answer：First, I need to identify the outcomes for each event. For event ( A ), which is "an even number greater than 2," the possible outcomes are 4 and 6. This gives event ( A ) two favorable outcomes.Next, for event ( B ), which is "an odd number less than 5," the possible outcomes are 1 and 3. This gives event ( B ) two favorable outcomes.Since a die has 6 faces, the probability of any specific outcome is ( frac{1}{6} ). Therefore, the probability of event ( A ) is ( frac{2}{6} = frac{1}{3} ), and the probability of event ( B ) is also ( frac{2}{6} = frac{1}{3} ).Events ( A ) and ( B ) are mutually exclusive, meaning they cannot occur simultaneously. Therefore, the probability of ( A cup B ) is the sum of their individual probabilities:[P(A cup B) = P(A) + P(B) = frac{1}{3} + frac{1}{3} = frac{2}{3}]

answer：Okay, so I need to figure out how much I would pay for 48 doughnuts based on the pricing given. The first doughnut costs 1, and then if I buy additional doughnuts in dozens, each dozen costs 6. Hmm, let's break this down step by step.First, I know that a dozen is 12 doughnuts. So, if I'm buying 48 doughnuts, I need to see how many dozens that is. To find that out, I can divide 48 by 12. Let me do that: 48 divided by 12 equals 4. So, 48 doughnuts make up 4 dozens.Now, according to the pricing, the first doughnut is 1. That means I have to pay 1 for the first doughnut, and then for the remaining doughnuts, I can buy them in dozens at 6 per dozen. Wait, but I need to make sure I understand this correctly. Is the first doughnut 1, and then every additional doughnut beyond that is priced at 6 per dozen? Or is it that the first doughnut is 1, and then any additional doughnuts purchased in dozens are 6 per dozen?I think it's the latter: the first doughnut is 1, and any additional doughnuts bought in dozens are 6 per dozen. So, if I want 48 doughnuts, I have to pay 1 for the first one, and then for the remaining 47 doughnuts, I need to see how many dozens that is.But wait, 47 doughnuts isn't a whole number of dozens. A dozen is 12, so 47 divided by 12 is... let's see, 12 times 3 is 36, and 12 times 4 is 48. So, 47 doughnuts would be 3 dozens and 11 doughnuts. But the pricing only allows me to buy additional doughnuts in dozens at 6 per dozen. Does that mean I have to buy another dozen to get the cheaper rate for the remaining 11 doughnuts? That would make it 4 dozens in total.So, if I buy 4 dozens, that's 48 doughnuts, but I already accounted for the first doughnut at 1. Wait, that doesn't seem right. If I buy 4 dozens, that's 48 doughnuts, but the first doughnut is already included in that count. So, maybe I need to adjust my calculation.Let me think again. The first doughnut is 1, and then any additional doughnuts beyond that can be bought in dozens at 6 per dozen. So, for 48 doughnuts, I have 1 doughnut at 1 and 47 doughnuts that need to be bought in dozens. But since 47 isn't a multiple of 12, I have to buy 4 dozens to cover all 48 doughnuts. That would mean 4 dozens at 6 per dozen, which is 4 times 6, so 24. Then, adding the first doughnut at 1, the total cost would be 24 plus 1, which is 25.Wait, but if I buy 4 dozens, that's 48 doughnuts, and the first doughnut is already included in that count. So, am I double-counting the first doughnut? That doesn't seem right. Maybe I need to subtract the first doughnut from the total count before calculating the dozens.So, if I have 48 doughnuts, and the first one is 1, then the remaining 47 doughnuts need to be bought in dozens. But 47 isn't a multiple of 12, so I have to buy 4 dozens to cover all 48 doughnuts. That means I'm paying for 4 dozens at 6 per dozen, which is 24, plus the first doughnut at 1, totaling 25.But wait, if I buy 4 dozens, that's 48 doughnuts, and the first doughnut is already part of that 48. So, I shouldn't have to pay an extra 1 for the first doughnut. Maybe the pricing is structured so that the first doughnut is 1, and then any additional doughnuts beyond that are priced at 6 per dozen. So, for 48 doughnuts, I have 1 doughnut at 1 and 47 doughnuts that need to be bought in dozens. But since 47 isn't a multiple of 12, I have to buy 4 dozens to cover all 48 doughnuts. That would mean 4 dozens at 6 per dozen, which is 24, plus the first doughnut at 1, totaling 25.But I'm still confused because if I buy 4 dozens, that's 48 doughnuts, and the first doughnut is already included in that count. So, I think the correct way is to consider that the first doughnut is 1, and then the remaining 47 doughnuts are bought in dozens. Since 47 isn't a multiple of 12, I have to buy 4 dozens to cover all 48 doughnuts. Therefore, the total cost is 4 dozens at 6 per dozen, which is 24, plus the first doughnut at 1, totaling 25.Alternatively, maybe the pricing is such that the first doughnut is 1, and then any additional doughnuts beyond that are priced at 6 per dozen. So, for 48 doughnuts, I have 1 doughnut at 1 and 47 doughnuts that need to be bought in dozens. Since 47 isn't a multiple of 12, I have to buy 4 dozens to cover all 48 doughnuts. That would mean 4 dozens at 6 per dozen, which is 24, plus the first doughnut at 1, totaling 25.Wait, but if I buy 4 dozens, that's 48 doughnuts, and the first doughnut is already part of that 48. So, I shouldn't have to pay an extra 1 for the first doughnut. Maybe the pricing is structured so that the first doughnut is 1, and then any additional doughnuts beyond that are priced at 6 per dozen. So, for 48 doughnuts, I have 1 doughnut at 1 and 47 doughnuts that need to be bought in dozens. But since 47 isn't a multiple of 12, I have to buy 4 dozens to cover all 48 doughnuts. That would mean 4 dozens at 6 per dozen, which is 24, plus the first doughnut at 1, totaling 25.I think I'm going in circles here. Let me try to approach it differently. If I need 48 doughnuts, and the first one is 1, then I have 47 left. Since I can only buy additional doughnuts in dozens, I need to see how many dozens are in 47. 47 divided by 12 is 3 with a remainder of 11. So, I can buy 3 dozens, which is 36 doughnuts, and then I have 11 doughnuts left. But I can't buy 11 doughnuts at the dozen rate, so I have to buy another dozen, making it 4 dozens in total. Therefore, I'm buying 4 dozens at 6 per dozen, which is 24, plus the first doughnut at 1, totaling 25.Yes, that makes sense. So, the total cost for 48 doughnuts would be 25.

answer：Okay, so I have this problem here with four different calculations labeled A, B, C, and D. I need to figure out which one is correct. Let me go through each option one by one and see if I can understand what's going on.Starting with option A: (a²)³ = a⁵. Hmm, I remember that when you have an exponent raised to another exponent, you multiply them. So, (a²)³ should be a^(2*3) which is a⁶. But here it's saying a⁵, so that doesn't seem right. Maybe I'm missing something, but I think A is incorrect.Moving on to option B: a³ + a² = a⁶. Okay, this one is adding two terms with exponents. I recall that you can't just add exponents like that unless they're multiplied. For addition, the exponents have to be the same to combine them, right? So, a³ and a² are not like terms, so you can't add them directly. Therefore, this equation doesn't hold, and B is also incorrect.Now, option C: a⁶ ÷ a³ = a³. Alright, division of exponents. I think the rule is that when you divide like bases, you subtract the exponents. So, a⁶ ÷ a³ should be a^(6-3) which is a³. That matches what's given here, so C seems correct.Lastly, option D: a³ · a² = a⁶. Multiplying exponents with the same base. I believe you add the exponents in this case. So, a³ · a² should be a^(3+2) which is a⁵. But here it's saying a⁶, so that's not right. D is incorrect.Wait, let me double-check my reasoning for each one to make sure I didn't make any mistakes.For A: (a²)³. Yes, exponents multiply, so 2*3=6, which is a⁶, not a⁵. So A is definitely wrong.For B: a³ + a². Since they're not like terms, you can't combine them into a single term with a higher exponent. So B is wrong.For C: a⁶ ÷ a³. Subtracting exponents gives 6-3=3, so a³. That matches, so C is correct.For D: a³ · a². Adding exponents gives 3+2=5, so a⁵, not a⁶. So D is wrong.Okay, I think I'm confident now that C is the correct calculation.

answer：First, I need to determine the width of Jesse's room. I know that the room is 20 feet long and that the length is 1 foot longer than the width.Let’s denote the width of the room as ( W ).According to the problem, the length of the room is ( W + 1 ) feet.Since the length is given as 20 feet, I can set up the equation:[W + 1 = 20]To find the width, I will solve for ( W ):[W = 20 - 1][W = 19]Therefore, the width of Jesse's room is 19 feet.

answer：First, I need to evaluate the expression ((a^b)^b - (b^a)^a) with the given values (a = 3) and (b = 4).I'll start by calculating ((3^4)^4). Calculating (3^4) gives me 81. Then, raising 81 to the power of 4, I get 43046721.Next, I'll calculate ((4^3)^3). Calculating (4^3) gives me 64. Raising 64 to the power of 3 results in 262144.Finally, I'll subtract the second result from the first: 43046721 minus 262144 equals 42792577.

answer：Okay, so I have this problem where a merchant has six bags with weights 13, 15, 16, 17, 21, and 24 kilograms. One of these bags contains turnips, and the rest are either onions or carrots. The key information is that the total weight of the carrots is twice the total weight of the onions. I need to figure out which bag could possibly contain the turnips.First, I think I should understand the total weight of all the bags combined. Let me add them up:13 + 15 + 16 + 17 + 21 + 24.Let me calculate that step by step:13 + 15 = 2828 + 16 = 4444 + 17 = 6161 + 21 = 8282 + 24 = 106.So, the total weight of all six bags is 106 kilograms.Now, one bag is turnips, and the remaining five are either onions or carrots. The total weight of the carrots is twice the total weight of the onions. Let me denote the total weight of onions as O and the total weight of carrots as C. According to the problem, C = 2O.Since one bag is turnips, the combined weight of onions and carrots is 106 minus the weight of the turnip bag. Let's denote the weight of the turnip bag as T. So, O + C = 106 - T.But since C = 2O, I can substitute that into the equation:O + 2O = 106 - TWhich simplifies to:3O = 106 - TSo, 106 - T must be divisible by 3 because O has to be an integer (since we're dealing with weights in whole kilograms). Therefore, T must be such that 106 - T is a multiple of 3.Let me list the possible weights for T and check which ones make 106 - T divisible by 3:1. If T = 13: 106 - 13 = 93. 93 divided by 3 is 31. That works.2. If T = 15: 106 - 15 = 91. 91 divided by 3 is approximately 30.333... Not an integer.3. If T = 16: 106 - 16 = 90. 90 divided by 3 is 30. That works.4. If T = 17: 106 - 17 = 89. 89 divided by 3 is approximately 29.666... Not an integer.5. If T = 21: 106 - 21 = 85. 85 divided by 3 is approximately 28.333... Not an integer.6. If T = 24: 106 - 24 = 82. 82 divided by 3 is approximately 27.333... Not an integer.So, the only possible weights for the turnip bag are 13 kg and 16 kg.Now, I need to verify if these actually work by checking if the remaining weights can be split into onions and carrots with the given ratio.First, let's check T = 13 kg:Total weight of onions and carrots = 106 - 13 = 93 kg.Given that C = 2O, we have:O + 2O = 93 => 3O = 93 => O = 31 kg, and C = 62 kg.Now, I need to see if the remaining bags (15, 16, 17, 21, 24) can be split into two groups where one group sums to 31 kg (onions) and the other to 62 kg (carrots).Let me try to find a combination that adds up to 31 kg:Looking at the weights: 15, 16, 17, 21, 24.15 + 16 = 31. Perfect, so onions could be 15 kg and 16 kg.Then, the remaining weights would be 17, 21, and 24. Let's check their sum:17 + 21 + 24 = 62 kg. That's exactly the required weight for carrots.So, T = 13 kg works.Now, let's check T = 16 kg:Total weight of onions and carrots = 106 - 16 = 90 kg.Again, C = 2O, so:O + 2O = 90 => 3O = 90 => O = 30 kg, and C = 60 kg.Now, I need to see if the remaining bags (13, 15, 17, 21, 24) can be split into two groups where one group sums to 30 kg (onions) and the other to 60 kg (carrots).Looking for a combination that adds up to 30 kg:13 + 17 = 30. Perfect, so onions could be 13 kg and 17 kg.Then, the remaining weights would be 15, 21, and 24. Let's check their sum:15 + 21 + 24 = 60 kg. That's exactly the required weight for carrots.So, T = 16 kg also works.Therefore, the turnip bag can be either 13 kg or 16 kg.