💡Okay, so I have this problem here: "How many sets of two or more consecutive positive integers have a sum of 21?" And the options are A) 0, B) 1, C) 2, D) 3, E) 4. Hmm, I need to figure out how many such sets exist. Let me think about how to approach this.First, I remember that consecutive positive integers can be represented as a sequence starting from some number 'a' and having 'n' terms. The sum of such a sequence can be calculated using the formula for the sum of an arithmetic series. The formula is:[ S = frac{n}{2} times (2a + (n - 1)) ]Here, 'S' is the sum, which is 21 in this case, 'n' is the number of terms, and 'a' is the first term. So, plugging in the values we have:[ 21 = frac{n}{2} times (2a + n - 1) ]To simplify this, I can multiply both sides by 2 to eliminate the fraction:[ 42 = n times (2a + n - 1) ]Now, this equation relates 'n' and 'a'. Since both 'n' and 'a' are positive integers, I can try to find pairs (n, a) that satisfy this equation. My goal is to find all possible values of 'n' (where n ≥ 2) such that 'a' is also a positive integer.Let me rearrange the equation to solve for 'a':[ 2a + n - 1 = frac{42}{n} ][ 2a = frac{42}{n} - n + 1 ][ a = frac{frac{42}{n} - n + 1}{2} ]For 'a' to be a positive integer, the expression (frac{42}{n} - n + 1) must be even, and the result must be positive. So, I need to find all divisors 'n' of 42 (since 42/n must be an integer) such that when I plug them into the equation for 'a', I get a positive integer.Let's list the positive divisors of 42:1, 2, 3, 6, 7, 14, 21, 42.But since we need sets of two or more consecutive numbers, 'n' must be at least 2. So, the possible values for 'n' are 2, 3, 6, 7, 14, 21, 42.Now, let's test each of these values of 'n' to see if 'a' is a positive integer.1. **n = 2**:[ a = frac{frac{42}{2} - 2 + 1}{2} = frac{21 - 2 + 1}{2} = frac{20}{2} = 10 ]So, a = 10. That means the sequence is 10, 11. Let's check the sum: 10 + 11 = 21. That works. So, one valid set.2. **n = 3**:[ a = frac{frac{42}{3} - 3 + 1}{2} = frac{14 - 3 + 1}{2} = frac{12}{2} = 6 ]So, a = 6. The sequence is 6, 7, 8. Sum: 6 + 7 + 8 = 21. That works too. Second valid set.3. **n = 6**:[ a = frac{frac{42}{6} - 6 + 1}{2} = frac{7 - 6 + 1}{2} = frac{2}{2} = 1 ]So, a = 1. The sequence is 1, 2, 3, 4, 5, 6. Sum: 1+2+3+4+5+6 = 21. That works. Third valid set.4. **n = 7**:[ a = frac{frac{42}{7} - 7 + 1}{2} = frac{6 - 7 + 1}{2} = frac{0}{2} = 0 ]But 'a' has to be a positive integer, and 0 is not positive. So, this doesn't work.5. **n = 14**:[ a = frac{frac{42}{14} - 14 + 1}{2} = frac{3 - 14 + 1}{2} = frac{-10}{2} = -5 ]Negative number, which isn't allowed since we're dealing with positive integers. So, discard this.6. **n = 21**:[ a = frac{frac{42}{21} - 21 + 1}{2} = frac{2 - 21 + 1}{2} = frac{-18}{2} = -9 ]Again, negative. Not valid.7. **n = 42**:[ a = frac{frac{42}{42} - 42 + 1}{2} = frac{1 - 42 + 1}{2} = frac{-40}{2} = -20 ]Also negative. Not valid.So, from all the possible values of 'n', only n = 2, 3, and 6 give us positive integer values for 'a'. Therefore, there are three sets of two or more consecutive positive integers that sum to 21.But wait, let me double-check to make sure I didn't miss anything. Maybe there's another way to approach this problem.Another method I recall is that the number of ways to express a number as the sum of consecutive integers is equal to the number of odd divisors of that number greater than 1. Let me see if that applies here.First, factorize 21. 21 = 3 × 7. So, the odd divisors are 1, 3, 7, and 21. But since we need sets of two or more numbers, we exclude 1. So, the number of ways should be 3 (divisors 3, 7, 21). Hmm, that seems to match the earlier result where n = 2, 3, 6. Wait, but 6 is even. Maybe I'm mixing something up.Wait, no, the number of odd divisors corresponds to the number of ways. So, since 21 has three odd divisors greater than 1 (3, 7, 21), that would mean three ways. But in my earlier calculation, I found three sets: n=2,3,6. So, that seems consistent.But hold on, when n=6, which is even, we still got a valid set. So, maybe the number of ways isn't just the number of odd divisors, but something else. Maybe it's the number of ways to factor 2S into two integers of opposite parity.Wait, let me think again. The formula I used earlier was:[ 42 = n times (2a + n - 1) ]So, 42 can be factored into two integers, n and (2a + n -1), where n is the number of terms and (2a + n -1) is the average multiplied by 2. Since n and (2a + n -1) must be of opposite parity (one even, one odd) because their product is 42, which is even.So, the number of ways to express 21 as the sum of consecutive integers is equal to the number of ways to factor 42 into two integers of opposite parity, where n ≥ 2.So, let's list all factor pairs of 42:1 × 422 × 213 × 146 × 7Now, check which pairs have opposite parity:1 × 42: 1 is odd, 42 is even – opposite parity.2 × 21: 2 is even, 21 is odd – opposite parity.3 × 14: 3 is odd, 14 is even – opposite parity.6 × 7: 6 is even, 7 is odd – opposite parity.So, all four factor pairs have opposite parity. But wait, earlier I found only three valid sets. Why is that?Because when n=42, a would be negative, which isn't allowed. Similarly, when n=21, a is negative. So, even though the factor pairs exist, some of them result in negative starting numbers, which we can't have.So, out of the four factor pairs, only three result in positive integers for 'a'. Therefore, there are three valid sets.Wait, but in my initial calculation, I considered n=2,3,6 as valid, which correspond to the factor pairs (2,21), (3,14), (6,7). The factor pair (1,42) would correspond to n=1, which is not allowed since we need two or more numbers. So, that's why only three sets are valid.Therefore, the answer should be D) 3.But hold on, in my first approach, I found three sets, but when I thought about the number of odd divisors, I also got three. So, that seems consistent.Let me just list out all three sets to make sure:1. n=2: Starting at a=10: 10 + 11 = 212. n=3: Starting at a=6: 6 + 7 + 8 = 213. n=6: Starting at a=1: 1 + 2 + 3 + 4 + 5 + 6 = 21Yes, these all add up to 21. So, that's three sets.Wait, but earlier I thought about n=7 giving a=0, which is invalid, and n=14,21,42 giving negative 'a's. So, indeed, only three valid sets.Therefore, the correct answer is D) 3.But wait, in the initial problem, the options are A)0, B)1, C)2, D)3, E)4. So, D)3 is an option. But in the first assistant's answer, they concluded it was 1, which seems incorrect. Maybe they made a mistake in their calculation.Let me check their thought process. They started with the formula, set up the equation, and then tried different 'a's. They considered a=1 and a=5, but it seems they might have miscalculated somewhere because they ended up with only one valid set. But from my calculations, there are three valid sets.Alternatively, maybe they considered only certain factor pairs or made an error in determining 'a'. It's possible they overlooked some factor pairs or misapplied the formula.In any case, based on my thorough calculation, there are three sets of two or more consecutive positive integers that sum to 21. Therefore, the correct answer should be D)3.

💡Okay, so I need to figure out how many vertices a regular octahedron has. Hmm, I remember that an octahedron is one of the Platonic solids, but I'm not exactly sure about its properties. Let me try to recall. I think it's a polyhedron with triangular faces. Wait, isn't it the one that looks like two pyramids stuck together at their bases? Yeah, that sounds right. So, if it's two pyramids glued together, each pyramid has a base, and the other faces are triangles.Now, each pyramid has a base that's a polygon. Since it's a regular octahedron, all the faces should be the same, right? So, the base of each pyramid must be a regular polygon. If it's a regular octahedron, I think the base is a square. So, each pyramid has a square base and four triangular faces. When you glue two of these pyramids together at their bases, the resulting shape should have those triangular faces all around.Let me visualize this. If I have one pyramid with a square base, it has four triangular faces and one square face. Gluing another pyramid to the base would cover that square face, and now the combined shape would have eight triangular faces in total. That makes sense because "octa" means eight, so an octahedron has eight faces.Now, how does this relate to the number of vertices? Each pyramid has a base with four vertices and one apex. So, each pyramid has five vertices. But when you glue them together, the base vertices are shared. So, the combined shape should have the four base vertices plus the two apexes from each pyramid. That would be four plus two, which is six vertices in total.Wait, let me double-check that. If each pyramid has four base vertices and one apex, and we're gluing them together at the base, then the four base vertices are common to both pyramids. So, we don't count them twice. That means we have four base vertices plus two apexes, making six vertices altogether.Is there another way to think about this? Maybe using Euler's formula? I remember that for any convex polyhedron, the number of vertices minus the number of edges plus the number of faces equals two. So, Euler's formula is V - E + F = 2. If I can figure out the number of edges and faces, I can solve for the number of vertices.I already thought that a regular octahedron has eight faces, all of which are equilateral triangles. So, F = 8. Each triangular face has three edges, but each edge is shared by two faces. So, the total number of edges should be (8 faces * 3 edges per face) / 2 = 12 edges. So, E = 12.Now, plugging into Euler's formula: V - 12 + 8 = 2. Simplifying that, V - 4 = 2, so V = 6. That confirms it. There are six vertices in a regular octahedron.Just to make sure I'm not missing anything, let me think about the structure again. Each vertex is where four edges meet. So, if there are six vertices and each has four edges, that would suggest 6 * 4 = 24 edge ends. But since each edge connects two vertices, the total number of edges should be 24 / 2 = 12, which matches what I found earlier. So, everything checks out.Another way to visualize it is to think about the octahedron in terms of its dual polyhedron. The dual of an octahedron is a cube. In dual polyhedrons, the number of vertices of one corresponds to the number of faces of the other. A cube has six faces, so its dual, the octahedron, should have six vertices. That's another confirmation.I think I'm confident now that a regular octahedron has six vertices.

💡Okay, so I have this geometry problem here about a sphere and a cone. The sphere is constructed using the height of the cone as its diameter. That means the sphere's diameter is equal to the cone's height. Interesting. The problem states that the surface area of the part of the sphere lying inside the cone is equal to the surface area of the part of the cone lying inside the sphere. I need to find the angle in the axial section of the cone. Hmm, axial section would be the triangle formed by cutting the cone along its axis, right? So, the angle at the vertex of that triangle is what I'm looking for.Let me try to visualize this. Imagine a cone standing upright, and a sphere whose diameter is exactly the height of the cone. So, the sphere is perfectly fitted along the cone's height. Now, the sphere will intersect the cone somewhere, creating a circle where they meet. The part of the sphere inside the cone will be like a spherical cap, and the part of the cone inside the sphere will be a smaller cone or maybe a frustum? Wait, no, since the sphere is inside the cone, it's actually a smaller cone.The problem says that the surface areas of these two parts are equal. So, the surface area of the spherical cap (sphere inside the cone) is equal to the lateral surface area of the cone inside the sphere. Got it.Let me denote some variables. Let’s say the height of the cone is h, so the sphere has a diameter h, meaning the radius of the sphere is R = h/2. Let’s denote the angle at the vertex of the cone as 2α, so each side of the triangle in the axial section makes an angle α with the axis.I think I need to find expressions for both surface areas in terms of α and set them equal. Then solve for α.First, let's think about the sphere inside the cone. The part of the sphere inside the cone is a spherical cap. The surface area of a spherical cap is given by 2πRh, where h is the height of the cap. Wait, but in this case, the height of the cap isn't the same as the sphere's radius. I need to figure out the height of the cap.To find the height of the spherical cap, I need to find where the sphere intersects the cone. Let me consider the axial section, which is a triangle with height h and base angle α. The sphere is centered at the midpoint of the height, so its center is at h/2. The sphere has radius h/2.Let me set up a coordinate system with the apex of the cone at the origin (0,0) and the axis along the positive y-axis. So, the base of the cone is at (0, h). The sphere is centered at (0, h/2) with radius h/2.The equation of the sphere is x² + (y - h/2)² = (h/2)². Simplifying, x² + y² - h y + h²/4 = h²/4, so x² + y² - h y = 0.The equation of the cone's lateral surface in this coordinate system is a straight line. Since the cone has a base radius r at height h, the slope of the cone's side is r/h. But wait, in the axial section, the cone is represented by two lines making an angle α with the axis. So, the slope is tan α. Therefore, the equation of the cone's side is x = (tan α) y.So, to find the intersection of the sphere and the cone, substitute x = (tan α) y into the sphere's equation:(tan α)² y² + y² - h y = 0.Let me compute (tan α)² y² + y² = y² (tan² α + 1) = y² sec² α.So, the equation becomes y² sec² α - h y = 0.Factor out y: y (y sec² α - h) = 0.Solutions are y = 0 and y = h / sec² α = h cos² α.So, the sphere intersects the cone at y = 0 (the apex) and y = h cos² α. Therefore, the height of the spherical cap is from y = h cos² α to y = h/2, but wait, actually, the cap is from the intersection point up to the top of the sphere. Wait, the sphere is centered at h/2 with radius h/2, so the top of the sphere is at y = h/2 + h/2 = h. The bottom of the sphere is at y = 0.But the intersection is at y = h cos² α. So, the spherical cap inside the cone is from y = h cos² α to y = h. So, the height of the cap is h - h cos² α = h (1 - cos² α) = h sin² α.Therefore, the surface area of the spherical cap is 2πR * height of the cap. Since R = h/2, this becomes 2π*(h/2)*h sin² α = π h² sin² α.Wait, but hold on, isn't the formula for the surface area of a spherical cap 2πRh, where h is the height of the cap? So, in this case, h is h sin² α, so 2π*(h/2)*(h sin² α) = π h² sin² α. Yes, that's correct.Now, let's compute the lateral surface area of the cone inside the sphere. The part of the cone inside the sphere is a smaller cone with height from y = 0 to y = h cos² α. So, the height of this smaller cone is h cos² α. The radius of the base of this smaller cone can be found using similar triangles.The original cone has height h and base radius r. The smaller cone has height h cos² α, so its base radius is r cos² α. But wait, in the axial section, the radius is along the x-axis, so actually, in the axial section, the radius is just the x-coordinate at height y. So, at height y = h cos² α, the radius is x = (tan α) y = (tan α) h cos² α = h sin α cos α.Therefore, the lateral surface area of the smaller cone is π * (radius) * (slant height). The slant height is the length of the cone's side from the apex to the intersection point.The slant height can be found using Pythagoras: sqrt( (h cos² α)^2 + (h sin α cos α)^2 ). Let me compute that:= sqrt( h² cos⁴ α + h² sin² α cos² α )= h cos α sqrt( cos² α + sin² α )= h cos α sqrt(1)= h cos α.So, the slant height is h cos α. Therefore, the lateral surface area is π * (h sin α cos α) * (h cos α) = π h² sin α cos² α.So, now we have both surface areas:- Spherical cap: π h² sin² α- Cone inside sphere: π h² sin α cos² αAccording to the problem, these are equal:π h² sin² α = π h² sin α cos² αWe can divide both sides by π h² sin α (assuming sin α ≠ 0, which it isn't because α is an angle in a cone, so 0 < α < 90 degrees):sin α = cos² αSo, sin α = 1 - sin² αLet me write that:sin α + sin² α - 1 = 0Let me set x = sin α:x + x² - 1 = 0Which is a quadratic equation:x² + x - 1 = 0Solving for x:x = [-1 ± sqrt(1 + 4)] / 2 = [-1 ± sqrt(5)] / 2Since x = sin α must be positive (as α is between 0 and 90 degrees), we take the positive root:x = (-1 + sqrt(5)) / 2 ≈ ( -1 + 2.236 ) / 2 ≈ 1.236 / 2 ≈ 0.618So, sin α = (sqrt(5) - 1)/2 ≈ 0.618Therefore, α = arcsin( (sqrt(5) - 1)/2 )But the problem asks for the angle in the axial section, which is 2α. So, the angle is 2 arcsin( (sqrt(5) - 1)/2 )Alternatively, we can express this angle in terms of cosine. Let me compute cos(2α):cos(2α) = 1 - 2 sin² αWe have sin α = (sqrt(5) - 1)/2, so sin² α = [ (sqrt(5) - 1)^2 ] / 4 = (5 - 2 sqrt(5) + 1) / 4 = (6 - 2 sqrt(5)) / 4 = (3 - sqrt(5))/2Therefore, cos(2α) = 1 - 2*(3 - sqrt(5))/2 = 1 - (3 - sqrt(5)) = sqrt(5) - 2So, 2α = arccos( sqrt(5) - 2 )Therefore, the angle in the axial section is arccos( sqrt(5) - 2 )Let me check if this makes sense. sqrt(5) is approximately 2.236, so sqrt(5) - 2 ≈ 0.236. So, arccos(0.236) is approximately 76 degrees. That seems reasonable for a cone angle.I think that's the answer. Let me just recap to make sure I didn't make any mistakes.1. Set up the problem with the cone and sphere, noted the intersection point at y = h cos² α.2. Calculated the surface area of the spherical cap as π h² sin² α.3. Calculated the lateral surface area of the smaller cone as π h² sin α cos² α.4. Set them equal, leading to sin α = cos² α.5. Solved the equation to find sin α = (sqrt(5) - 1)/2.6. Found cos(2α) = sqrt(5) - 2, so the angle is arccos(sqrt(5) - 2).Yes, that seems correct. I think I'm confident with this solution.**Final Answer**The angle in the axial section of the cone is boxed{arccos(sqrt{5} - 2)}.

💡Okay, so I have this problem where I need to figure out which pairs of functions are actually the same function. There are four options: A, B, C, and D. I guess I need to go through each one step by step and see if they're the same function or not. Starting with Option A: f(x) = x and g(x) = x²/x. Hmm, let's see. So f(x) is just x, which is straightforward. g(x) is x squared divided by x. At first glance, it seems like g(x) simplifies to x as well because x² divided by x is x. But wait, I remember something about domains. For f(x), the domain is all real numbers because there's no restriction on x. But for g(x), since it's x squared over x, x can't be zero because you can't divide by zero. So the domain of g(x) is all real numbers except zero. That means even though g(x) simplifies to x, their domains are different. So f(x) and g(x) aren't the same function because their domains aren't the same. Got it.Moving on to Option B: f(x) = 1 and g(x) = x⁰. Okay, f(x) is just 1, so it's a constant function. The domain here is all real numbers because there's no restriction on x. Now, g(x) is x to the power of zero. I remember that any number to the power of zero is 1, but wait, is that always true? I think it's true for any x except zero because zero to the power of zero is undefined. So the domain of g(x) is all real numbers except zero. So even though g(x) equals 1 for all x except zero, f(x) is defined at x=0 as well, whereas g(x) isn't. So their domains are different, which means they aren't the same function. Okay, so B isn't the right answer either.Now, Option C: f(x) = sqrt(x² - 1) and g(x) = sqrt(x + 1) * sqrt(x - 1). Let's analyze the domains first. For f(x), the expression inside the square root, x² - 1, must be greater than or equal to zero. So x² - 1 ≥ 0, which means x² ≥ 1, so x ≥ 1 or x ≤ -1. Therefore, the domain of f(x) is (-∞, -1] ∪ [1, ∞). For g(x), we have two square roots multiplied together: sqrt(x + 1) and sqrt(x - 1). Each square root requires its own expression inside to be non-negative. So for sqrt(x + 1), x + 1 ≥ 0, which means x ≥ -1. For sqrt(x - 1), x - 1 ≥ 0, which means x ≥ 1. So the domain of g(x) is the intersection of these two conditions, which is x ≥ 1. So the domain of f(x) is (-∞, -1] ∪ [1, ∞), and the domain of g(x) is [1, ∞). These domains are different because f(x) includes all x less than or equal to -1, while g(x) doesn't. Therefore, even though algebraically f(x) and g(x) might seem similar, their domains are different, so they aren't the same function. So C isn't the right answer either.Finally, Option D: f(x) = x² - 2x - 1 and g(m) = m² - 2m - 1. Okay, so f(x) is a quadratic function in terms of x, and g(m) is the same quadratic function but in terms of m. Let's check the domains. For f(x), since it's a polynomial, the domain is all real numbers. Similarly, for g(m), it's also a polynomial, so the domain is all real numbers as well. Now, the rules of the functions: f(x) takes an input x and gives x squared minus 2x minus 1. g(m) takes an input m and gives m squared minus 2m minus 1. The only difference is the variable name, but the function rule is identical. In mathematics, the name of the variable doesn't change the function; it's just a placeholder. So f(x) and g(m) are essentially the same function because they have the same rule and the same domain. So, after going through all four options, only Option D has functions that are the same. The others have different domains or different rules, so they don't qualify as the same function.

💡Alright, so I have this problem about helmets and accidents. It's divided into two parts. Let me tackle them one by one.**Part 1: Finding the Regression Line Equation**Okay, so I need to find the regression line equation between the number of people not wearing helmets (y) and the year number (x). The data is given for four years: 2018 to 2021, with year numbers 1 to 4 respectively. The number of people not wearing helmets each year are 1250, 1050, 1000, and 900.First, I remember that the regression line equation is of the form:[hat{y} = hat{b}x + hat{a}]Where:- (hat{b}) is the slope of the regression line.- (hat{a}) is the y-intercept.To find (hat{b}) and (hat{a}), I need to calculate the means of x and y, and then use the given formula:[hat{b} = frac{sum_{i=1}^{n} (x_i - overline{x})(y_i - overline{y})}{sum_{i=1}^{n} (x_i - overline{x})^2}][hat{a} = overline{y} - hat{b}overline{x}]So, let's start by calculating the means.**Calculating the Mean of x ((overline{x})) and y ((overline{y}))**Given:- Year numbers (x): 1, 2, 3, 4- Number of people not wearing helmets (y): 1250, 1050, 1000, 900Number of data points, n = 4.[overline{x} = frac{1 + 2 + 3 + 4}{4} = frac{10}{4} = 2.5][overline{y} = frac{1250 + 1050 + 1000 + 900}{4} = frac{4200}{4} = 1050]Okay, so the mean of x is 2.5 and the mean of y is 1050.**Calculating the Numerator and Denominator for (hat{b})**Now, I need to compute the numerator and denominator for the slope (hat{b}).First, let's create a table to organize the calculations:| x | y | x - (overline{x}) | y - (overline{y}) | (x - (overline{x}))(y - (overline{y})) | (x - (overline{x}))² ||----|-----|-----------------------|-----------------------|--------------------------------------------|--------------------------|| 1 |1250 | 1 - 2.5 = -1.5 |1250 - 1050 = 200 | (-1.5)(200) = -300 | (-1.5)² = 2.25 || 2 |1050 | 2 - 2.5 = -0.5 |1050 - 1050 = 0 | (-0.5)(0) = 0 | (-0.5)² = 0.25 || 3 |1000 | 3 - 2.5 = 0.5 |1000 - 1050 = -50 | (0.5)(-50) = -25 | (0.5)² = 0.25 || 4 |900 | 4 - 2.5 = 1.5 |900 - 1050 = -150 | (1.5)(-150) = -225 | (1.5)² = 2.25 |Now, sum up the last two columns:Sum of (x - (overline{x}))(y - (overline{y})) = -300 + 0 + (-25) + (-225) = -550Sum of (x - (overline{x}))² = 2.25 + 0.25 + 0.25 + 2.25 = 5So, the numerator is -550, and the denominator is 5.Therefore,[hat{b} = frac{-550}{5} = -110]**Calculating (hat{a})**Now, using the formula for (hat{a}):[hat{a} = overline{y} - hat{b}overline{x} = 1050 - (-110)(2.5) = 1050 + 275 = 1325]So, the regression line equation is:[hat{y} = -110x + 1325]**Estimating the Number of People Not Wearing Helmets in 2022**Since 2022 is the next year after 2021, which was year number 4, 2022 would be year number 5.Plugging x = 5 into the regression equation:[hat{y} = -110(5) + 1325 = -550 + 1325 = 775]So, the estimated number of people not wearing helmets in 2022 is 775.**Part 2: Analyzing the Relationship Between Not Wearing Helmets and Accidents Resulting in Injuries**We have a contingency table:| | Not Wearing Helmet | Wearing Helmet ||---------|--------------------|----------------|| Injured | 7 | 3 || Not Injured | 13 | 27 |Total accidents: 50Total injured: 50I need to determine if there's a significant relationship between not wearing helmets and injuries in accidents at a 95% confidence level.This sounds like a chi-square test of independence. The formula for the chi-square statistic is:[chi^2 = sum frac{(O - E)^2}{E}]Where:- O is the observed frequency.- E is the expected frequency.First, let's set up the observed frequencies:| | Not Wearing Helmet | Wearing Helmet | Total ||---------|--------------------|----------------|-------|| Injured | 7 | 3 | 10 || Not Injured | 13 | 27 | 40 || Total | 20 | 30 | 50 |Now, calculate the expected frequencies for each cell.The formula for expected frequency is:[E = frac{text{Row Total} times text{Column Total}}{text{Grand Total}}]So,- For Injured and Not Wearing Helmet:[E = frac{10 times 20}{50} = frac{200}{50} = 4]- For Injured and Wearing Helmet:[E = frac{10 times 30}{50} = frac{300}{50} = 6]- For Not Injured and Not Wearing Helmet:[E = frac{40 times 20}{50} = frac{800}{50} = 16]- For Not Injured and Wearing Helmet:[E = frac{40 times 30}{50} = frac{1200}{50} = 24]Now, set up the expected frequencies table:| | Not Wearing Helmet | Wearing Helmet ||---------|--------------------|----------------|| Injured | 4 | 6 || Not Injured | 16 | 24 |Now, calculate the chi-square statistic:[chi^2 = frac{(7 - 4)^2}{4} + frac{(3 - 6)^2}{6} + frac{(13 - 16)^2}{16} + frac{(27 - 24)^2}{24}]Calculating each term:- (frac{(7 - 4)^2}{4} = frac{9}{4} = 2.25)- (frac{(3 - 6)^2}{6} = frac{9}{6} = 1.5)- (frac{(13 - 16)^2}{16} = frac{9}{16} = 0.5625)- (frac{(27 - 24)^2}{24} = frac{9}{24} = 0.375)Adding them up:[chi^2 = 2.25 + 1.5 + 0.5625 + 0.375 = 4.6875]Now, determine the degrees of freedom (df):[df = (text{Number of Rows} - 1) times (text{Number of Columns} - 1) = (2 - 1) times (2 - 1) = 1]Next, find the critical value for a chi-square distribution with 1 degree of freedom at a 95% confidence level. From the chi-square table, the critical value is approximately 3.841.Compare the calculated chi-square statistic (4.6875) with the critical value (3.841):Since 4.6875 > 3.841, we reject the null hypothesis. This means there is a significant association between not wearing helmets and injuries in accidents at the 95% confidence level.**Final Answer**(1) The regression line equation is (hat{y} = -110x + 1325), and the estimated number of people not wearing helmets in 2022 is (boxed{775}).(2) Yes, we can be 95% confident that not wearing helmets is related to injuries in accidents. The answer is (boxed{text{Yes}}).

💡Okay, so I have this sequence problem here, and I need to tackle three parts. Let me start by understanding what each part is asking for.First, the sequence is defined as {a_n} where a₁ = t, a₂ = t², and for n ≥ 2, a_{n+1} = (t + 1)a_n - t a_{n-1}. The first part asks me to prove that if t ≠ 1, then the sequence {a_{n+1} - a_n} is a geometric progression. Hmm, okay. So I need to show that the difference between consecutive terms forms a geometric sequence.Let me write down the recurrence relation again: a_{n+1} = (t + 1)a_n - t a_{n-1}. Maybe I can manipulate this to find a relation for a_{n+1} - a_n.Let me compute a_{n+1} - a_n:a_{n+1} - a_n = (t + 1)a_n - t a_{n-1} - a_n= (t + 1 - 1)a_n - t a_{n-1}= t a_n - t a_{n-1}= t(a_n - a_{n-1})Oh, that's neat! So, the difference a_{n+1} - a_n is equal to t times the previous difference a_n - a_{n-1}. That means the ratio between consecutive differences is t, which is constant. So, that's the definition of a geometric progression. Therefore, {a_{n+1} - a_n} is a geometric progression with common ratio t.But wait, the problem specifies t ≠ 1. I guess if t were 1, the recurrence might behave differently? Let me check that. If t = 1, then the recurrence becomes a_{n+1} = 2a_n - a_{n-1}, which is a linear recurrence with characteristic equation r² - 2r + 1 = 0, which has a repeated root r = 1. So, in that case, the solution would be linear, not geometric. So, yeah, t ≠ 1 is important here.Okay, so part (1) seems manageable. I think I can write that up.Moving on to part (2), it asks for the general formula for the sequence {a_n}. So, given the recurrence relation and the initial terms, I need to find a closed-form expression for a_n.Given that in part (1), we found that the differences form a geometric progression, maybe we can use that to find a_n. Let me recall that if the differences form a geometric progression, then the sequence itself can be expressed as the sum of a geometric series.From part (1), a_{n+1} - a_n = t(a_n - a_{n-1}). So, starting from a₁ and a₂, we can write each term as the previous term plus a geometrically increasing difference.Let me write out the first few differences:a₂ - a₁ = t² - t = t(t - 1)a₃ - a₂ = t(a₂ - a₁) = t(t² - t) = t³ - t²a₄ - a₃ = t(a₃ - a₂) = t(t³ - t²) = t⁴ - t³...a_{n} - a_{n-1} = t^{n} - t^{n-1}So, each difference is t^{k} - t^{k-1} for k from 2 to n.Therefore, to find a_n, we can sum up all these differences starting from a₁:a_n = a₁ + (a₂ - a₁) + (a₃ - a₂) + ... + (a_n - a_{n-1})= t + (t² - t) + (t³ - t²) + ... + (t^{n} - t^{n-1})This is a telescoping series. Let's see:= t + t² - t + t³ - t² + ... + t^{n} - t^{n-1}= t^{n}Wait, that's interesting. All the intermediate terms cancel out, leaving just t^{n}.But hold on, does this hold for all n? Let me check with n=1,2,3.For n=1: a₁ = t, which is t¹, so that's correct.For n=2: a₂ = t², which is t², correct.For n=3: a₃ = (t + 1)a₂ - t a₁ = (t + 1)t² - t * t = t³ + t² - t² = t³, correct.Similarly, a₄ = (t + 1)a₃ - t a₂ = (t + 1)t³ - t * t² = t⁴ + t³ - t³ = t⁴, correct.So, indeed, a_n = t^{n} for all n ≥ 1, regardless of t ≠ 1. But wait, what if t = 1? Then a_n would be 1 for all n, which is consistent with t^{n} = 1^{n} = 1. So, even when t = 1, the formula a_n = t^{n} holds.Wait, but in part (1), we assumed t ≠ 1. So, does that mean that when t = 1, the differences are zero? Let me check.If t = 1, then a_{n+1} - a_n = 1*(a_n - a_{n-1}). So, a_{n+1} - a_n = a_n - a_{n-1}, which is a constant difference. But since a₂ - a₁ = 1 - 1 = 0, all differences are zero. Therefore, a_n = 1 for all n, which is consistent with a_n = 1^{n} = 1.So, actually, the general formula a_n = t^{n} holds for all t > 0, regardless of whether t is 1 or not. So, part (2) is solved.Now, onto part (3). It says, if 1/2 < t < 2, and b_n = (2a_n)/(1 + a_n²), compare the sum 1/b₁ + 1/b₂ + ... + 1/b_n with 2^{n} - 2^{-n/2}.First, let me express b_n in terms of t. Since a_n = t^{n}, then:b_n = (2t^{n}) / (1 + t^{2n})So, 1/b_n = (1 + t^{2n}) / (2t^{n}) = (1)/(2t^{n}) + (t^{2n})/(2t^{n}) = (1)/(2t^{n}) + t^{n}/2Therefore, 1/b_n = (t^{n} + 1/t^{n}) / 2So, the sum S = sum_{k=1}^{n} 1/b_k = sum_{k=1}^{n} (t^{k} + 1/t^{k}) / 2 = (1/2) sum_{k=1}^{n} (t^{k} + 1/t^{k})So, S = (1/2)(sum_{k=1}^{n} t^{k} + sum_{k=1}^{n} 1/t^{k})These are two geometric series. Let me compute each sum.First, sum_{k=1}^{n} t^{k} = t(1 - t^{n}) / (1 - t) if t ≠ 1.Similarly, sum_{k=1}^{n} 1/t^{k} = (1/t)(1 - (1/t)^{n}) / (1 - 1/t) = (1 - t^{-n}) / (t - 1)Wait, let me verify that:sum_{k=1}^{n} r^{k} = r(1 - r^{n}) / (1 - r)So, for sum_{k=1}^{n} t^{k} = t(1 - t^{n}) / (1 - t)Similarly, sum_{k=1}^{n} (1/t)^{k} = (1/t)(1 - (1/t)^{n}) / (1 - 1/t) = (1 - t^{-n}) / (t - 1)Yes, that's correct.So, putting it together:S = (1/2)[ t(1 - t^{n}) / (1 - t) + (1 - t^{-n}) / (t - 1) ]Note that (1 - t^{-n}) / (t - 1) = -(1 - t^{-n}) / (1 - t) = (t^{-n} - 1) / (1 - t)So, S = (1/2)[ t(1 - t^{n}) / (1 - t) + (t^{-n} - 1) / (1 - t) ]Combine the terms:= (1/2)[ (t(1 - t^{n}) + t^{-n} - 1) / (1 - t) ]Let me compute the numerator:t(1 - t^{n}) + t^{-n} - 1 = t - t^{n+1} + t^{-n} - 1So, numerator = (t - 1) - t^{n+1} + t^{-n}Therefore,S = (1/2)[ ( (t - 1) - t^{n+1} + t^{-n} ) / (1 - t) ]Simplify the numerator:(t - 1) = -(1 - t), so:= (1/2)[ ( - (1 - t) - t^{n+1} + t^{-n} ) / (1 - t) ]Split the fraction:= (1/2)[ - (1 - t)/(1 - t) - t^{n+1}/(1 - t) + t^{-n}/(1 - t) ]Simplify:= (1/2)[ -1 - t^{n+1}/(1 - t) + t^{-n}/(1 - t) ]Factor out 1/(1 - t):= (1/2)[ -1 + ( - t^{n+1} + t^{-n} ) / (1 - t) ]Hmm, this seems a bit messy. Maybe there's another approach.Alternatively, perhaps instead of computing S directly, I can find an inequality to compare S with 2^{n} - 2^{-n/2}.Given that 1/2 < t < 2, so t is between 1/2 and 2.Let me consider t and 1/t. If t is between 1/2 and 2, then 1/t is also between 1/2 and 2. So, t and 1/t are both in (1/2, 2).Wait, but if t is in (1/2, 2), then t can be greater than 1 or less than 1. Hmm.But in the expression for S, we have t^{k} + 1/t^{k} over 2. So, for each k, 1/b_k = (t^{k} + 1/t^{k}) / 2.I wonder if I can find an inequality for each term 1/b_k.Given that t is in (1/2, 2), let me see if I can bound t^{k} + 1/t^{k}.Let me consider the function f(x) = x + 1/x for x > 0. The function f(x) has a minimum at x=1, where f(1)=2. For x ≠ 1, f(x) > 2.Moreover, since t is in (1/2, 2), t^{k} is in (1/2^{k}, 2^{k}).But perhaps more importantly, for each k, t^{k} + 1/t^{k} ≤ 2^{k} + 2^{-k}.Wait, is that true?Let me test for k=1:t + 1/t ≤ 2 + 1/2 = 2.5. But t is in (1/2, 2), so t + 1/t can be as large as 2 + 1/2 = 2.5, but actually, when t approaches 2, t + 1/t approaches 2 + 1/2 = 2.5, and when t approaches 1/2, t + 1/t approaches 1/2 + 2 = 2.5. So, the maximum of t + 1/t is 2.5, which is equal to 2 + 1/2.Similarly, for k=2:t² + 1/t². Let me compute the maximum of t² + 1/t² when t is in (1/2, 2).Compute derivative: f(t) = t² + 1/t², f’(t) = 2t - 2/t³. Setting f’(t)=0: 2t = 2/t³ => t⁴=1 => t=1. So, the minimum is at t=1, f(1)=2. The maximum occurs at the endpoints.At t=2: f(2)=4 + 1/4=4.25At t=1/2: f(1/2)=1/4 + 4=4.25So, t² + 1/t² ≤ 4.25=17/4=4.25.Similarly, 2² + 1/2²=4 + 1/4=4.25.So, in general, for each k, t^{k} + 1/t^{k} ≤ 2^{k} + 2^{-k}.Therefore, 1/b_k = (t^{k} + 1/t^{k}) / 2 ≤ (2^{k} + 2^{-k}) / 2.Thus, sum_{k=1}^{n} 1/b_k ≤ sum_{k=1}^{n} (2^{k} + 2^{-k}) / 2 = (1/2) sum_{k=1}^{n} 2^{k} + (1/2) sum_{k=1}^{n} 2^{-k}Compute these sums:sum_{k=1}^{n} 2^{k} = 2(2^{n} - 1)/(2 - 1) = 2^{n+1} - 2sum_{k=1}^{n} 2^{-k} = (1/2)(1 - (1/2)^{n}) / (1 - 1/2) ) = (1/2)(1 - 2^{-n}) / (1/2) ) = 1 - 2^{-n}Therefore, sum_{k=1}^{n} 1/b_k ≤ (1/2)(2^{n+1} - 2) + (1/2)(1 - 2^{-n}) = (2^{n} - 1) + (1/2 - 2^{-n-1}) = 2^{n} - 1 + 1/2 - 2^{-n-1} = 2^{n} - 1/2 - 2^{-n-1}Wait, but the problem asks to compare the sum with 2^{n} - 2^{-n/2}. So, my upper bound is 2^{n} - 1/2 - 2^{-n-1}, which is less than 2^{n} - 2^{-n/2} because 1/2 + 2^{-n-1} > 2^{-n/2} for n ≥1.Wait, let me check for n=1:sum 1/b₁ = (t + 1/t)/2. For t in (1/2, 2), this is between (1 +1)/2=1 and (2 + 1/2)/2=1.25.2^{1} - 2^{-1/2}=2 - 1/√2≈2 -0.707≈1.293. So, 1.25 <1.293, so the sum is less than 2 - 2^{-1/2}.Similarly, for n=2:sum 1/b₁ +1/b₂= (t +1/t)/2 + (t² +1/t²)/2. Let's compute for t=1: sum=1 +1=2. 2^{2} -2^{-1}=4 -0.5=3.5. So, 2 <3.5.For t approaching 2: 1/b₁ approaches (2 +1/2)/2=1.25, 1/b₂ approaches (4 +1/4)/2=2.125. So, sum≈1.25 +2.125=3.375 <4 - 2^{-1}=3.5.Similarly, for t approaching 1/2: 1/b₁ approaches (1/2 +2)/2=1.25, 1/b₂ approaches (1/4 +4)/2=2.125. Sum≈3.375 <3.5.So, for n=2, the sum is less than 4 - 2^{-1}=3.5.Similarly, for n=3:sum=1/b₁ +1/b₂ +1/b₃. For t=1, sum=1 +1 +1=3. 2^{3} -2^{-3/2}=8 - 1/(2√2)≈8 -0.353≈7.647. So, 3 <7.647.For t approaching 2: 1/b₁≈1.25, 1/b₂≈2.125, 1/b₃≈(8 +1/8)/2≈4.0625. Sum≈1.25 +2.125 +4.0625≈7.4375 <8 - 0.353≈7.647.Similarly, for t approaching 1/2: same as above.So, it seems that the sum is less than 2^{n} -2^{-n/2}.But wait, in my earlier computation, I found that sum 1/b_k ≤2^{n} -1/2 -2^{-n-1}, which is less than 2^{n} -2^{-n/2} because 1/2 +2^{-n-1} >2^{-n/2} for n≥1.Wait, let me check for n=1:1/2 +2^{-2}=0.5 +0.25=0.75 >2^{-0.5}=≈0.707. So, yes.For n=2:1/2 +2^{-3}=0.5 +0.125=0.625 >2^{-1}=0.5.For n=3:1/2 +2^{-4}=0.5 +0.0625=0.5625 >2^{-1.5}=≈0.353.So, indeed, 1/2 +2^{-n-1} >2^{-n/2} for n≥1.Therefore, sum 1/b_k ≤2^{n} - (1/2 +2^{-n-1}) <2^{n} -2^{-n/2}.Hence, the sum is less than 2^{n} -2^{-n/2}.Therefore, the conclusion is that the sum 1/b₁ +1/b₂ +...+1/b_n <2^{n} -2^{-n/2}.I think that's the comparison required. So, the sum is less than 2^{n} -2^{-n/2}.**Final Answer**(1) boxed{a_{n+1} - a_n} is a geometric progression.(2) boxed{a_n = t^n}(3) boxed{frac{1}{b_1} + frac{1}{b_2} + cdots + frac{1}{b_n} < 2^n - 2^{-frac{n}{2}}}