💡First, I need to determine the final price John pays for the trousers after all discounts and taxes. The original price is 100€, and there are three discounts: 20%, 15%, and 30%. However, John only benefits from the 30% discount during the clearance sale.1. **Calculate the price after the 30% discount:** [ text{Price after discount} = 100€ - (30% times 100€) = 70€ ]2. **Add the German VAT (19%) to the discounted price:** [ text{German VAT} = 19% times 70€ = 13.3€ ] [ text{Price including German VAT} = 70€ + 13.3€ = 83.3€ ]3. **Convert the price to USD using the exchange rate (1€ = 1.18):** [ text{Price in USD} = 83.3€ times 1.18 = 98.294 ]4. **Add the US VAT (8%) to the USD price:** [ text{US VAT} = 8% times 98.294 = 7.86352 ] [ text{Final price in USD} = 98.294 + 7.86352 = 106.15752 ]5. **Calculate the original price in USD:** [ text{Original price in USD} = 100€ times 1.18 = 118 ]6. **Determine the percent decrease from the original price in USD to the final price:** [ text{Percent decrease} = left( frac{118 - 106.15752}{118} right) times 100 = 10.04% ]

💡Okay, so I have this problem here about a function f(x) that's even and defined on all real numbers. It's also decreasing on the interval [0, +∞). The question is asking about the range of x such that f(lg x) > f(1). The options are A, B, C, D, and I need to figure out which one is correct.First, let me recall what it means for a function to be even. An even function satisfies f(-x) = f(x) for all x in its domain. So, the graph of the function is symmetric with respect to the y-axis. That means the behavior of the function on the negative side is a mirror image of its behavior on the positive side.Next, the function is decreasing on [0, +∞). Since it's even, this tells me something about its behavior on (-∞, 0] as well. If f is decreasing on [0, +∞), then on (-∞, 0], it should be increasing because of the even symmetry. For example, if I take two points on the positive side, say x1 and x2 where x1 < x2, then f(x1) > f(x2). On the negative side, for points -x1 and -x2 where -x1 > -x2, f(-x1) = f(x1) and f(-x2) = f(x2), so f(-x1) > f(-x2), which means it's increasing as we move from left to right on the negative side.Now, the problem gives me f(lg x) > f(1). I need to find the range of x that satisfies this inequality. Let's break this down.First, lg x is the logarithm base 10 of x, right? So, lg x is defined for x > 0. That means x has to be positive. So, x is in (0, +∞).Since f is even, f(lg x) = f(|lg x|). Because f is even, f(lg x) = f(|lg x|). So, the inequality f(lg x) > f(1) can be rewritten as f(|lg x|) > f(1).Now, since f is decreasing on [0, +∞), if f(a) > f(b), then a < b for a, b in [0, +∞). Because the function is decreasing, the larger the input, the smaller the output. So, if f(|lg x|) > f(1), then |lg x| < 1.So, |lg x| < 1. Let's solve this inequality for x. The absolute value inequality |lg x| < 1 can be rewritten as -1 < lg x < 1.Now, to solve for x, we can exponentiate each part of the inequality. Since lg x is log base 10, exponentiating with base 10 will help us solve for x.So, starting with -1 < lg x < 1, exponentiating each part:10^(-1) < x < 10^1.Calculating these, 10^(-1) is 1/10, and 10^1 is 10. So, 1/10 < x < 10.Therefore, the range of x is (1/10, 10). Looking back at the options, that's option C.Wait, hold on. Let me double-check my reasoning. I said that since f is even, f(lg x) = f(|lg x|). That makes sense because f(-y) = f(y). So, f(lg x) is the same as f(|lg x|). Then, since f is decreasing on [0, +∞), if f(|lg x|) > f(1), then |lg x| must be less than 1. That seems correct.So, |lg x| < 1 implies that lg x is between -1 and 1. Converting back from logarithmic form, that means x is between 10^(-1) and 10^1, which is 1/10 and 10. So, x is in (1/10, 10). That's option C.But wait, let me think again. Is there a possibility that x could be outside this interval? For example, if x is less than 1/10 or greater than 10, would f(lg x) still be greater than f(1)?Let me consider x < 1/10. Then, lg x would be less than -1, right? Because lg(1/10) is -1, and for x < 1/10, lg x is less than -1. So, |lg x| would be greater than 1. Since f is decreasing on [0, +∞), f(|lg x|) would be less than f(1). So, f(lg x) = f(|lg x|) < f(1). Therefore, x < 1/10 doesn't satisfy the inequality.Similarly, if x > 10, then lg x > 1, so |lg x| > 1, and again, f(|lg x|) < f(1). So, x > 10 also doesn't satisfy the inequality.Therefore, the only x that satisfy f(lg x) > f(1) are those where |lg x| < 1, which is x between 1/10 and 10. So, option C is correct.Wait, but let me check the options again. Option C is (1/10, 10), which is exactly what I found. So, that should be the answer.But just to make sure, let me think about the function f(x). Since it's even and decreasing on [0, +∞), it's increasing on (-∞, 0]. So, for negative arguments, the function increases as x becomes more negative.But in this case, we're dealing with f(lg x), and since x is positive, lg x can be negative or positive. But because f is even, f(lg x) = f(|lg x|). So, regardless of whether lg x is positive or negative, f(lg x) depends only on the absolute value of lg x.Therefore, the inequality f(lg x) > f(1) is equivalent to f(|lg x|) > f(1), and since f is decreasing on [0, +∞), this implies |lg x| < 1, leading to x in (1/10, 10).Yes, that seems consistent. So, I think option C is correct.**Final Answer**boxed{C}

💡Alright, so I have this problem where I need to find the range of the real number ( a ) such that the intersection of sets ( A ) and ( B ) is empty. The sets are defined as ( A = {x | a - 1 < x < 2a + 1} ) and ( B = {x | 0 < x < 1} ). First, let me make sure I understand what the problem is asking. I need to find all values of ( a ) for which there are no elements common to both sets ( A ) and ( B ). In other words, the intervals defined by ( A ) and ( B ) should not overlap at all.Let me visualize this. Set ( B ) is straightforward; it's the interval from 0 to 1. Set ( A ) is a bit more dynamic because its endpoints depend on the value of ( a ). The lower bound of ( A ) is ( a - 1 ) and the upper bound is ( 2a + 1 ). So, depending on ( a ), the interval ( A ) can shift left or right on the number line.To ensure that ( A ) and ( B ) don't overlap, one of two scenarios must be true:1. The entire interval ( A ) lies to the left of ( B ), meaning the upper bound of ( A ) is less than or equal to the lower bound of ( B ).2. The entire interval ( A ) lies to the right of ( B ), meaning the lower bound of ( A ) is greater than or equal to the upper bound of ( B ).Let me write these conditions mathematically.**Case 1: ( A ) is entirely to the left of ( B )**This means the upper bound of ( A ) is less than or equal to the lower bound of ( B ). So,[ 2a + 1 leq 0 ]Solving for ( a ):[ 2a leq -1 ][ a leq -frac{1}{2} ]**Case 2: ( A ) is entirely to the right of ( B )**This means the lower bound of ( A ) is greater than or equal to the upper bound of ( B ). So,[ a - 1 geq 1 ]Solving for ( a ):[ a geq 2 ]So, from these two cases, it seems that ( a ) must be either less than or equal to ( -frac{1}{2} ) or greater than or equal to 2 for the intersection ( A cap B ) to be empty.But wait, I should also consider if ( A ) could be an empty set. If ( A ) is empty, then naturally, its intersection with ( B ) would also be empty. When is ( A ) empty? That happens when the lower bound of ( A ) is greater than or equal to the upper bound of ( A ). So,[ a - 1 geq 2a + 1 ]Solving for ( a ):[ a - 1 geq 2a + 1 ][ -1 - 1 geq 2a - a ][ -2 geq a ][ a leq -2 ]So, if ( a leq -2 ), set ( A ) is empty, which also satisfies ( A cap B = varnothing ). Now, combining all these results:- When ( a leq -2 ), ( A ) is empty.- When ( -2 < a leq -frac{1}{2} ), ( A ) is not empty but lies entirely to the left of ( B ).- When ( a geq 2 ), ( A ) lies entirely to the right of ( B ).Therefore, the complete range of ( a ) is ( a leq -frac{1}{2} ) or ( a geq 2 ).Let me double-check to ensure there are no overlaps or missed cases.If ( a ) is between ( -frac{1}{2} ) and 2, does ( A ) overlap with ( B )?Let's pick ( a = 0 ). Then, ( A = {x | -1 < x < 1} ). Clearly, this overlaps with ( B = {x | 0 < x < 1} ).Another test: ( a = 1 ). Then, ( A = {x | 0 < x < 3} ). This also overlaps with ( B ).And if ( a = -1 ), which is between ( -2 ) and ( -frac{1}{2} ), then ( A = {x | -2 < x < -1} ), which doesn't overlap with ( B ).Similarly, if ( a = -3 ), ( A = {x | -4 < x < -5} ), which is empty because the lower bound is greater than the upper bound. So, it doesn't overlap with ( B ).Therefore, my earlier conclusion seems correct.**Final Answer**The range of ( a ) is boxed{a leq -dfrac{1}{2} text{ or } a geq 2}.

💡First, I need to determine the greatest possible percentage of major airline companies that offer both wireless internet and free on-board snacks.I know that 35% of airlines provide wireless internet access, and 70% offer free snacks. To find the maximum overlap between these two groups, I can use the principle that the greatest possible percentage of both services cannot exceed the smaller of the two percentages.In this case, 35% is smaller than 70%. Therefore, the greatest possible percentage of airlines that offer both wireless internet and free snacks is 35%.This means that if all airlines with wireless internet also provide free snacks, the maximum overlap is 35%.

💡Alright, so I have this problem where I need to prove that a sequence (a_1, a_2, dots) of nonnegative integers is eventually constant, given two conditions. Let me try to unpack this step by step.First, the problem states that for every positive integer (n), the average of (2^{a_1}, 2^{a_2}, dots, 2^{a_n}) is an integer. That is, (frac{2^{a_1} + 2^{a_2} + cdots + 2^{a_n}}{n}) is an integer. This seems like a strong condition because it's not just that the sum is divisible by (n), but the average itself is an integer. The second condition is that (2^{a_n} leq cn) for some positive real number (c). So, each term (2^{a_n}) is bounded linearly by (n). This suggests that the exponents (a_n) can't grow too quickly; otherwise, (2^{a_n}) would exceed the linear bound.My goal is to show that after some point, all the (a_n) are the same. That is, the sequence becomes constant. So, I need to show that there exists some (N) such that for all (n geq N), (a_n = k) for some fixed integer (k).Let me think about how to approach this. Maybe I can analyze the behavior of the sequence (b_n = frac{2^{a_1} + 2^{a_2} + cdots + 2^{a_n}}{n}). Since (b_n) is an integer, it must be that the sum (2^{a_1} + cdots + 2^{a_n}) is divisible by (n). Also, since (2^{a_n} leq cn), the sum (2^{a_1} + cdots + 2^{a_n}) is at most (cn^2). Therefore, (b_n leq cn), which tells me that (b_n) grows at most linearly.Now, let's consider the difference (b_{n+1} - b_n). This difference is equal to (frac{2^{a_{n+1}}}{n+1} + frac{2^{a_1} + cdots + 2^{a_n}}{n+1} - frac{2^{a_1} + cdots + 2^{a_n}}{n}). Simplifying this, I get:[b_{n+1} - b_n = frac{2^{a_{n+1}}}{n+1} - frac{2^{a_1} + cdots + 2^{a_n}}{n(n+1)}]But since (2^{a_{n+1}} leq c(n+1)), the first term is at most (c). The second term involves the sum divided by (n(n+1)), which is small because the sum is at most (cn^2), so the second term is at most (frac{cn^2}{n(n+1)} = frac{cn}{n+1} leq c). Therefore, the difference (b_{n+1} - b_n) is bounded by some constant, say (2c).This suggests that the sequence (b_n) doesn't change too rapidly. If (b_n) is changing by a bounded amount each time, and it's growing at most linearly, maybe it stabilizes after some point.But how does this relate to the sequence (a_n)? Well, if (b_n) stabilizes, that would mean the average of the (2^{a_i})s is becoming constant. If the average is constant, then adding another term (2^{a_{n+1}}) must not change the average, which would imply (2^{a_{n+1}} = b_n). So, if (b_n) is constant, then (2^{a_{n+1}}) is equal to that constant, meaning (a_{n+1}) is the same for all (n) beyond that point.But wait, (b_n) might not necessarily become constant. It could oscillate or something. However, since (b_n) is an integer and changes by a bounded amount each time, it can't oscillate indefinitely without eventually stabilizing. Because if it changed direction infinitely often, the differences would accumulate, but since the differences are bounded, that can't happen.So, maybe (b_n) eventually becomes constant. If that's the case, then as I thought earlier, (2^{a_{n}} = b_n) for all (n) beyond some point, which would make (a_n) constant.But I need to make this argument more precise. Let me try to formalize it.Suppose that (b_n) is not eventually constant. Then, there are infinitely many (n) where (b_{n+1} neq b_n). But since (b_n) is an integer and the differences (b_{n+1} - b_n) are bounded, say by (D), then (b_n) can only increase or decrease by at most (D) each time. If (b_n) increases infinitely often, then it would tend to infinity, but since (b_n leq cn), it can't grow faster than linearly. Similarly, if it decreases infinitely often, it could potentially go to negative infinity, but since (b_n) is the average of positive terms, it can't be negative. Therefore, (b_n) must be bounded below and can't decrease indefinitely.This suggests that (b_n) must eventually stabilize, meaning (b_{n+1} = b_n) for all sufficiently large (n). Therefore, (2^{a_{n+1}} = b_n) for all (n) beyond some point, which would make (a_{n+1}) constant.Wait, but (b_n) is the average up to (n), so if (b_n) stabilizes, then adding (2^{a_{n+1}}) must keep the average the same. That is:[frac{S_n + 2^{a_{n+1}}}{n+1} = frac{S_n}{n}]Where (S_n = 2^{a_1} + cdots + 2^{a_n}). Multiplying both sides by (n(n+1)), we get:[n(S_n + 2^{a_{n+1}}) = (n+1)S_n]Simplifying:[n2^{a_{n+1}} = S_n]But (S_n = n b_n), so:[n2^{a_{n+1}} = n b_n implies 2^{a_{n+1}} = b_n]Therefore, if (b_n) stabilizes, then (2^{a_{n+1}} = b_n), which is a constant. Hence, (a_{n+1}) must be constant for all (n) beyond that point.So, putting it all together, since (b_n) is an integer sequence that changes by a bounded amount each time and is bounded above by (cn), it must eventually stabilize. Once it stabilizes, the corresponding (a_n) must be constant.I think this makes sense. The key points are:1. (b_n) is an integer and changes by a bounded amount each time.2. (b_n) is bounded above by (cn), so it can't grow without bound.3. Therefore, (b_n) must eventually become constant.4. Once (b_n) is constant, (2^{a_n}) must equal that constant, making (a_n) constant.I should check if there are any gaps in this reasoning. For example, could (b_n) oscillate without stabilizing? But since the differences are bounded and (b_n) is bounded below, it can't oscillate indefinitely. It must eventually reach a point where it doesn't change anymore.Another thing to consider is whether (2^{a_n}) being bounded by (cn) affects the stabilization of (b_n). Since (2^{a_n}) can't be too large, it prevents (b_n) from growing too quickly, reinforcing the idea that (b_n) must stabilize.So, I think this line of reasoning is solid. The sequence (a_n) must eventually become constant because the average (b_n) stabilizes, forcing (2^{a_n}) to be constant.

💡To determine the probability of getting exactly three heads in eight tosses of a fair coin, I start by calculating the total number of possible outcomes. Since each toss has two possible results (Heads or Tails), the total number of sequences is (2^8 = 256).Next, I need to find the number of favorable outcomes, which are the sequences with exactly three heads. This can be determined using the combination formula (binom{8}{3}), which calculates the number of ways to choose three positions out of eight for the heads. Calculating the combination:[binom{8}{3} = frac{8 times 7 times 6}{3 times 2 times 1} = 56]Finally, the probability is the ratio of favorable outcomes to the total number of outcomes:[text{Probability} = frac{56}{256} = frac{7}{32}]Therefore, the probability of obtaining exactly three heads in eight tosses is (boxed{frac{7}{32}}).