiplication, division, and raising powers to powers.

💡Okay, so I'm trying to figure out this story about Ivan Tsarevich and Leshy. It seems like a classic Russian fairy tale, but with some math involved. Let me try to break it down step by step.First, Leshy tells Ivan that he traveled for four days and four nights to reach Koschei's kingdom. On the first day, he walked straight north and covered one-third of the journey. Then, on the second day, he turned west and walked through the forest, covering half of the remaining distance. On the third day, he went south through the forest and reached a straight road leading east. He walked 100 versts on that road in a day and arrived in Koschei's kingdom. Leshy suggests that Ivan, being as nimble a traveler, can reach the kingdom by the fifth day. But Ivan says he'll see Vasilisa tomorrow. So, the question is: Is Ivan right? How many versts did Leshy travel, and how many versts does Ivan expect to travel?Alright, let's start by understanding the journey. Leshy's journey is divided into four days, each with a specific direction and distance. The key here is to figure out the total distance to Koschei's kingdom and then determine how much Ivan needs to travel.Let's denote the total distance to Koschei's kingdom as ( D ) versts.On the first day, Leshy walked north and covered one-third of the journey. So, the distance he covered on the first day is:[ frac{D}{3} ]This means the remaining distance after the first day is:[ D - frac{D}{3} = frac{2D}{3} ]On the second day, he turned west and covered half of the remaining distance. So, the distance covered on the second day is:[ frac{1}{2} times frac{2D}{3} = frac{D}{3} ]Now, the remaining distance after the second day is:[ frac{2D}{3} - frac{D}{3} = frac{D}{3} ]On the third day, he went south and covered the same distance as the first day, which is:[ frac{D}{3} ]This brings him to a straight road leading east. On the fourth day, he walked 100 versts on that road and arrived in Koschei's kingdom. So, the distance covered on the fourth day is 100 versts.Now, let's sum up all the distances Leshy traveled:[ frac{D}{3} + frac{D}{3} + frac{D}{3} + 100 = D + 100 ]Wait a minute, that doesn't make sense because the total distance should be ( D ), not ( D + 100 ). I must have made a mistake here.Let me re-examine the problem. On the third day, Leshy went south through the forest and reached a straight road leading east. He walked 100 versts on that road in a day and arrived in Koschei's kingdom. So, the 100 versts is the remaining distance after the first three days.So, after three days, the remaining distance is 100 versts. Let's adjust the calculations accordingly.After the first day:[ text{Distance covered} = frac{D}{3} ][ text{Remaining distance} = frac{2D}{3} ]After the second day:[ text{Distance covered} = frac{D}{3} ][ text{Remaining distance} = frac{2D}{3} - frac{D}{3} = frac{D}{3} ]After the third day:[ text{Distance covered} = frac{D}{3} ][ text{Remaining distance} = frac{D}{3} - frac{D}{3} = 0 ]Wait, that can't be right because he still has 100 versts to go on the fourth day. So, perhaps the distance covered on the third day is not ( frac{D}{3} ), but something else.Let me think again. On the third day, he went south through the forest and reached a straight road leading east. So, the distance he covered on the third day is the same as the distance he covered on the first day, which is ( frac{D}{3} ). But this brings him to the straight road, and then he walks 100 versts east to reach the kingdom.So, the total distance ( D ) is the sum of the distances covered on each day:[ frac{D}{3} + frac{D}{3} + frac{D}{3} + 100 = D + 100 ]But this implies that ( D = D + 100 ), which is impossible. There must be a misunderstanding in how the distances are being added.Perhaps the 100 versts on the fourth day is part of the total distance ( D ). So, the total distance ( D ) is:[ frac{D}{3} + frac{D}{3} + frac{D}{3} + 100 = D ]But this simplifies to:[ D + 100 = D ]Which again is impossible.I think the mistake is in assuming that the distance covered on the third day is ( frac{D}{3} ). Maybe it's not. Let's re-examine the problem.On the first day, Leshy walked north and covered one-third of the journey:[ frac{D}{3} ]Remaining distance:[ frac{2D}{3} ]On the second day, he turned west and covered half of the remaining distance:[ frac{1}{2} times frac{2D}{3} = frac{D}{3} ]Remaining distance:[ frac{2D}{3} - frac{D}{3} = frac{D}{3} ]On the third day, he went south through the forest and reached a straight road leading east. The distance he covered on the third day is the same as the distance he covered on the first day, which is ( frac{D}{3} ). But this brings him to the straight road, so the remaining distance is:[ frac{D}{3} - frac{D}{3} = 0 ]But he still has to walk 100 versts on the fourth day. This suggests that the 100 versts is part of the total distance ( D ).Wait, maybe the total distance ( D ) is the sum of the distances covered on each day, including the 100 versts on the fourth day. So:[ frac{D}{3} + frac{D}{3} + frac{D}{3} + 100 = D ]But this simplifies to:[ D + 100 = D ]Which is impossible. Therefore, my initial assumption must be wrong.Perhaps the distance covered on the third day is not ( frac{D}{3} ), but something else. Let's denote the distance covered on the third day as ( x ). Then, the total distance ( D ) is:[ frac{D}{3} + frac{D}{3} + x + 100 = D ]Simplifying:[ frac{2D}{3} + x + 100 = D ][ x + 100 = frac{D}{3} ][ x = frac{D}{3} - 100 ]But we also know that on the third day, he went south through the forest and reached a straight road leading east. This implies that the distance he covered on the third day is equal to the distance he covered on the first day, which is ( frac{D}{3} ). Therefore:[ x = frac{D}{3} ]Substituting back:[ frac{D}{3} = frac{D}{3} - 100 ][ 0 = -100 ]Which is impossible.Hmm, I'm stuck here. Maybe I need to approach this differently. Let's consider the journey as a series of movements on a coordinate system.Let's assume that the starting point is at the origin (0,0). Leshy travels north on the first day, west on the second day, south on the third day, and east on the fourth day.Let's denote:- North as positive y-direction- West as negative x-direction- South as negative y-direction- East as positive x-directionLet’s denote the total distance as ( D ).On the first day, he travels north and covers ( frac{D}{3} ). So, his position after day 1 is (0, ( frac{D}{3} )).On the second day, he turns west and covers half of the remaining distance. The remaining distance after day 1 is ( D - frac{D}{3} = frac{2D}{3} ). Half of this is ( frac{D}{3} ). So, he travels west ( frac{D}{3} ). His position after day 2 is (-( frac{D}{3} ), ( frac{D}{3} )).On the third day, he goes south and covers the same distance as the first day, which is ( frac{D}{3} ). So, he travels south ( frac{D}{3} ). His position after day 3 is (-( frac{D}{3} ), 0).On the fourth day, he reaches a straight road leading east and walks 100 versts to reach the kingdom. So, he travels east 100 versts. His position after day 4 is (-( frac{D}{3} + 100 ), 0).But since he has reached the kingdom, his final position should be the same as the starting point, which is (0,0). Wait, that doesn't make sense because he started at (0,0) and ended at (-( frac{D}{3} + 100 ), 0). To return to (0,0), he would need to travel ( frac{D}{3} - 100 ) versts east on the fourth day. But the problem states he walked 100 versts east and arrived in Koschei's kingdom. This suggests that his final position is (-( frac{D}{3} + 100 ), 0), which is the location of Koschei's kingdom.But we need to find the total distance ( D ). Let's consider the distance from the starting point to Koschei's kingdom. The distance from (0,0) to (-( frac{D}{3} + 100 ), 0) is | -( frac{D}{3} + 100 ) | versts. But this should be equal to the total distance ( D ).Wait, that doesn't make sense because the total distance ( D ) is the sum of all the distances he traveled, not the straight-line distance from start to finish. So, perhaps I'm overcomplicating it by using coordinates.Let me try a different approach. Let's denote the total distance as ( D ).On day 1: ( frac{D}{3} ) north.Remaining distance: ( D - frac{D}{3} = frac{2D}{3} ).On day 2: half of the remaining distance, which is ( frac{D}{3} ) west.Remaining distance: ( frac{2D}{3} - frac{D}{3} = frac{D}{3} ).On day 3: he goes south and covers the same distance as day 1, which is ( frac{D}{3} ). So, he covers ( frac{D}{3} ) south.Remaining distance: ( frac{D}{3} - frac{D}{3} = 0 ).But he still has to walk 100 versts east on day 4. This suggests that the 100 versts is part of the total distance ( D ). Therefore, the total distance ( D ) is:[ frac{D}{3} + frac{D}{3} + frac{D}{3} + 100 = D ][ D + 100 = D ]Which is impossible. Therefore, my initial assumption that the distance covered on day 3 is ( frac{D}{3} ) must be incorrect.Perhaps the distance covered on day 3 is not ( frac{D}{3} ), but something else. Let's denote the distance covered on day 3 as ( x ). Then:[ frac{D}{3} + frac{D}{3} + x + 100 = D ][ frac{2D}{3} + x + 100 = D ][ x + 100 = frac{D}{3} ][ x = frac{D}{3} - 100 ]But we also know that on day 3, he went south through the forest and reached a straight road leading east. This implies that the distance he covered on day 3 is equal to the distance he covered on day 1, which is ( frac{D}{3} ). Therefore:[ x = frac{D}{3} ]Substituting back:[ frac{D}{3} = frac{D}{3} - 100 ][ 0 = -100 ]Which is impossible.I'm clearly missing something here. Maybe the total distance ( D ) is not the sum of all the segments, but rather the straight-line distance from start to finish. Let's consider that.If we think of the journey as a series of movements on a coordinate system, the total distance ( D ) would be the straight-line distance from the starting point to Koschei's kingdom. However, Leshy's path is not straight; it's a series of north, west, south, and east movements.Let's denote:- North as positive y-direction- West as negative x-direction- South as negative y-direction- East as positive x-directionLet’s denote the total straight-line distance as ( D ).On day 1: he travels north ( frac{D}{3} ). So, position after day 1: (0, ( frac{D}{3} )).On day 2: he travels west ( frac{D}{3} ). Position after day 2: (-( frac{D}{3} ), ( frac{D}{3} )).On day 3: he travels south ( frac{D}{3} ). Position after day 3: (-( frac{D}{3} ), 0).On day 4: he travels east 100 versts. Position after day 4: (-( frac{D}{3} + 100 ), 0).Since he has reached Koschei's kingdom, the straight-line distance from the starting point to the kingdom is the distance from (0,0) to (-( frac{D}{3} + 100 ), 0), which is | -( frac{D}{3} + 100 ) |.But this should be equal to the total straight-line distance ( D ). Therefore:[ | -frac{D}{3} + 100 | = D ]Since distance can't be negative, we can drop the absolute value:[ -frac{D}{3} + 100 = D ][ 100 = D + frac{D}{3} ][ 100 = frac{4D}{3} ][ D = frac{100 times 3}{4} ][ D = 75 ]Wait, that can't be right because Leshy traveled for four days, and on the fourth day, he walked 100 versts. If the total straight-line distance is 75 versts, but he walked 100 versts on the fourth day, that doesn't make sense.I think I'm confusing the total distance traveled with the straight-line distance. The total distance Leshy traveled is the sum of all the segments, which is:[ frac{D}{3} + frac{D}{3} + frac{D}{3} + 100 = D + 100 ]But this implies that the total distance traveled is ( D + 100 ), which contradicts the definition of ( D ) as the total distance. Therefore, my initial approach is flawed.Let me try to think differently. Maybe the total distance ( D ) is the straight-line distance from start to finish, and the path Leshy took is a detour. So, the total distance Leshy traveled is more than ( D ).Let’s denote:- ( D ) as the straight-line distance from start to Koschei's kingdom.- ( L ) as the total distance Leshy traveled.From the problem, Leshy traveled:- Day 1: ( frac{L}{3} ) north.- Day 2: ( frac{L}{6} ) west.- Day 3: ( frac{L}{3} ) south.- Day 4: 100 versts east.So, the total distance Leshy traveled is:[ frac{L}{3} + frac{L}{6} + frac{L}{3} + 100 = L ][ frac{2L}{6} + frac{L}{6} + frac{2L}{6} + 100 = L ][ frac{5L}{6} + 100 = L ][ 100 = L - frac{5L}{6} ][ 100 = frac{L}{6} ][ L = 600 ]So, Leshy traveled a total of 600 versts.Now, the straight-line distance ( D ) can be found using the coordinates. Let's calculate the net displacement.From the journey:- North: ( frac{L}{3} = 200 ) versts.- West: ( frac{L}{6} = 100 ) versts.- South: ( frac{L}{3} = 200 ) versts.- East: 100 versts.Net north-south displacement:[ 200 text{ north} - 200 text{ south} = 0 ]Net east-west displacement:[ 100 text{ west} - 100 text{ east} = 0 ]Wait, that can't be right because he ended up at Koschei's kingdom, which should be a certain distance away. But according to this, his net displacement is zero, meaning he returned to the starting point, which contradicts the story.I must have made a mistake in interpreting the directions. Let's re-examine the problem.On the first day, he went north ( frac{L}{3} ).On the second day, he went west ( frac{L}{6} ).On the third day, he went south ( frac{L}{3} ).On the fourth day, he went east 100 versts.So, net north-south displacement:[ frac{L}{3} - frac{L}{3} = 0 ]Net east-west displacement:[ frac{L}{6} - 100 ]But since he ended up at Koschei's kingdom, which is a certain distance away, the net displacement should be equal to the straight-line distance ( D ).Therefore:[ sqrt{( frac{L}{6} - 100 )^2 + 0^2} = D ][ | frac{L}{6} - 100 | = D ]But we already found that ( L = 600 ) versts. So:[ | frac{600}{6} - 100 | = D ][ | 100 - 100 | = D ][ 0 = D ]This is impossible because Koschei's kingdom is not at the starting point.I think the mistake is in assuming that the straight-line distance ( D ) is the net displacement. Instead, the total distance Leshy traveled is 600 versts, and the straight-line distance ( D ) is the distance from the starting point to Koschei's kingdom, which is the same as the distance from the starting point to the point where he turned east on the fourth day.Wait, let's think about it differently. After three days, Leshy is at a point where he can go straight east to Koschei's kingdom. So, the distance from that point to Koschei's kingdom is 100 versts east. Therefore, the straight-line distance from the starting point to Koschei's kingdom is the hypotenuse of a right triangle where one leg is the net east-west displacement after three days, and the other leg is the net north-south displacement after three days.From the journey:- North: ( frac{L}{3} = 200 ) versts.- West: ( frac{L}{6} = 100 ) versts.- South: ( frac{L}{3} = 200 ) versts.Net north-south displacement:[ 200 - 200 = 0 ]Net east-west displacement:[ 100 text{ west} ]So, after three days, he is 100 versts west of the starting point. Then, he goes east 100 versts to reach Koschei's kingdom. Therefore, the straight-line distance from the starting point to Koschei's kingdom is 100 versts east.Wait, that can't be right because he traveled 600 versts in total, but the straight-line distance is only 100 versts. That would mean he took a very long detour.But the problem states that Ivan Tsarevich is as nimble a traveler as Leshy, meaning he can travel the same distance per day. Leshy took four days to travel 600 versts, so he travels 150 versts per day. Therefore, Ivan can travel 150 versts per day.But Ivan says he will see Vasilisa tomorrow, meaning he thinks he can reach Koschei's kingdom in one day. If the straight-line distance is 100 versts, and Ivan can travel 150 versts per day, he can indeed reach it in less than a day, but since he can only travel in whole days, he would reach it on the next day, which is tomorrow.Wait, but Leshy's total distance is 600 versts, and Ivan can travel 150 versts per day. So, Ivan would need:[ frac{600}{150} = 4 text{ days} ]But Ivan says he will reach it tomorrow, which is in one day. This contradicts the previous conclusion.I think the confusion arises from whether the total distance is 600 versts or the straight-line distance is 100 versts. Let's clarify.If Leshy traveled 600 versts in four days, that's 150 versts per day. Therefore, Ivan, being as nimble, can also travel 150 versts per day. If the straight-line distance to Koschei's kingdom is 100 versts, Ivan can reach it in:[ frac{100}{150} approx 0.67 text{ days} ]Which is less than a day, so he would reach it on the same day. But since he starts tomorrow, he would see Vasilisa the next day, which is tomorrow.However, if the total distance is 600 versts, Ivan would need four days to travel that distance, which contradicts his statement.Therefore, the key is to determine whether the total distance is 600 versts or the straight-line distance is 100 versts.From the problem, Leshy traveled 600 versts in four days, and Ivan can travel at the same speed. Therefore, Ivan can travel 150 versts per day. If the straight-line distance is 100 versts, Ivan can reach it in less than a day, but since he starts tomorrow, he would arrive the same day, which is tomorrow.However, if the total distance is 600 versts, Ivan would need four days, which contradicts his statement.Therefore, the correct interpretation is that the straight-line distance is 100 versts, and Leshy's total distance is 600 versts due to his detour. Therefore, Ivan can reach Koschei's kingdom in one day by taking the straight road, which is 100 versts.So, Ivan is right. Leshy traveled 600 versts, and Ivan expects to travel 100 versts.

💡First, I need to solve for the unknown value represented by "?" in the equation:? × 24 = 173 × 240To isolate "?", I will divide both sides of the equation by 24:? = (173 × 240) / 24Next, I calculate the product of 173 and 240:173 × 240 = 41,520Now, I divide 41,520 by 24 to find the value of "?":41,520 ÷ 24 = 1,730Therefore, the value of "?" is 1,730.

💡First, identify the values of ( a ) and ( b ) from the given hyperbola equation. The equation is in the standard form ( frac{x^2}{a^2} - frac{y^2}{b^2} = 1 ), so ( a^2 = 4 ) and ( b^2 = 2 ). Therefore, ( a = 2 ) and ( b = sqrt{2} ).Next, calculate ( c ) using the relationship ( c^2 = a^2 + b^2 ). Substituting the known values:[c^2 = 4 + 2 = 6][c = sqrt{6}]Finally, determine the eccentricity ( e ) using the formula ( e = frac{c}{a} ):[e = frac{sqrt{6}}{2}]Thus, the eccentricity of the hyperbola is ( frac{sqrt{6}}{2} ).

💡To determine the initial number of pretzels, let's work backwards from the end of the problem. 1. **After the Cheshire Cat ate:** - The Cheshire Cat ate half of the remaining pretzels and then one more, leaving no pretzels. - Let ( C ) be the number of pretzels before the Cheshire Cat ate. - The Cat ate ( frac{C}{2} + 1 ), so: [ C - left( frac{C}{2} + 1 right) = 0 Rightarrow frac{C}{2} - 1 = 0 Rightarrow C = 2 ]2. **After the Dormouse ate:** - The Dormouse ate half of the remaining pretzels and then one more, leaving ( C = 2 ) pretzels. - Let ( D ) be the number of pretzels before the Dormouse ate. - The Dormouse ate ( frac{D}{2} + 1 ), so: [ D - left( frac{D}{2} + 1 right) = 2 Rightarrow frac{D}{2} - 1 = 2 Rightarrow frac{D}{2} = 3 Rightarrow D = 6 ]3. **After the March Hare ate:** - The March Hare ate half of the remaining pretzels and then one more, leaving ( D = 6 ) pretzels. - Let ( M ) be the number of pretzels before the March Hare ate. - The March Hare ate ( frac{M}{2} + 1 ), so: [ M - left( frac{M}{2} + 1 right) = 6 Rightarrow frac{M}{2} - 1 = 6 Rightarrow frac{M}{2} = 7 Rightarrow M = 14 ]4. **After the Hatter ate:** - The Hatter ate half of the remaining pretzels and then one more, leaving ( M = 14 ) pretzels. - Let ( H ) be the initial number of pretzels. - The Hatter ate ( frac{H}{2} + 1 ), so: [ H - left( frac{H}{2} + 1 right) = 14 Rightarrow frac{H}{2} - 1 = 14 Rightarrow frac{H}{2} = 15 Rightarrow H = 30 ]Thus, the initial number of pretzels was **30**.

💡Alright, so I have this problem about an environmental protection department conducting a survey on waste classification knowledge. They've given me some data about the scores of 1000 participants, and there are two parts to the problem. Let me try to understand and solve each part step by step.**Part (1): Finding the Probability P(36 ≤ Z ≤ 79.5)**First, they mention that the scores Z follow a normal distribution N(μ, 210), where μ is the average score of the 1000 participants. They also say that μ is approximated by the midpoint value of each interval for the data in the same group. So, I need to calculate μ first.Looking at the table:| Group | [30,40] | [40,50] | [50,60] | [60,70] | [70,80] | [80,90] | [90,100] ||-------------|---------|---------|---------|---------|---------|---------|----------|| Frequency | 25 | 150 | 200 | 250 | 225 | 100 | 50 |Each group is a score interval, and the frequency is the number of participants who scored in that interval. To find μ, I need to calculate the midpoint of each interval and then take the weighted average using the frequencies.Let me list the midpoints:- [30,40]: midpoint is 35- [40,50]: midpoint is 45- [50,60]: midpoint is 55- [60,70]: midpoint is 65- [70,80]: midpoint is 75- [80,90]: midpoint is 85- [90,100]: midpoint is 95Now, I'll calculate the weighted average:μ = (35 * 25 + 45 * 150 + 55 * 200 + 65 * 250 + 75 * 225 + 85 * 100 + 95 * 50) / 1000Let me compute each term:- 35 * 25 = 875- 45 * 150 = 6750- 55 * 200 = 11,000- 65 * 250 = 16,250- 75 * 225 = 16,875- 85 * 100 = 8,500- 95 * 50 = 4,750Now, sum all these up:875 + 6750 = 76257625 + 11,000 = 18,62518,625 + 16,250 = 34,87534,875 + 16,875 = 51,75051,750 + 8,500 = 60,25060,250 + 4,750 = 65,000So, μ = 65,000 / 1000 = 65Okay, so the mean score μ is 65.Now, the scores follow a normal distribution N(65, 210). Wait, in the notation N(μ, σ²), so the variance is 210, which means the standard deviation σ is sqrt(210). Let me calculate that:σ = sqrt(210) ≈ 14.49Now, I need to find P(36 ≤ Z ≤ 79.5). Since Z is a normal variable, I can standardize it to find the probability.First, let's find the z-scores for 36 and 79.5.For Z = 36:z1 = (36 - μ) / σ = (36 - 65) / 14.49 ≈ (-29) / 14.49 ≈ -2.00For Z = 79.5:z2 = (79.5 - 65) / 14.49 ≈ 14.5 / 14.49 ≈ 1.00So, we need to find P(-2.00 ≤ Z ≤ 1.00), where Z is the standard normal variable.From standard normal distribution tables, P(Z ≤ 1.00) ≈ 0.8413 and P(Z ≤ -2.00) ≈ 0.0228.Therefore, P(-2.00 ≤ Z ≤ 1.00) = P(Z ≤ 1.00) - P(Z ≤ -2.00) ≈ 0.8413 - 0.0228 = 0.8185So, approximately 0.8185 or 81.85%.Wait, but let me double-check my calculations. The standard deviation was sqrt(210) ≈ 14.49. So, 36 is about 29 points below the mean, which is roughly 2 standard deviations. Similarly, 79.5 is about 14.5 points above the mean, which is roughly 1 standard deviation. So, the probability between -2 and +1 standard deviations is indeed around 81.85%.I think that's correct.**Part (2): Distribution and Expectation of X**Now, part (2) is about the reward scheme. Resident A wants to participate, and X represents the phone credit received.The reward scheme is as follows:1. Those with a score not less than μ (which is 65) receive 2 chances of random phone credit top-up.2. Those with a score lower than μ receive 1 chance.3. Each top-up has a probability of 3/4 to receive 20 yuan and 1/4 to receive 40 yuan.So, first, I need to find the distribution of X, which is the total credit received.Let me break it down.First, determine the probability that resident A scores ≥65 or <65.From the original data, the frequencies are:- [30,40]: 25- [40,50]: 150- [50,60]: 200- [60,70]: 250- [70,80]: 225- [80,90]: 100- [90,100]: 50Total participants: 1000Scores ≥65 would be in the intervals [60,70], [70,80], [80,90], [90,100]. Let's sum their frequencies:250 (60-70) + 225 (70-80) + 100 (80-90) + 50 (90-100) = 625Scores <65 would be in [30,40], [40,50], [50,60]. Their frequencies sum to:25 + 150 + 200 = 375Therefore, the probability that resident A scores ≥65 is 625/1000 = 0.625, and <65 is 375/1000 = 0.375.So, P(score ≥65) = 0.625, P(score <65) = 0.375.Now, for each case:- If score ≥65: 2 top-up chances.- If score <65: 1 top-up chance.Each top-up is independent, with P(20 yuan) = 3/4 and P(40 yuan) = 1/4.So, X can be the sum of 1 or 2 top-ups, each of which can be 20 or 40 yuan.Let me consider both cases.**Case 1: Score <65 (P = 0.375)**Only 1 top-up.Possible outcomes:- 20 yuan with probability 3/4- 40 yuan with probability 1/4So, X can be 20 or 40.**Case 2: Score ≥65 (P = 0.625)**2 top-ups. Each can be 20 or 40, so possible total credits:- 20 + 20 = 40- 20 + 40 = 60- 40 + 20 = 60- 40 + 40 = 80But since the top-ups are independent, the probabilities need to be calculated accordingly.Let me list all possible outcomes and their probabilities:1. First top-up: 20 (3/4), Second top-up: 20 (3/4) → Total: 40 yuan Probability: (3/4)*(3/4) = 9/162. First top-up: 20 (3/4), Second top-up: 40 (1/4) → Total: 60 yuan Probability: (3/4)*(1/4) = 3/163. First top-up: 40 (1/4), Second top-up: 20 (3/4) → Total: 60 yuan Probability: (1/4)*(3/4) = 3/164. First top-up: 40 (1/4), Second top-up: 40 (1/4) → Total: 80 yuan Probability: (1/4)*(1/4) = 1/16So, combining the probabilities for the same total:- 40 yuan: 9/16- 60 yuan: 3/16 + 3/16 = 6/16 = 3/8- 80 yuan: 1/16Therefore, for Case 2 (score ≥65), X can be 40, 60, or 80 yuan with probabilities 9/16, 3/8, and 1/16 respectively.Now, combining both cases, the overall distribution of X is:- For Case 1 (score <65, P=0.375): - X=20 with P=3/4 - X=40 with P=1/4- For Case 2 (score ≥65, P=0.625): - X=40 with P=9/16 - X=60 with P=3/8 - X=80 with P=1/16To find the overall distribution, we need to multiply the probabilities by the probability of being in each case.So, let's compute the overall probabilities for each possible X:1. X=20: - Only occurs in Case 1 - P(X=20) = P(Case 1) * P(X=20 | Case 1) = 0.375 * (3/4) = 0.375 * 0.75 = 0.281252. X=40: - Occurs in both Case 1 and Case 2 - P(X=40) = P(Case 1) * P(X=40 | Case 1) + P(Case 2) * P(X=40 | Case 2) - = 0.375 * (1/4) + 0.625 * (9/16) - = 0.375 * 0.25 + 0.625 * 0.5625 - = 0.09375 + 0.3515625 - = 0.44531253. X=60: - Only occurs in Case 2 - P(X=60) = P(Case 2) * P(X=60 | Case 2) = 0.625 * (3/8) = 0.625 * 0.375 = 0.2343754. X=80: - Only occurs in Case 2 - P(X=80) = P(Case 2) * P(X=80 | Case 2) = 0.625 * (1/16) = 0.625 * 0.0625 = 0.0390625Let me verify that all probabilities sum to 1:0.28125 (X=20) + 0.4453125 (X=40) + 0.234375 (X=60) + 0.0390625 (X=80) =0.28125 + 0.4453125 = 0.72656250.7265625 + 0.234375 = 0.96093750.9609375 + 0.0390625 = 1Good, they sum to 1. So, the distribution is correct.Now, to find the expectation E[X], we can compute the sum of each possible X multiplied by its probability.E[X] = 20 * P(X=20) + 40 * P(X=40) + 60 * P(X=60) + 80 * P(X=80)Plugging in the values:E[X] = 20 * 0.28125 + 40 * 0.4453125 + 60 * 0.234375 + 80 * 0.0390625Let's compute each term:1. 20 * 0.28125 = 5.6252. 40 * 0.4453125 = 17.81253. 60 * 0.234375 = 14.06254. 80 * 0.0390625 = 3.125Now, sum these up:5.625 + 17.8125 = 23.437523.4375 + 14.0625 = 37.537.5 + 3.125 = 40.625So, E[X] = 40.625 yuan.Alternatively, since 40.625 is equal to 325/8, but it's more straightforward to present it as 40.625.Wait, let me double-check the calculations:20 * 0.28125 = 5.62540 * 0.4453125 = 17.812560 * 0.234375 = 14.062580 * 0.0390625 = 3.125Adding them up:5.625 + 17.8125 = 23.437523.4375 + 14.0625 = 37.537.5 + 3.125 = 40.625Yes, that's correct.So, the expectation is 40.625 yuan.Alternatively, since 0.625 is 5/8, and 40.625 is 325/8, but 40.625 is more straightforward.Alternatively, since 40.625 is equal to 325/8, but 40.625 is more straightforward.Wait, 40.625 is equal to 40 and 5/8, which is 325/8. But 40.625 is fine.So, summarizing:The distribution of X is:- X = 20 yuan with probability 0.28125- X = 40 yuan with probability 0.4453125- X = 60 yuan with probability 0.234375- X = 80 yuan with probability 0.0390625And the expectation E[X] is 40.625 yuan.Alternatively, to express the probabilities as fractions:- P(X=20) = 9/32- P(X=40) = 14.25/32 ≈ but it's better to keep as decimals or fractions as computed.But since the question asks for the distribution and expectation, I think the decimal probabilities are acceptable.Alternatively, to express them as fractions:- 0.28125 = 9/32- 0.4453125 = 14.25/32, but that's not a clean fraction. Alternatively, 0.4453125 = 71/160- 0.234375 = 15/64- 0.0390625 = 1/25.6, but that's not a clean fraction. Alternatively, 0.0390625 = 1/25.6 ≈ 1/25.6, but better to keep as decimals.Alternatively, since 0.28125 = 9/32, 0.4453125 = 29/64, 0.234375 = 15/64, 0.0390625 = 1/25.6, but 1/25.6 is not a standard fraction. Alternatively, 0.0390625 = 1/25.6 ≈ 1/25.6, but it's better to keep as decimals.Alternatively, since 0.0390625 = 1/25.6, but 25.6 is not a whole number. Alternatively, 0.0390625 = 1/25.6 ≈ 0.0390625.Alternatively, to express all probabilities with denominator 32:- 0.28125 = 9/32- 0.4453125 = 14.25/32, but that's not a whole number.- 0.234375 = 7.5/32, not a whole number.- 0.0390625 = 1.25/32, not a whole number.So, perhaps it's better to keep the probabilities as decimals.Therefore, the distribution is:- X = 20 with P = 0.28125- X = 40 with P = 0.4453125- X = 60 with P = 0.234375- X = 80 with P = 0.0390625And the expectation is 40.625 yuan.Alternatively, to express the expectation as a fraction:40.625 = 40 + 0.625 = 40 + 5/8 = 325/8So, E[X] = 325/8 yuan.But 325/8 is 40.625, so both are correct.I think it's better to present the expectation as 325/8 yuan or 40.625 yuan.So, summarizing:The distribution of X is:- 20 yuan with probability 9/32- 40 yuan with probability 29/64- 60 yuan with probability 15/64- 80 yuan with probability 1/25.6But since 29/64 is approximately 0.453125 and 15/64 is approximately 0.234375, and 1/25.6 is approximately 0.0390625, it's better to present the probabilities as decimals for clarity.Therefore, the distribution is:- X = 20 yuan with P = 0.28125- X = 40 yuan with P = 0.4453125- X = 60 yuan with P = 0.234375- X = 80 yuan with P = 0.0390625And the expectation E[X] = 40.625 yuan.Alternatively, since 40.625 is equal to 325/8, which is 40 and 5/8, but 40.625 is more straightforward.So, I think that's the solution.