answer：First, I need to determine Cary's hourly wage after the initial 20% raise. Her starting wage is 10/hour. A 20% raise means she earns an additional 20% of 10, which is 2. Therefore, her wage after the raise is 12/hour.In the second year, her wage is reduced to 9/hour. To find out what percentage 9 is of her previous wage of 12, I can set up the equation:9 = (x/100) * 12To solve for x, I divide both sides by 12:9/12 = x/100Simplifying 9/12 gives 0.75. Therefore:0.75 = x/100Multiplying both sides by 100 gives:75 = xSo, Cary is now making 75% of her previous wage.

answer：Alright, so I've got this problem about moving counters on an infinite grid. It's divided into two parts, and I need to tackle both. Let me start by understanding what's being asked.First, the setup: There's an n x n square grid, each cell has a counter. A move is sliding n adjacent counters in a row or column by one space. A returning sequence is a series of moves that brings all counters back to their original n x n square. Part (a) says that if all counters are distinguishable except two, which are indistinct, then any distinguishable arrangement can be reached by a returning sequence. Hmm, so even though two counters are the same, we can still rearrange everything else as needed.Part (b) is trickier: If all counters are distinguishable, then there's no returning sequence that just swaps two counters and leaves the rest as they were. So, you can't just swap two and have everything else back to normal.Let me think about part (b) first because it might give me some insight into part (a). Maybe it's about permutations and parity? If I consider each move as a permutation of the counters, then a returning sequence is a permutation that's a product of these moves and brings everything back. Each move slides n counters, so it's a cyclic shift of n elements. A cyclic shift of n elements is an (n-1)-cycle in permutation terms. For example, if n=2, sliding two counters is a transposition, which is a 2-cycle. For n=3, it's a 3-cycle, which is even or odd depending on n.Wait, the parity of a permutation is important here. If n is even, an (n-1)-cycle is an odd permutation because it can be written as n-2 transpositions. If n is odd, it's even. But the problem says n > 1, so n can be either even or odd.But in part (b), all counters are distinguishable, so we need to consider the entire permutation group. The key is that a returning sequence must be an even permutation because each move contributes a certain parity, and to return to the original position, the total number of moves must result in an even permutation.Wait, actually, each move is an (n-1)-cycle. The parity of an (n-1)-cycle is (n-2), which is even if n is even, and odd if n is odd. So, if n is even, each move is an even permutation, and composing even permutations gives an even permutation. If n is odd, each move is an odd permutation, and composing two moves gives an even permutation.But regardless, a returning sequence must result in the identity permutation, which is even. So, if you have a sequence of moves that brings everything back, the total permutation is even. Therefore, you can't get an odd permutation like a single transposition (swapping two counters) because that's an odd permutation.So, for part (b), it's impossible to have a returning sequence that just swaps two counters because that would require an odd permutation, but any returning sequence must result in an even permutation.Now, for part (a), since two counters are indistinct, maybe the permutation group is different. If two counters are the same, then swapping them doesn't change the arrangement. So, the group of permutations is the symmetric group modulo the subgroup generated by swapping those two counters. In other words, the group is S_{n^2}/⟨(i j)⟩, where i and j are the two indistinct counters. Since we're quotienting out by a transposition, which is an odd permutation, the group becomes smaller. But wait, actually, the group of distinguishable arrangements is S_{n^2 - 2} × S_2, but since the two are indistinct, it's S_{n^2 - 2} × C_2, where C_2 is the cyclic group of order 2. But I'm not sure if that's the right way to think about it.Alternatively, since two counters are indistinct, the number of distinguishable arrangements is (n^2)! / 2. So, the group is S_{n^2} with two elements identified. But regardless, since in part (b) we saw that only even permutations are achievable, in part (a), since two counters are indistinct, maybe we can achieve any permutation, even odd ones, because swapping the two indistinct counters is allowed without changing the arrangement.Wait, no. Because if you have two indistinct counters, you can't distinguish between their positions, so swapping them doesn't change the overall arrangement. Therefore, any permutation that swaps these two counters is equivalent to the identity in terms of distinguishable arrangements.So, the group of distinguishable arrangements is S_{n^2} / ⟨(i j)⟩, which is a group of order (n^2)! / 2. But in part (b), we saw that only even permutations are achievable because each move contributes an even permutation. So, in part (a), since we can now swap the two indistinct counters, which is an odd permutation, maybe we can generate any permutation.Wait, but actually, the two indistinct counters can be swapped, but the rest must still follow the even permutation rule. Hmm, maybe not. Let me think again.Each move is an (n-1)-cycle, which is an odd permutation if n is odd, and even if n is even. So, if n is even, each move is even, so the group generated is a subgroup of the alternating group. If n is odd, each move is odd, so composing two moves gives an even permutation, so the group is also a subgroup of the alternating group.But in part (a), since two counters are indistinct, maybe the group is larger. Because swapping the two indistinct counters is allowed, which is a transposition, an odd permutation. So, if we can generate both even and odd permutations, we can reach any arrangement.Wait, but the moves themselves are still even or odd depending on n. So, if n is even, each move is even, and swapping the two indistinct counters is odd, so together, we can generate both even and odd permutations. Similarly, if n is odd, each move is odd, so composing two moves gives even, but swapping the two counters is also odd, so we can generate both even and odd permutations.Therefore, in part (a), since we can generate both even and odd permutations, we can reach any distinguishable arrangement.But I'm not entirely sure. Maybe I need to think about it differently. Let's consider the effect of the moves on the permutation parity.If n is even, each move is an even permutation, so the group generated is a subgroup of the alternating group. But since we can swap two indistinct counters, which is an odd permutation, we can generate the entire symmetric group.Similarly, if n is odd, each move is an odd permutation, so composing two moves gives an even permutation. But since we can also swap the two indistinct counters, which is odd, we can generate both even and odd permutations, hence the entire symmetric group.Therefore, in part (a), we can reach any distinguishable arrangement because we can generate any permutation, considering the two indistinct counters allow us to break the parity restriction.So, to summarize:For part (b), since all counters are distinguishable, any returning sequence must result in an even permutation, so you can't just swap two counters (an odd permutation).For part (a), since two counters are indistinct, swapping them is allowed without changing the arrangement, which effectively allows us to generate both even and odd permutations, thus reaching any distinguishable arrangement.I think that's the gist of it. Maybe I should try to formalize it more, but intuitively, the parity argument seems to hold.

answer：Okay, so I have this problem here about sets and their complements. Let me try to understand it step by step. The universal set is U, and there are two subsets, M and N. The condition given is that the intersection of M and N equals N, which is written as M ∩ N = N. I need to figure out which of the options A, B, C, or D is true based on this condition.First, I remember that the intersection of two sets, M ∩ N, consists of all elements that are common to both M and N. So if M ∩ N = N, that means every element in N is also in M. In other words, N is a subset of M. I think this is a key point because it tells me about the relationship between M and N.So, if N is a subset of M, that means all elements of N are contained within M, but M might have additional elements that are not in N. This is important because it affects the complements of these sets.Now, let's recall what a complement of a set is. The complement of a set, say C_U M, is the set of all elements in the universal set U that are not in M. Similarly, C_U N is the set of all elements in U that are not in N.Given that N is a subset of M, I can think about how their complements relate. If N is entirely contained within M, then the complement of M should be a subset of the complement of N. Wait, let me think about that again. If N is inside M, then everything not in M would also not be in N, right? So, if something is not in M, it can't be in N either because N is entirely within M. That means the complement of M is contained within the complement of N. So, C_U M is a subset of C_U N.Let me write that down to make it clearer:- Since N ⊆ M, then C_U M ⊆ C_U N.Looking back at the options:A. C_U M ⊇ C_U NB. M ⊆ C_U NC. C_U M ⊆ C_U ND. M ⊇ C_U NFrom my reasoning, option C seems to be correct because C_U M is a subset of C_U N. Let me double-check the other options to make sure I'm not missing anything.Option A says that the complement of M is a superset of the complement of N. That would mean C_U M contains all elements of C_U N, which isn't necessarily true because we've established that C_U M is actually a subset of C_U N. So, A is incorrect.Option B claims that M is a subset of the complement of N. But if M were a subset of C_U N, that would mean M and N have no elements in common, which contradicts our initial condition that M ∩ N = N. So, B is definitely wrong.Option D states that M is a superset of the complement of N. This would mean that M contains all elements not in N. However, since N is a subset of M, M could have elements outside of N, but it doesn't necessarily include all elements not in N. For example, if U has elements that are neither in M nor in N, M wouldn't include those. So, D isn't necessarily true either.Therefore, after analyzing each option, I'm confident that option C is the correct answer because the complement of M is indeed a subset of the complement of N when N is a subset of M.

answer：First, I need to determine the total number of balls in the box, which is 10. These balls are divided into 7 red balls, 2 yellow balls, and 1 black ball.Next, I calculate the probability of drawing each color:- Probability of drawing a red ball: 7 out of 10, which is 7/10.- Probability of drawing a yellow ball: 2 out of 10, which is 2/10.- Probability of drawing a black ball: 1 out of 10, which is 1/10.Comparing these probabilities, the red ball has the highest probability (7/10), followed by the yellow ball (2/10), and the black ball has the lowest probability (1/10).Now, I evaluate each statement:- **A: It must be a red ball.** This is incorrect because while red balls are the most probable, it's not guaranteed.- **B: The probability of drawing a red ball is the highest.** This is correct based on the calculated probabilities.- **C: It cannot be a black ball.** This is incorrect because there is a small probability (1/10) of drawing the black ball.- **D: The probability of drawing a yellow ball is the lowest.** This is incorrect because the black ball has the lowest probability.Therefore, the correct statement is B.

answer：Okay, so I have this problem here with three sets: A, B, and C. Let me try to understand what each set represents. Set A is defined as all real numbers x such that x is greater than 1. So, A = {x | x > 1}. That means A includes every number on the number line starting just above 1 and going to infinity. Set B is {x | -2 < x < 2}. So, B includes all numbers between -2 and 2, but not including -2 and 2 themselves. That's a finite interval from -2 to 2.Set C is {x | -3 < x < 5}. So, C includes all numbers between -3 and 5, again not including -3 and 5. That's a wider interval than B.The first part of the problem asks me to find (A ∪ B) ∩ C. Let me break this down step by step. First, I need to find the union of A and B, which is A ∪ B. The union of two sets includes all elements that are in either set. So, A is everything greater than 1, and B is everything between -2 and 2. If I combine these two, what do I get?Well, A starts at 1 and goes to infinity, and B starts at -2 and goes up to 2. So, combining these, the union would start at -2 and go to infinity because B covers from -2 to 2, and A covers from 1 to infinity. But wait, there's some overlap between A and B. Specifically, from 1 to 2, both sets overlap. So, the union would be from -2 to infinity.But let me double-check that. If I take all numbers greater than 1 (A) and all numbers between -2 and 2 (B), then the smallest number in the union is -2 (from B), and the largest is infinity (from A). So yes, A ∪ B is all real numbers greater than -2.Now, I need to find the intersection of this union with set C. The intersection of two sets includes only the elements that are common to both sets. So, (A ∪ B) is all numbers greater than -2, and C is all numbers between -3 and 5. So, the intersection will be the overlap between these two sets. Since (A ∪ B) starts at -2 and goes to infinity, and C starts at -3 and goes to 5, the overlap will be from -2 to 5. Because from -2 to 5, both sets include those numbers. Beyond 5, only (A ∪ B) includes those numbers, but C stops at 5. Below -2, C includes numbers down to -3, but (A ∪ B) doesn't include anything below -2. So, the intersection is from -2 to 5.Therefore, (A ∪ B) ∩ C is the interval (-2, 5).Now, moving on to the second part of the problem, which is to compute the expression:(2 1/4)^(1/2) - (-9.6)^0 - (3 3/8)^(-2/3) + (1.5)^(-2)Let me parse this expression step by step.First, (2 1/4)^(1/2). I think 2 1/4 is a mixed number, which is equal to 2 + 1/4 = 9/4. So, (9/4)^(1/2) is the square root of 9/4, which is 3/2 or 1.5.Next, (-9.6)^0. Any non-zero number raised to the power of 0 is 1. So, this term is 1.Then, (3 3/8)^(-2/3). Again, 3 3/8 is a mixed number, which is 3 + 3/8 = 27/8. So, (27/8)^(-2/3). Negative exponents mean reciprocals, so this is the same as 1 / (27/8)^(2/3). Now, (27/8)^(2/3) can be broken down. The exponent 2/3 means we take the cube root first and then square it. The cube root of 27 is 3, and the cube root of 8 is 2. So, (27/8)^(1/3) is 3/2. Then, squaring that gives (3/2)^2 = 9/4. Therefore, (27/8)^(2/3) is 9/4, and the reciprocal is 4/9.So, (3 3/8)^(-2/3) is 4/9.Lastly, (1.5)^(-2). 1.5 is the same as 3/2. So, (3/2)^(-2) is the reciprocal of (3/2)^2. (3/2)^2 is 9/4, so the reciprocal is 4/9.Putting it all together:(2 1/4)^(1/2) = 3/2(-9.6)^0 = 1(3 3/8)^(-2/3) = 4/9(1.5)^(-2) = 4/9So, the expression becomes:3/2 - 1 - 4/9 + 4/9Let's compute this step by step.First, 3/2 - 1. 3/2 is 1.5, and 1 is 1.0, so 1.5 - 1.0 = 0.5 or 1/2.Next, subtracting 4/9: 1/2 - 4/9. To subtract these, I need a common denominator. The least common denominator of 2 and 9 is 18.1/2 = 9/184/9 = 8/18So, 9/18 - 8/18 = 1/18.Then, adding 4/9: 1/18 + 4/9. Again, convert to common denominators.4/9 = 8/18So, 1/18 + 8/18 = 9/18 = 1/2.Wait, that seems a bit confusing. Let me double-check.Original expression:3/2 - 1 - 4/9 + 4/9Compute 3/2 - 1 first: 3/2 - 1 = 1/2.Then, 1/2 - 4/9 + 4/9.Notice that -4/9 + 4/9 cancels out to 0.So, 1/2 - 0 = 1/2.Ah, that's simpler. So, the entire expression simplifies to 1/2.Therefore, the value of the expression is 1/2.So, summarizing:(A ∪ B) ∩ C = (-2, 5)And the computed expression equals 1/2.

answer：Okay, so I have this problem here: If the domain of the function ( f(x+1) ) is ([0,1]), then I need to find the domain of the function ( f(2^x - 2) ). Hmm, let me try to figure this out step by step.First, I remember that the domain of a function is the set of all possible input values (x-values) for which the function is defined. So, if the domain of ( f(x+1) ) is ([0,1]), that means ( x ) can be any value between 0 and 1, inclusive. Wait, but ( f(x+1) ) is a transformation of the function ( f(x) ). Specifically, it's a horizontal shift. If I recall correctly, replacing ( x ) with ( x + 1 ) shifts the graph of ( f(x) ) to the left by 1 unit. So, if the domain of ( f(x+1) ) is ([0,1]), that tells me something about the domain of the original function ( f(x) ).Let me think. If ( x ) is in ([0,1]) for ( f(x+1) ), then ( x + 1 ) must be in ([1,2]). So, the input to the function ( f ) is between 1 and 2. That means the domain of ( f(x) ) itself is ([1,2]). Okay, that makes sense.Now, I need to find the domain of ( f(2^x - 2) ). So, for this function, the input to ( f ) is ( 2^x - 2 ). Since we already established that ( f ) is only defined for inputs between 1 and 2, we need to find all ( x ) such that ( 2^x - 2 ) is between 1 and 2.Let me write that down as an inequality:[ 1 leq 2^x - 2 leq 2 ]Now, I need to solve this inequality for ( x ). Let me break it down into two separate inequalities:1. ( 2^x - 2 geq 1 )2. ( 2^x - 2 leq 2 )Starting with the first inequality:[ 2^x - 2 geq 1 ][ 2^x geq 3 ]To solve for ( x ), I can take the logarithm base 2 of both sides:[ x geq log_2{3} ]Okay, so that's the lower bound for ( x ). Now, the second inequality:[ 2^x - 2 leq 2 ][ 2^x leq 4 ]Again, taking the logarithm base 2 of both sides:[ x leq log_2{4} ]But ( log_2{4} ) is equal to 2, since ( 2^2 = 4 ). So, the upper bound for ( x ) is 2.Putting it all together, the values of ( x ) that satisfy both inequalities are:[ log_2{3} leq x leq 2 ]Therefore, the domain of ( f(2^x - 2) ) is the interval ([ log_2{3}, 2 ]).Wait, let me double-check my steps to make sure I didn't make any mistakes. 1. The domain of ( f(x+1) ) is ([0,1]), so ( x+1 ) is in ([1,2]), meaning the domain of ( f(x) ) is ([1,2]). That seems right.2. For ( f(2^x - 2) ), the input must be within ([1,2]), so ( 1 leq 2^x - 2 leq 2 ). That makes sense.3. Solving ( 2^x - 2 geq 1 ) gives ( 2^x geq 3 ), so ( x geq log_2{3} ). Correct.4. Solving ( 2^x - 2 leq 2 ) gives ( 2^x leq 4 ), so ( x leq 2 ). Also correct.I think I did everything correctly. So, the domain should indeed be from ( log_2{3} ) to 2.Just to be thorough, let me approximate ( log_2{3} ) to get a sense of the interval. Since ( 2^1 = 2 ) and ( 2^{1.58496} approx 3 ), so ( log_2{3} ) is approximately 1.58496. So, the domain is roughly from 1.585 to 2. That seems reasonable.I don't think I missed anything here. The key was understanding how the domain of ( f(x+1) ) relates to the domain of ( f(x) ), and then applying that to the transformed function ( f(2^x - 2) ). It's all about figuring out the valid inputs for the inner function that fall within the domain of ( f ).Yeah, I'm confident with this answer now.**Final Answer**The domain of the function ( f(2^x - 2) ) is boxed{[log_2{3}, 2]}.