💡First, I recognize that the random variable ( X ) can take the values ( 1, 2, 3, ldots, n ).Since the problem states that ( P(X < 4) = 0.3 ), this means the probability of ( X ) being 1, 2, or 3 is 0.3.Assuming each value of ( X ) has an equal probability, the probability of ( X = k ) is ( frac{1}{n} ).Therefore, the probability ( P(X < 4) ) is the sum of the probabilities for ( X = 1 ), ( X = 2 ), and ( X = 3 ), which is ( frac{3}{n} ).Setting up the equation:[frac{3}{n} = 0.3]Solving for ( n ):[n = frac{3}{0.3} = 10]Thus, the value of ( n ) is 10.

💡First, I need to find the complex number ( z ) that satisfies the equation ( z(1+i) = 3+i ).To isolate ( z ), I will divide both sides of the equation by ( 1+i ):[z = frac{3+i}{1+i}]Next, I'll simplify this expression by multiplying the numerator and the denominator by the conjugate of the denominator, which is ( 1-i ):[z = frac{(3+i)(1-i)}{(1+i)(1-i)} = frac{4-2i}{2} = 2 - i]Now that I have ( z = 2 - i ), I can calculate its magnitude using the formula for the modulus of a complex number:[|z| = sqrt{2^2 + (-1)^2} = sqrt{4 + 1} = sqrt{5}]Therefore, the magnitude of ( z ) is ( sqrt{5} ).

💡Alright, so I've got this geometry problem here, and I need to find the value of ( x ). The question gives me a diagram with some angles marked, and I need to figure out what ( x ) is. The options are (A) 20, (B) 80, (C) 100, (D) 120, and (E) 60. First, I need to recall some basic geometry principles that might help me here. Since the problem mentions a diagram with angles, I'm thinking it's probably a triangle or some polygon where the sum of angles is involved. I remember that in a triangle, the sum of the interior angles is always 180 degrees. That seems like a useful piece of information.Looking at the options, they're all pretty standard angle measures, so it's likely that ( x ) is part of a triangle where the other angles are known or can be deduced. Maybe the diagram shows a triangle with two known angles and ( x ) as the third angle. If that's the case, I can use the triangle angle sum theorem to find ( x ).Let me try to visualize the diagram. If there's a triangle with angles 40 degrees, 60 degrees, and ( x ), then ( x ) would be 80 degrees because 40 + 60 + 80 equals 180. That seems straightforward, but I need to make sure I'm not missing anything.Wait, could there be more to it? Maybe the diagram isn't just a simple triangle. Perhaps it's a more complex figure with intersecting lines or multiple triangles. If that's the case, I might need to use other properties like the exterior angle theorem or properties of parallel lines.But since the options are relatively simple, and the problem is likely designed for someone at a basic geometry level, I think it's safe to assume it's a straightforward triangle problem. So, going back to the triangle angle sum theorem, if two angles are 40 and 60, then the third angle ( x ) must be 80 degrees.Let me double-check my reasoning. If I add 40 and 60, that's 100. Subtracting that from 180 gives me 80. That seems correct. I don't see any reason to doubt this conclusion unless there's additional information in the diagram that I'm not accounting for.Maybe the diagram has some parallel lines, and I need to use the concept of corresponding angles or alternate interior angles. If that's the case, I might need to find some other angles first before I can determine ( x ). But again, without seeing the actual diagram, it's hard to be certain.However, given the options and the typical structure of such problems, I'm going to stick with my initial thought. The value of ( x ) is 80 degrees. It fits neatly into the triangle angle sum theorem, and the other options don't seem to make as much sense in this context.Just to be thorough, let's consider if ( x ) could be something else. If ( x ) were 20 degrees, then the other two angles would have to add up to 160 degrees, which doesn't align with the given angles of 40 and 60. Similarly, 100 degrees would mean the other angles add up to 80, which again doesn't fit. 120 and 60 degrees also don't seem to work out when considering the sum of angles in a triangle.Therefore, after carefully considering the possibilities and verifying my calculations, I'm confident that the correct answer is 80 degrees.

💡Alright, so I have this problem where Hen Hao randomly selects two distinct squares on a standard 8x8 chessboard. The condition is that the two squares touch each other, either at a side or a vertex. I need to find the probability that these two squares are the same color. The answer should be in the form of a fraction m/n where m and n are coprime, and then compute 100m + n.First, let me visualize the chessboard. It's an 8x8 grid, so there are 64 squares in total. Each square is colored alternately black and white, right? So, no two adjacent squares (sharing a side) are the same color. But wait, the problem says the squares can touch at a vertex or a side. So, that means they can be diagonally adjacent as well.Hmm, so if two squares are touching at a side, they are definitely different colors. But if they touch at a vertex, meaning they are diagonally adjacent, are they the same color or different? Let me think. On a standard chessboard, diagonally adjacent squares are actually the same color. For example, if you have a white square at (1,1), then the square at (2,2) is also white. So, diagonally adjacent squares are the same color.Wait, so if two squares touch at a side, they are different colors, but if they touch at a vertex, they are the same color. Interesting. So, the problem is asking for the probability that two touching squares are the same color, given that they touch. So, I need to consider both cases: touching at a side and touching at a vertex.But first, let me figure out how many pairs of squares touch each other on the chessboard. There are two types of touching: sharing a side (which are adjacent horizontally or vertically) and sharing a vertex (which are adjacent diagonally). I need to calculate the total number of such pairs.Starting with squares that share a side. In each row of 8 squares, there are 7 horizontal adjacent pairs. Since there are 8 rows, the total number of horizontal adjacent pairs is 8 * 7 = 56. Similarly, for vertical adjacent pairs, each column has 7 vertical adjacent pairs, and there are 8 columns, so 8 * 7 = 56. So, the total number of squares sharing a side is 56 + 56 = 112.Now, for squares that share a vertex, which are diagonally adjacent. Let's think about how many such pairs there are. For each square that's not on the edge, it can have four diagonal neighbors. But to count the number of diagonal pairs, let's consider that each 2x2 block of squares has two diagonal pairs. For example, in a 2x2 block, the top-left and bottom-right form one diagonal pair, and the top-right and bottom-left form another.Since the chessboard is 8x8, the number of 2x2 blocks is (8 - 1) * (8 - 1) = 7 * 7 = 49. Each 2x2 block contributes two diagonal pairs, so the total number of diagonal pairs is 49 * 2 = 98.So, the total number of pairs of squares that touch each other (either at a side or a vertex) is 112 (side) + 98 (vertex) = 210.Now, out of these 210 pairs, how many of them are the same color? As I thought earlier, squares that share a side are different colors, so none of the 112 side-sharing pairs are the same color. However, squares that share a vertex (diagonal pairs) are the same color. So, all 98 vertex-sharing pairs are same-color pairs.Therefore, the number of favorable outcomes (same color) is 98, and the total number of possible outcomes (touching pairs) is 210. So, the probability is 98/210.Let me simplify this fraction. Both numerator and denominator are divisible by 14. 98 ÷ 14 = 7, and 210 ÷ 14 = 15. So, the simplified fraction is 7/15.Wait, but let me double-check my counts. For the side-sharing pairs, 56 horizontal and 56 vertical, that's 112. For the diagonal pairs, 49 blocks each contributing 2 pairs, that's 98. So, total 210. That seems correct.And since side-sharing pairs are different colors, and diagonal pairs are same colors, so 98 same-color pairs out of 210 total touching pairs. So, 98/210 reduces to 7/15. Yes, that seems right.So, the probability is 7/15. Therefore, m is 7 and n is 15. They are coprime since 7 is prime and doesn't divide 15. So, 100m + n is 100*7 + 15 = 700 + 15 = 715.Wait, hold on, I just realized something. When I counted the diagonal pairs, I considered each 2x2 block contributing two pairs. But does that overcount? Let me think. Each diagonal pair is unique to a 2x2 block. For example, the pair (1,1) and (2,2) is only in the 2x2 block starting at (1,1). Similarly, (1,2) and (2,1) is in the 2x2 block starting at (1,1). So, each diagonal pair is counted once per 2x2 block, and since each diagonal pair is only in one 2x2 block, the count should be correct.Alternatively, another way to count diagonal pairs is: for each square, count the number of diagonal neighbors it has, then divide by 2 to avoid double-counting. Let's see. On the chessboard, corner squares have 2 diagonal neighbors, edge squares (but not corners) have 3 diagonal neighbors, and inner squares have 4 diagonal neighbors.But maybe that's more complicated. Alternatively, since each diagonal pair is determined by their top-left square, so for each square except the last row and last column, there is a diagonal pair to the bottom-right. Similarly, for each square except the last row and first column, there is a diagonal pair to the bottom-left. So, the number of diagonal pairs is 7*7 (for bottom-right diagonals) plus 7*7 (for bottom-left diagonals), which is 49 + 49 = 98. So, that's consistent with my earlier count.Therefore, my count of 98 diagonal pairs is correct. So, the probability is indeed 98/210, which simplifies to 7/15.Therefore, the final answer is 715.

💡Okay, so I need to find the domain of the function ( g(x) = frac{f(2x)}{x - 1} ), given that the domain of ( f(x) = sqrt{x(2 - x)} ) is ([0, 2]). Hmm, let's break this down step by step.First, I know that the domain of a function is the set of all possible input values (x-values) for which the function is defined. For ( f(x) ), it's given as ([0, 2]). That means ( f(x) ) is only defined when ( x ) is between 0 and 2, inclusive. Now, looking at ( g(x) ), it's defined as ( frac{f(2x)}{x - 1} ). So, ( g(x) ) depends on ( f(2x) ) and also has a denominator ( x - 1 ). I need to consider both parts when determining the domain of ( g(x) ).Let me start with the ( f(2x) ) part. Since ( f(x) ) is defined for ( x ) in ([0, 2]), then ( 2x ) must also lie within ([0, 2]) for ( f(2x) ) to be defined. So, I can set up the inequality:[ 0 leq 2x leq 2 ]To solve for ( x ), I'll divide all parts of the inequality by 2:[ 0 leq x leq 1 ]So, ( x ) must be between 0 and 1, inclusive, for ( f(2x) ) to be defined.Next, I need to consider the denominator ( x - 1 ). A denominator cannot be zero because division by zero is undefined. So, I need to find the value of ( x ) that makes ( x - 1 = 0 ):[ x - 1 = 0 ][ x = 1 ]This means ( x = 1 ) is not allowed in the domain of ( g(x) ).Putting these two conditions together, ( x ) must satisfy both ( 0 leq x leq 1 ) and ( x neq 1 ). So, combining these, the domain of ( g(x) ) is all real numbers from 0 to 1, including 0 but excluding 1.Therefore, the domain of ( g(x) ) is ([0, 1)).Wait, let me double-check to make sure I didn't miss anything. The function ( f(2x) ) requires ( 2x ) to be in ([0, 2]), which gives ( x ) in ([0, 1]). Then, the denominator ( x - 1 ) can't be zero, so ( x ) can't be 1. So, combining these, it's indeed ([0, 1)). Yeah, that seems right.I don't think I missed any other restrictions. The numerator ( f(2x) ) only imposes the condition on ( x ) to be between 0 and 1, and the denominator just excludes 1. So, the domain should be all numbers from 0 up to but not including 1.Just to visualize, if I were to graph ( g(x) ), it would be defined starting at ( x = 0 ) and going up to, but not including, ( x = 1 ). At ( x = 1 ), there would be a vertical asymptote because the denominator becomes zero, making the function undefined there.So, yeah, I'm confident that the domain of ( g(x) ) is ([0, 1)).**Final Answer**The domain of the function ( g(x) ) is boxed{[0, 1)}.

💡Okay, so I have this problem where I need to prove that if ( a ) is an even number, then three different expressions involving ( a ) are divisible by 8. The expressions are ( a(a^2 + 20) ), ( a(a^2 - 20) ), and ( a(a^2 - 4) ). Hmm, let me try to figure this out step by step.First, since ( a ) is even, I know that it can be written as ( a = 2k ) where ( k ) is an integer. That makes sense because any even number is just twice some integer. So, substituting ( a = 2k ) into each of these expressions might help simplify things and see if they are divisible by 8.Let's start with the first expression: ( a(a^2 + 20) ). If I substitute ( a = 2k ), it becomes ( 2k((2k)^2 + 20) ). Let me compute that step by step. First, ( (2k)^2 ) is ( 4k^2 ), so the expression inside the parentheses becomes ( 4k^2 + 20 ). Then, the whole expression is ( 2k(4k^2 + 20) ). Now, I can factor out a 4 from the terms inside the parentheses: ( 4(k^2 + 5) ). So, the expression becomes ( 2k times 4(k^2 + 5) ), which simplifies to ( 8k(k^2 + 5) ). Ah, okay, so this expression is clearly a multiple of 8 because it's 8 times some integer ( k(k^2 + 5) ). So, ( a(a^2 + 20) ) is divisible by 8.Alright, moving on to the second expression: ( a(a^2 - 20) ). Again, substituting ( a = 2k ), it becomes ( 2k((2k)^2 - 20) ). Let's compute that. ( (2k)^2 ) is still ( 4k^2 ), so the expression inside the parentheses is ( 4k^2 - 20 ). Then, the whole expression is ( 2k(4k^2 - 20) ).Factoring out a 4 from inside the parentheses gives ( 4(k^2 - 5) ). So, the expression becomes ( 2k times 4(k^2 - 5) ), which simplifies to ( 8k(k^2 - 5) ). Just like before, this is 8 times some integer ( k(k^2 - 5) ), so ( a(a^2 - 20) ) is also divisible by 8.Now, the third expression: ( a(a^2 - 4) ). Substituting ( a = 2k ) gives ( 2k((2k)^2 - 4) ). Calculating inside the parentheses, ( (2k)^2 ) is ( 4k^2 ), so it becomes ( 4k^2 - 4 ). Therefore, the entire expression is ( 2k(4k^2 - 4) ).Factoring out a 4 from inside the parentheses gives ( 4(k^2 - 1) ). So, the expression becomes ( 2k times 4(k^2 - 1) ), which simplifies to ( 8k(k^2 - 1) ). Again, this is 8 times some integer ( k(k^2 - 1) ), meaning ( a(a^2 - 4) ) is divisible by 8.Wait a second, all three expressions simplified to 8 times something, which means they are all divisible by 8. That seems consistent. But let me double-check to make sure I didn't make any mistakes in my substitutions or factoring.For the first expression: ( a(a^2 + 20) = 2k(4k^2 + 20) = 8k(k^2 + 5) ). Yep, that looks right.Second expression: ( a(a^2 - 20) = 2k(4k^2 - 20) = 8k(k^2 - 5) ). Also correct.Third expression: ( a(a^2 - 4) = 2k(4k^2 - 4) = 8k(k^2 - 1) ). Yep, that's accurate too.I think I've covered all the steps correctly. Each time, by expressing ( a ) as ( 2k ), I was able to factor out an 8, showing that each expression is indeed divisible by 8. So, I feel confident that this proof is solid.Just to be thorough, maybe I should test with a specific even number to see if it holds. Let's pick ( a = 2 ).First expression: ( 2(2^2 + 20) = 2(4 + 20) = 2(24) = 48 ). 48 divided by 8 is 6, so yes, divisible by 8.Second expression: ( 2(2^2 - 20) = 2(4 - 20) = 2(-16) = -32 ). -32 divided by 8 is -4, so also divisible by 8.Third expression: ( 2(2^2 - 4) = 2(4 - 4) = 2(0) = 0 ). Zero is divisible by any number, including 8.Okay, that worked. Let's try another even number, say ( a = 4 ).First expression: ( 4(4^2 + 20) = 4(16 + 20) = 4(36) = 144 ). 144 divided by 8 is 18, so yes.Second expression: ( 4(4^2 - 20) = 4(16 - 20) = 4(-4) = -16 ). -16 divided by 8 is -2, so yes.Third expression: ( 4(4^2 - 4) = 4(16 - 4) = 4(12) = 48 ). 48 divided by 8 is 6, so yes.Alright, that seems to confirm it. Maybe one more test with a larger even number, like ( a = 6 ).First expression: ( 6(6^2 + 20) = 6(36 + 20) = 6(56) = 336 ). 336 divided by 8 is 42, so yes.Second expression: ( 6(6^2 - 20) = 6(36 - 20) = 6(16) = 96 ). 96 divided by 8 is 12, so yes.Third expression: ( 6(6^2 - 4) = 6(36 - 4) = 6(32) = 192 ). 192 divided by 8 is 24, so yes.Okay, so all these tests are confirming that the expressions are indeed divisible by 8 when ( a ) is even. That gives me more confidence in my proof.I think I've covered all the necessary steps and even tested with specific examples to ensure the proof holds. So, I can conclude that if ( a ) is an even number, then ( a(a^2 + 20) ), ( a(a^2 - 20) ), and ( a(a^2 - 4) ) are all divisible by 8.