💡First, I recognize that the distribution is symmetric about the mean ( m ), and 68% of the data lies within one standard deviation ( d ) of the mean. This implies that 34% of the data is between ( m ) and ( m + d ), and another 34% is between ( m ) and ( m - d ).Since the distribution is symmetric, the percentage of the data that is less than ( m + d ) includes all the data below the mean ( m ) plus the 34% between ( m ) and ( m + d ). Therefore, the total percentage of the distribution that is less than ( m + d ) is ( 50% + 34% = 84% ).

💡Okay, so I have this problem where I need to find the ratio of the wealth of a citizen from Country X to that of a citizen from Country Y. Let me try to understand the problem step by step.First, the problem states that Country X has g% of the world's population and h% of the world's wealth. Similarly, Country Y has i% of the population and j% of the wealth. Country Z also has some percentages, but since the question is only about X and Y, maybe I don't need to worry about Z for now.The key here is that the wealth is equally distributed within each country. So, I need to figure out how much wealth each citizen in Country X has compared to each citizen in Country Y.Let me think about how to approach this. If I consider the total world population as P and the total world wealth as W, then:- Country X's population would be (g/100) * P.- Country X's wealth would be (h/100) * W.- Similarly, Country Y's population is (i/100) * P.- Country Y's wealth is (j/100) * W.Now, to find the wealth per citizen in each country, I can divide the country's wealth by its population.For Country X:Wealth per citizen = (h/100 * W) / (g/100 * P)The 100s cancel out, so it's (h * W) / (g * P)For Country Y:Wealth per citizen = (j/100 * W) / (i/100 * P)Again, the 100s cancel, so it's (j * W) / (i * P)Now, I need the ratio of Country X's wealth per citizen to Country Y's wealth per citizen. So, I'll divide the two expressions:Ratio = [(h * W) / (g * P)] / [(j * W) / (i * P)]Let me simplify this. The W and P terms are in both the numerator and denominator, so they cancel out:Ratio = (h / g) / (j / i) = (h / g) * (i / j) = (h * i) / (g * j)So, the ratio is (h * i) divided by (g * j). Looking at the options, that's option B: hi/gj.Wait, let me double-check to make sure I didn't make a mistake. I converted the percentages to fractions by dividing by 100, then canceled them out. Then I divided the wealth per citizen of X by that of Y, which gave me the ratio. The W and P terms canceled, leaving me with h*i over g*j. Yeah, that seems right.I think I got it. The answer should be B.

💡Okay, so I need to verify these properties of the symmetric difference operation, which is denoted by the triangle symbol, △. I remember that the symmetric difference between two sets A and B, written as A △ B, is the set of elements that are in either A or B but not in both. So, it's like the union of A minus B and B minus A. First, let me write down the definitions and properties I know:1. **Definition of Symmetric Difference**: A △ B = (A B) ∪ (B A)2. **Associative Property**: (A △ B) △ C = A △ (B △ C)3. **Another Property**: (A △ B) △ (B △ C) = A △ C4. **Equivalence Relation**: A △ B = C ⇨ A = B △ CAlright, let's tackle each property one by one.**1. Associative Property: (A △ B) △ C = A △ (B △ C)**I need to show that the symmetric difference is associative, meaning the way we group the sets doesn't affect the result. Let me start by expanding both sides using the definition.Left-hand side (LHS): (A △ B) △ C= [(A B) ∪ (B A)] △ C= [(A B) ∪ (B A) C] ∪ [C (A B) ∪ (B A)]Hmm, this is getting a bit complicated. Maybe I should use another approach. I remember that symmetric difference can also be expressed using union and intersection:A △ B = (A ∪ B) (A ∩ B)So, maybe rewriting it this way will help.Let me try that.LHS: (A △ B) △ C= [(A ∪ B) (A ∩ B)] △ C= [( (A ∪ B) (A ∩ B) ) ∪ C] [ ( (A ∪ B) (A ∩ B) ) ∩ C ]Similarly, the right-hand side (RHS): A △ (B △ C)= A △ [(B ∪ C) (B ∩ C)]= [A ∪ ( (B ∪ C) (B ∩ C) ) ] [ A ∩ ( (B ∪ C) (B ∩ C) ) ]This still looks messy. Maybe there's a better way. I recall that symmetric difference can be expressed using the formula:A △ B = (A ∪ B) (A ∩ B)But also, it's equivalent to:A △ B = (A B) ∪ (B A)Maybe I can use Venn diagrams to visualize this. If I draw three sets A, B, and C, and shade the regions corresponding to (A △ B) △ C and A △ (B △ C), they should look the same if the property holds.Alternatively, maybe using algebraic manipulation with set operations. Let's recall some properties:- Associativity of union and intersection.- Distributive laws.Wait, symmetric difference is associative, but I need to prove it, not just state it.Perhaps I can use the fact that symmetric difference is equivalent to addition modulo 2 in some algebraic structure, but that might be too advanced for now.Let me try expanding both sides step by step.Starting with LHS: (A △ B) △ CFirst compute A △ B:= (A B) ∪ (B A)Then, take that result and symmetric difference with C:= [(A B) ∪ (B A)] △ C= [ ( (A B) ∪ (B A) ) C ] ∪ [ C ( (A B) ∪ (B A) ) ]Simplify each part:First part: ( (A B) ∪ (B A) ) C= (A B C) ∪ (B A C)Second part: C ( (A B) ∪ (B A) )= C (A B) ∩ C (B A)= (C ∩ B) ∩ (C ∩ A^c) ∩ (C ∩ A) ∩ (C ∩ B^c)Wait, this is getting too convoluted. Maybe I should use the formula for symmetric difference in terms of union and intersection:A △ B = (A ∪ B) (A ∩ B)So, let's try that.LHS: (A △ B) △ C= [(A ∪ B) (A ∩ B)] △ C= [ ( (A ∪ B) (A ∩ B) ) ∪ C ] [ ( (A ∪ B) (A ∩ B) ) ∩ C ]Similarly, RHS: A △ (B △ C)= A △ [(B ∪ C) (B ∩ C)]= [ A ∪ ( (B ∪ C) (B ∩ C) ) ] [ A ∩ ( (B ∪ C) (B ∩ C) ) ]This still seems complicated. Maybe instead of expanding, I can use the fact that symmetric difference is associative, which I think is a known property, but I need to prove it.Alternatively, maybe using characteristic functions. For each element, check if it belongs to both sides.Let me think about an element x. For x to be in (A △ B) △ C, it must be in exactly one of (A △ B) or C. Similarly, for x to be in A △ (B △ C), it must be in exactly one of A or (B △ C).Let me see:If x is in (A △ B) △ C, then:- If x is in (A △ B) but not in C, then x is in A △ B, meaning x is in exactly one of A or B.- If x is in C but not in (A △ B), then x is in C and in both A and B or in neither.Similarly, for x in A △ (B △ C):- If x is in A but not in (B △ C), then x is in A and in both B and C or in neither.- If x is in (B △ C) but not in A, then x is in exactly one of B or C.Wait, this seems similar to the previous case. So, if x is in exactly one of A, B, or C, then it will be in both (A △ B) △ C and A △ (B △ C). Therefore, the two expressions are equal.So, the first property holds.**2. (A △ B) △ (B △ C) = A △ C**Alright, moving on to the second property. This one looks interesting. Let me try expanding both sides.First, compute (A △ B) △ (B △ C).Using the definition:= [(A B) ∪ (B A)] △ [(B C) ∪ (C B)]Again, this seems messy. Maybe using the characteristic function approach again.For an element x, x is in (A △ B) △ (B △ C) if and only if x is in exactly one of (A △ B) or (B △ C).Similarly, x is in A △ C if and only if x is in exactly one of A or C.Let me see if these are equivalent.If x is in exactly one of A or C, then:- If x is in A but not C, then x is in A △ C.Now, let's see if x is in (A △ B) △ (B △ C):- If x is in A but not C, then depending on B, x could be in A △ B or not, and similarly in B △ C.Wait, maybe it's better to consider cases.Case 1: x is in A only.Then, x is in A △ B if x is not in B.If x is not in B, then x is in A △ B.Then, x is in (A △ B) △ (B △ C) if x is not in (B △ C).But x is not in B, so x is in B △ C only if x is in C.Since x is in A only, if x is not in C, then x is not in B △ C, so x is in (A △ B) △ (B △ C).If x is in C, then x is in B △ C (since x is not in B), so x is not in (A △ B) △ (B △ C).Wait, but x is in A only, so if x is in C, it's in A and C, but not B.So, x is in A △ C only if x is not in C.Hmm, this is getting confusing. Maybe another approach.Let me use the fact that symmetric difference is associative and commutative.So, (A △ B) △ (B △ C) = A △ B △ B △ CSince symmetric difference is commutative and associative, we can rearrange:= A △ (B △ B) △ CBut B △ B is the empty set, since symmetric difference of a set with itself is empty.So, = A △ ∅ △ CAnd A △ ∅ = A, so = A △ CTherefore, (A △ B) △ (B △ C) = A △ CThat's a much cleaner proof! I should have thought of using the properties of symmetric difference being associative and commutative, and that A △ A = ∅.**3. A △ B = C ⇨ A = B △ C**Alright, the last property. It's an equivalence, so I need to show both directions:If A △ B = C, then A = B △ C, and conversely, if A = B △ C, then A △ B = C.Let me start with the forward direction: Assume A △ B = C, show that A = B △ C.Using the definition:A △ B = (A B) ∪ (B A) = CWe need to show that A = B △ C.Let me express A in terms of B and C.Starting from A △ B = C:A = (A △ B) △ BBecause symmetric difference with B twice cancels out.So, A = C △ BBut symmetric difference is commutative, so A = B △ CThat's the forward direction.Now, the converse: Assume A = B △ C, show that A △ B = C.Starting from A = B △ C, compute A △ B:A △ B = (B △ C) △ BAgain, using the property that (X △ Y) △ Y = XSo, (B △ C) △ B = CTherefore, A △ B = CHence, the equivalence holds.So, all three properties are verified.

💡First, I need to determine the number of dogs Cassie has based on the total number of nails she needs to cut.I'll start by calculating the total number of nails from the parrots. Each parrot has 3 claws on each leg, and since they have 2 legs, each parrot has 6 claws. However, one parrot has an extra toe, making it 7 claws for that parrot.With 8 parrots, the total number of claws is:7 parrots × 6 claws = 42 claws1 parrot × 7 claws = 7 clawsTotal claws from parrots = 42 + 7 = 49 clawsNext, I'll subtract the parrots' claws from the total nails Cassie needs to cut to find out how many nails belong to the dogs:Total nails to cut = 113Nails from parrots = 49Nails from dogs = 113 - 49 = 64Each dog has 4 nails on each foot, and since dogs have 4 feet, each dog has:4 nails/foot × 4 feet = 16 nailsFinally, I'll determine the number of dogs by dividing the total number of dog nails by the number of nails per dog:Number of dogs = 64 nails ÷ 16 nails/dog = 4 dogs

💡Okay, so I have this problem where Julio is cutting off the four corners, or vertices, of a regular tetrahedron. I need to figure out how many vertices the remaining shape will have. The options are A) 8, B) 9, C) 11, D) 12, and E) 15.First, let me recall what a regular tetrahedron is. A regular tetrahedron is a three-dimensional shape with four triangular faces, each of which is an equilateral triangle. It has four vertices and six edges. Each vertex is connected to the other three vertices.Now, the problem says that Julio is cutting off the four corners, or vertices, of this tetrahedron. I think this process is called truncation. When you truncate a polyhedron, you're essentially cutting off its corners, which changes the shape of the original faces and introduces new faces in place of the truncated corners.I need to visualize what happens when you truncate a tetrahedron. When you cut off a vertex, you're replacing that vertex with a new face. Since a tetrahedron has triangular faces, cutting off a vertex should replace that vertex with a new triangular face. Each original vertex is connected to three edges, so cutting it off would create a new triangular face where the vertex used to be.So, if I start with a tetrahedron that has four vertices, and I cut off each of these four vertices, each truncation will add a new triangular face. But how does this affect the number of vertices?Each original vertex is removed, but in its place, the truncation process creates new vertices. Specifically, when you cut off a vertex, you're introducing new vertices along the edges that were connected to the original vertex. For a tetrahedron, each vertex is connected to three edges. If I cut off a vertex, each of these three edges will now have a new vertex where the truncation occurred.So, for each original vertex that's cut off, we're adding three new vertices. Since there are four original vertices, each being cut off, that would be 4 times 3, which is 12 new vertices.But wait, I need to make sure I'm not double-counting any vertices. When you truncate a polyhedron, the new vertices introduced by truncating one vertex are distinct from those introduced by truncating another vertex. So, in this case, each of the four original vertices is being truncated independently, and each truncation adds three unique new vertices.Therefore, the total number of new vertices added is 12. But what about the original vertices? Since we've cut them off, they no longer exist. So, the remaining shape doesn't have the original four vertices anymore.So, the remaining shape has only the new vertices that were introduced by the truncation. That would be 12 vertices in total.Let me double-check this reasoning. A regular tetrahedron has four vertices. Cutting off each vertex replaces it with a new face and introduces new vertices. Each truncation adds three new vertices, and since there are four truncations, that's 4 times 3, which is 12. The original vertices are removed, so they don't contribute to the final count.Yes, that seems correct. The remaining shape after truncating all four vertices of a regular tetrahedron would have 12 vertices.

💡Alright, so I have this problem about an n-sided pyramid with an inscribed sphere. The task is to prove two things: first, that if all the lateral faces are rotated around their respective base edges to lie flat on the base plane, then all the points where these faces were tangent to the sphere will coincide at one point H. Second, that the vertices of these faces will lie on a single circle centered at H.Hmm, okay. Let me try to visualize this. I imagine a pyramid with a base that's an n-sided polygon, and the sides are triangles connecting each base edge to the apex. Since it has an inscribed sphere, that means the sphere touches each face exactly once. So, for each lateral face, there's a point where the sphere is tangent to that face.Now, if I rotate each lateral face around its base edge until it lies flat on the base plane, what happens to the point of tangency? Intuitively, since the face is being rotated around the edge, the point where it was tangent to the sphere should move along some path. But the problem says they all coincide at a single point H. That suggests that all these points of tangency, when rotated, end up at the same spot.I think the key here is to consider the properties of the inscribed sphere. The sphere is tangent to all the faces, so the distance from the center of the sphere to each face is equal to the radius. When we rotate a face around its base edge, the point of tangency moves, but maybe the center of the sphere relates to this movement in a way that makes all these points converge.Let me try to break it down step by step.First, let's denote the pyramid as having a base ABCD...E (for n sides) and an apex S. The inscribed sphere has its center at point O. The sphere is tangent to each lateral face at some point, say P for face SAB, Q for face SBC, and so on.Since the sphere is tangent to each face, the line from O to P is perpendicular to face SAB, and similarly for O to Q and other points. Also, the distance from O to each face is equal to the radius r of the sphere.Now, when we rotate face SAB around edge AB, the point P will move in some circular path around AB. Similarly, when we rotate face SBC around BC, point Q will move around BC, and so on.But the problem states that all these points of tangency will coincide at a single point H after rotation. So, H must be a special point related to the sphere and the pyramid.Wait, since the sphere is tangent to the base as well, let's denote the point of tangency on the base as H. That might be the point where all the rotated tangency points converge.Let me think about that. If H is the point where the sphere is tangent to the base, then it's the foot of the perpendicular from O to the base. So, OH is perpendicular to the base.Now, if we rotate face SAB around AB, the point P is moving in a circle around AB. Similarly, when we rotate SBC around BC, Q moves around BC. But since all these rotations are happening around edges of the base, which are all in the base plane, the movement of P, Q, etc., is constrained to circles in planes perpendicular to the base edges.But how does this lead them all to H? Maybe because H is equidistant from all these edges in some way.Wait, since the sphere is tangent to all the faces, the distances from O to each face are equal. When we rotate the face around its edge, the point P moves such that its distance from O remains the same, but its position changes.But H is the point where the sphere is tangent to the base, so it's the closest point from O to the base. So, when we rotate the face SAB around AB, the point P, which was at a distance r from O, moves along a circle until it reaches H, which is also at distance r from O but in the base plane.Is that correct? Let me see.If I consider the rotation of P around AB, since P is on face SAB and we're rotating the face around AB, the point P should trace a circle in a plane perpendicular to AB. The center of this circle would be the projection of O onto the plane of SAB, but since SAB is being rotated, maybe the projection changes.Wait, maybe I need to think in terms of the sphere's properties. Since the sphere is tangent to all the faces, the point P is where the sphere touches SAB. When we rotate SAB around AB, the sphere remains fixed, so the point P moves along the sphere's surface until it reaches the base.But the sphere is also tangent to the base at H, so when SAB is rotated down to the base, the point P must coincide with H because it's the only point where the sphere is tangent to the base.That makes sense. So, rotating each lateral face around its base edge brings the tangency point on that face to the common tangency point H on the base.Okay, that seems to explain the first part. Now, for the second part: the vertices of the faces will lie on a single circle centered at H.The vertices of the faces are the apex S and the base vertices. Wait, no, when we rotate the lateral faces around their base edges, the apex S will move. So, after rotation, each apex S is moved to a new position, and all these new positions lie on a circle centered at H.Hmm, I need to visualize this. So, originally, all the lateral faces meet at apex S. When we rotate each face around its base edge, the apex S moves to a new position for each face. So, each rotation moves S to a different point, but all these points lie on a circle centered at H.Why is that?Well, since each rotation is around a different base edge, the movement of S is constrained by the rotation axis. The distance from S to each base edge is the same because the pyramid is convex and has an inscribed sphere.Wait, actually, in a pyramid with an inscribed sphere, the distances from the apex to each base edge are equal? Or is it the distances from the center of the sphere to each face?Yes, the distances from the center O to each face are equal because the sphere is tangent to each face. So, the distance from O to each lateral face is equal to the radius r.But how does that relate to the apex S?Hmm, maybe I need to consider the movement of S when each face is rotated. Since each face is rotated around its base edge, the apex S will move along a circular path for each rotation. The set of all these positions of S should lie on a circle.But why centered at H?Perhaps because H is the common point where all the tangency points converge, and the distances from H to each rotated apex are equal.Wait, when we rotate a face around its base edge, the apex S moves such that its distance to the base edge remains the same, but its position changes. Since all these movements are constrained by the sphere, maybe the distances from H to each moved apex are equal.Alternatively, maybe the original apex S is equidistant from all the base edges, and after rotation, these distances translate into equal distances from H.I'm not entirely sure, but I think the key idea is that since the sphere is tangent to all faces, the movement of each face's apex during rotation preserves some symmetry, resulting in all moved apexes lying on a circle centered at H.I need to formalize this a bit more.Let me consider two adjacent lateral faces, say SAB and SBC. When I rotate SAB around AB, the apex S moves to a new point S'. Similarly, rotating SBC around BC moves S to another point S''. I need to show that S' and S'' lie on a circle centered at H.Since the rotations are around AB and BC respectively, the points S' and S'' are images of S under these rotations. The circle centered at H should pass through all such images.Given that the sphere is tangent to the base at H, and the rotations bring all tangency points to H, it's plausible that the moved apexes lie on a circle around H.Maybe the radius of this circle is equal to the distance from H to S, but adjusted by the rotation.Wait, no. The distance from H to S is fixed, but when we rotate S around AB, the distance from H to S' might change.Alternatively, perhaps the circle has a radius equal to the slant height of the pyramid or something related to the sphere's radius.I think I need to consider the properties of the inscribed sphere more carefully.Since the sphere is tangent to the base at H, and tangent to each lateral face at some point P_i, the line from O to P_i is perpendicular to the face, and OP_i = r.When we rotate the face around its base edge, the point P_i moves to H, as established earlier. So, the rotation effectively maps P_i to H.Similarly, the apex S is connected to P_i on the face. When we rotate the face, S moves to a new position S', but the line SP_i is rotated to S'H.Since SP_i is tangent to the sphere at P_i, the length SP_i is equal to the length from S to the point of tangency on the sphere. Similarly, after rotation, S'H is equal to SP_i.Therefore, all these lengths S'H are equal, meaning that all S' lie on a sphere centered at H with radius equal to SP_i.But wait, the problem states that the vertices lie on a single circle, not a sphere. So, perhaps the movement is constrained such that all S' lie on a circle in the base plane.Wait, no, the rotated faces lie on the base plane, so the apex S' after rotation is in the base plane. Therefore, all S' lie on the base plane, and specifically, on a circle centered at H.So, the distance from H to each S' is equal because SP_i is equal for all faces (since the sphere is tangent to each face at the same radius). Therefore, all S' lie on a circle centered at H with radius equal to SP_i.That seems to make sense.So, putting it all together:1. The inscribed sphere is tangent to each lateral face at a point P_i.2. Rotating each lateral face around its base edge brings P_i to the common tangency point H on the base.3. The apex S, when rotated around each base edge, moves to a new position S' such that the distance from H to S' is equal for all S', meaning they lie on a circle centered at H.Therefore, both assertions are proven.I think that covers it. I might have skipped some steps, but overall, the key ideas are the properties of the inscribed sphere, the rotation of the faces bringing tangency points to H, and the equal distances from H to the moved apexes ensuring they lie on a circle.**Final Answer**boxed{H}