💡Alright, so I have this probability problem here about a bus and Laura arriving at a bus stop. Let me try to understand it step by step. The bus arrives randomly between 1:00 and 2:30 PM, waits for 20 minutes, and then leaves. Laura also arrives randomly between 1:00 and 2:30 PM. I need to find the probability that Laura arrives while the bus is still there.First, I think it's helpful to visualize this problem. Maybe I can represent the arrival times on a coordinate system where one axis is the bus's arrival time and the other axis is Laura's arrival time. Since both arrivals are between 1:00 and 2:30 PM, which is a 90-minute window, I can consider each axis to range from 0 to 90 minutes, with 0 representing 1:00 PM and 90 representing 2:30 PM.So, if I plot Laura's arrival time on the x-axis and the bus's arrival time on the y-axis, each point in this 90x90 square represents a possible scenario of their arrivals. The total area of this square is 90*90 = 8100 square minutes. This will be the denominator in my probability calculation because it represents all possible outcomes.Now, I need to find the area where Laura arrives while the bus is still at the stop. The bus waits for 20 minutes, so if the bus arrives at time y, it will be at the stop from y to y+20 minutes. Laura arrives at time x. For Laura to catch the bus, her arrival time x must be between y and y+20. So, the condition is y ≤ x ≤ y+20.But I also need to consider the boundaries. If the bus arrives after 2:10 PM (which is 70 minutes after 1:00 PM), it can't wait the full 20 minutes because it would exceed 2:30 PM. So, in that case, the bus would only wait until 2:30 PM. Therefore, for y > 70, the condition becomes y ≤ x ≤ 90.To find the area where Laura catches the bus, I can divide the problem into two parts:1. When the bus arrives between 0 and 70 minutes (from 1:00 PM to 2:10 PM), the bus waits until y+20 minutes.2. When the bus arrives between 70 and 90 minutes (from 2:10 PM to 2:30 PM), the bus waits until 2:30 PM.For the first part, when y is between 0 and 70, the area where Laura catches the bus is a band from y to y+20 on the x-axis. This forms a trapezoid on the graph. The area of this trapezoid can be calculated as the average of the two parallel sides multiplied by the height. The two parallel sides are both 20 minutes long (from y to y+20), and the height is 70 minutes. So, the area is (20 + 20)/2 * 70 = 1400 square minutes.For the second part, when y is between 70 and 90, the bus can only wait until 90 minutes. So, the area where Laura catches the bus is a triangle. The base and height of this triangle are both 20 minutes (from y to 90). The area of this triangle is (20 * 20)/2 = 200 square minutes.Adding these two areas together gives the total area where Laura catches the bus: 1400 + 200 = 1600 square minutes.Finally, to find the probability, I divide this area by the total area of the square: 1600 / 8100. Simplifying this fraction gives 16/81.So, the probability that Laura arrives while the bus is still there is 16/81.

💡Alright, so I have this geometry problem here about three circles of the same radius intersecting each other pairwise, and one of the intersection points is common to all. I need to prove that the circle passing through the three other intersection points is equal to the three given circles. Hmm, okay, let me try to visualize this.First, let me draw three circles with the same radius. Let's say they all have radius r. They intersect each other pairwise, meaning each pair of circles intersects at two points. But one of these intersection points is common to all three circles. So, there's a single point where all three circles meet, and then each pair also meets at another distinct point. Let me label the common intersection point as O. Then, the other intersection points can be labeled as A, B, and C, each being the other intersection point for each pair of circles.So, circle 1 and circle 2 intersect at O and A, circle 2 and circle 3 intersect at O and B, and circle 3 and circle 1 intersect at O and C. Now, I need to consider the circle passing through points A, B, and C. The claim is that this circle is equal to the three given circles, meaning it has the same radius r.Okay, so I need to show that the circle passing through A, B, and C has radius r. To do this, maybe I can find the circumradius of triangle ABC and show that it's equal to r.But before that, perhaps I should explore the properties of points A, B, and C. Since all three original circles have the same radius and intersect at O, maybe there's some symmetry here.Let me consider the distances from O to each of the centers of the circles. Let's denote the centers of the three circles as O1, O2, and O3. Since all circles have the same radius r, the distance between any two centers should be equal to the distance between their centers, which is 2r times the sine of half the angle between them, but I'm not sure if that's the right approach.Wait, maybe I should think about the triangle formed by the centers O1, O2, and O3. Since all circles have the same radius and intersect at O, the distances from O to each center are equal to r. So, triangle O1O2O3 is an equilateral triangle? Because each side is equal to the distance between centers, which, if all circles have radius r and intersect at O, the distance between centers should be 2r times the sine of 60 degrees, but I'm getting confused.Hold on, maybe I should use coordinate geometry. Let me place point O at the origin (0,0). Then, let me place the centers of the three circles at points (a,0), (b,c), and (d,e), ensuring that each center is at a distance r from O. So, the distance from O to each center is r, meaning sqrt(a^2) = r, sqrt(b^2 + c^2) = r, and sqrt(d^2 + e^2) = r. Therefore, a = r or -r, and similarly for the others.But this might get too complicated. Maybe there's a better way.Let me think about the power of point O with respect to each circle. Since O lies on all three circles, the power of O with respect to each circle is zero. But I'm not sure if that helps.Alternatively, maybe I can use the fact that points A, B, and C lie on the radical axes of the pairs of circles. The radical axis of two circles is the set of points with equal power with respect to both circles. Since A is the other intersection point of circles 1 and 2, it lies on the radical axis of circles 1 and 2. Similarly for B and C.But how does that help me find the circumradius of triangle ABC?Wait, maybe I can use the property that the circumradius R of triangle ABC can be found using the formula:R = frac{abc}{4Δ}where a, b, c are the lengths of the sides of the triangle, and Δ is the area of the triangle.But to use this, I need to find the lengths of AB, BC, and AC, as well as the area of triangle ABC. Hmm, that might be a bit involved, but let's see.Alternatively, maybe I can use the fact that the circumradius of triangle ABC is equal to the distance from the circumcenter to any of the vertices A, B, or C. So, if I can find the circumcenter of triangle ABC and show that it's at a distance r from each of A, B, and C, then I'm done.But how do I find the circumcenter? It's the intersection of the perpendicular bisectors of the sides of the triangle. So, if I can show that the perpendicular bisectors intersect at a point that is at distance r from A, B, and C, then that point would be the circumcenter, and the radius would be r.Hmm, this seems promising. Let me try to find the perpendicular bisectors.But wait, I don't have specific coordinates for A, B, and C. Maybe I need to use vector geometry or some other method.Alternatively, maybe I can use the fact that points A, B, and C are the other intersection points of the circles, and since all circles have radius r, there might be some symmetry or congruence that I can exploit.Wait, another idea: since all three original circles have the same radius and intersect at O, the triangle formed by the centers O1, O2, O3 is equilateral. Because each center is at distance r from O, and the distance between any two centers is 2r sin(60°) = r√3. So, triangle O1O2O3 is equilateral with side length r√3.If that's the case, then the centers form an equilateral triangle, which might imply that the points A, B, and C form another equilateral triangle. If triangle ABC is equilateral, then its circumradius can be calculated, and maybe it's equal to r.But is triangle ABC equilateral? Let me think.Since each pair of circles intersects at O and another point, and the centers form an equilateral triangle, the other intersection points A, B, and C might also form an equilateral triangle. Because of the symmetry, each angle at O is 60 degrees, so the arcs between O and A, O and B, and O and C are equal, making triangle ABC equilateral.If triangle ABC is equilateral, then its circumradius R is given by R = frac{a}{√3}, where a is the length of each side. So, if I can find the length of AB, BC, or AC, I can compute R.But how do I find the length of AB? Since A and B are intersection points of different circles, maybe I can use the distance formula or some geometric properties.Wait, another approach: since O is the common intersection point, and A, B, C are the other intersection points, perhaps OA, OB, and OC are all equal to r, because O is on all three circles. But A, B, and C are also on the circles, so OA = OB = OC = r.Wait, but A is on circle 1 and circle 2, so OA = r (since O is the center of circle 1) and OA is also equal to the radius of circle 2, which is r. Similarly for OB and OC.So, points A, B, and C all lie on a circle centered at O with radius r. But that's the original circles. However, the circle passing through A, B, and C is different. Wait, is it?Wait, no. If A, B, and C all lie on a circle centered at O with radius r, then the circle passing through A, B, and C is just one of the original circles. But that can't be, because A, B, and C are distinct points on different circles.Hmm, maybe I'm getting confused. Let me clarify.Each pair of circles intersects at O and another point. So, circle 1 and circle 2 intersect at O and A; circle 2 and circle 3 intersect at O and B; circle 3 and circle 1 intersect at O and C. So, points A, B, and C are distinct from O and from each other.Now, I need to consider the circle passing through A, B, and C. Is this circle the same as the original three circles? Or is it a different circle with the same radius?Wait, the problem says "the circle passing through the three other intersection points is equal to the three given circles." So, it's not necessarily the same circle, but a circle with the same radius.So, I need to show that the radius of the circle passing through A, B, and C is equal to r.Okay, so maybe I can use the fact that OA = OB = OC = r, and then find the circumradius of triangle ABC.But how?Wait, perhaps I can use the formula for the circumradius in terms of the sides and the area. But I need to find the lengths of AB, BC, and AC.Alternatively, maybe I can use the fact that the circumradius R of triangle ABC is related to the distances from O to the sides of the triangle.Wait, another idea: since O is the common point, and A, B, C are points on the circles, maybe OA, OB, and OC are all radii, so OA = OB = OC = r.If I can show that the circumradius of triangle ABC is equal to r, then I'm done.But how?Wait, maybe I can use the fact that in triangle ABC, the distances from O to each vertex are equal to r, and O is the circumcenter. But is O the circumcenter of triangle ABC?Wait, if O is equidistant from A, B, and C, then O is the circumcenter of triangle ABC. Therefore, the circumradius of triangle ABC is equal to OA = OB = OC = r.But is that correct? If O is equidistant from A, B, and C, then yes, O is the circumcenter, and the circumradius is r.Wait, but in that case, the circle passing through A, B, and C is centered at O with radius r, which is one of the original circles. But the problem says "the circle passing through the three other intersection points is equal to the three given circles." So, does that mean it's the same circle? Or just equal in radius?Wait, maybe I'm misinterpreting. The problem says "the circle passing through the three other intersection points is equal to the three given circles." So, it's saying that this new circle is equal in size, i.e., has the same radius, as the three given circles.But if O is the circumcenter of triangle ABC, and OA = OB = OC = r, then the circumradius is r, so the circle passing through A, B, and C has radius r, same as the given circles.Therefore, the circle passing through A, B, and C is equal to the three given circles in terms of radius.Wait, but is O the circumcenter of triangle ABC? Because O is equidistant from A, B, and C, yes, so O is the circumcenter. Therefore, the circumradius is r.Therefore, the circle passing through A, B, and C is centered at O with radius r, same as the original circles.But wait, the original circles are centered at O1, O2, O3, not at O. So, the circle passing through A, B, and C is a different circle, centered at O, with the same radius r.So, it's equal in radius, but not necessarily the same circle.Therefore, the circle passing through A, B, and C is equal to the three given circles in terms of radius.So, that's the proof.But let me double-check.Given three circles of radius r intersecting at O, and pairwise at A, B, and C. Then, OA = OB = OC = r, since O is on all three circles. Therefore, the circumradius of triangle ABC is r, since O is equidistant from A, B, and C. Therefore, the circle passing through A, B, and C has radius r, same as the original circles.Yes, that seems correct.So, the key idea is that since O is the common intersection point, and A, B, and C are points on the circles, OA = OB = OC = r, making O the circumcenter of triangle ABC with circumradius r. Therefore, the circle passing through A, B, and C has radius r, equal to the given circles.I think that's the proof.

💡Okay, so I have this probability problem here about selecting tags from a box. Let me try to understand it step by step. The box has 5 tags labeled 1 through 5. I need to calculate the probability that the numbers on the two selected tags are consecutive integers. There are two scenarios: one where the selection is without replacement, and another where it's with replacement.Starting with the first part, without replacement. Hmm, without replacement means once I pick a tag, I can't pick it again. So, when I select two tags, they have to be different. I think this is a combination problem because the order doesn't matter here. So, the total number of possible outcomes is the number of ways to choose 2 tags out of 5. The formula for combinations is n choose k, which is n! / (k!(n - k)!). Plugging in the numbers, that would be 5 choose 2, which is 5! / (2!(5 - 2)!) = (5 × 4 × 3!) / (2 × 1 × 3!) = (20) / 2 = 10. So, there are 10 possible outcomes.Now, I need to find how many of these outcomes result in consecutive integers. Let me list all the possible pairs and see which ones are consecutive. The pairs are {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}. Looking at these, the consecutive pairs are {1,2}, {2,3}, {3,4}, and {4,5}. That's 4 pairs. So, the number of favorable outcomes is 4.Therefore, the probability without replacement should be the number of favorable outcomes divided by the total number of outcomes. That is 4/10, which simplifies to 2/5. Okay, that seems straightforward.Now, moving on to the second part, with replacement. With replacement means that after picking a tag, I put it back before picking the next one. So, it's possible to pick the same tag twice. This changes things because now the total number of possible outcomes increases. Since each pick is independent, the number of possible outcomes is 5 × 5 = 25. So, there are 25 possible ordered pairs.But wait, in the first part, we considered combinations where order didn't matter, but here, since we're replacing, the order might matter. Or does it? Let me think. If we're just looking for two numbers that are consecutive, regardless of the order, then maybe we still need to consider unordered pairs. But actually, in the case of replacement, the pair (1,2) is different from (2,1), but both are consecutive. Hmm, so do I need to count both?Wait, no, because in the first part, without replacement, {1,2} and {2,1} are the same, but with replacement, they are different. So, does the problem consider ordered pairs or unordered? The problem says "the numbers on the two selected tags are consecutive integers." It doesn't specify order, so maybe it's still about the set of numbers, regardless of the order they were picked in. But in the case of replacement, you can have duplicates, like (1,1), which aren't consecutive. So, perhaps I need to adjust my approach.Alternatively, maybe I should think of it as ordered pairs because with replacement, the two selections are independent, so each outcome is an ordered pair. So, the total number of possible outcomes is 25. Now, how many of these ordered pairs consist of consecutive integers?Let me list all the ordered pairs where the two numbers are consecutive. Starting with 1, the next number can be 2, so (1,2). Similarly, starting with 2, the next can be 1 or 3, so (2,1) and (2,3). For 3, it can be 2 or 4, so (3,2) and (3,4). For 4, it can be 3 or 5, so (4,3) and (4,5). For 5, the next can be 4, so (5,4). So, listing these out: (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4). That's 8 ordered pairs.So, the number of favorable outcomes is 8, and the total number of possible outcomes is 25. Therefore, the probability with replacement is 8/25.Wait, but in the first part, we had 4 favorable outcomes out of 10, which simplified to 2/5. In the second part, it's 8/25. That makes sense because with replacement, there are more possible outcomes, but also more ways to get consecutive numbers since order matters.Let me double-check my reasoning. For without replacement, we have 10 possible pairs, 4 of which are consecutive. So, 4/10 = 2/5. For with replacement, we have 25 possible ordered pairs, and 8 of them are consecutive. So, 8/25. That seems correct.I think I got it. So, summarizing:1. Without replacement: Probability = 2/5.2. With replacement: Probability = 8/25.Yeah, that makes sense. I don't think I made any mistakes here. The key was recognizing that without replacement, we're dealing with combinations, while with replacement, it's permutations with possible duplicates, leading to more total outcomes and more favorable outcomes due to considering order.

💡First, I need to simplify each term in the expression separately.For the term (0.064^{- frac{1}{3}}), I recognize that (0.064) is equal to (0.4^3). Using the power of a power property, this simplifies to (0.4^{-1}), which is (frac{1}{0.4}) or (frac{10}{4}).Next, the term ((- frac{7}{8})^{0}) is straightforward because any non-zero number raised to the power of 0 is 1.For the term (16^{0.75}), I know that (16) is (2^4). Applying the power of a power property, this becomes (2^{4 times 0.75}), which simplifies to (2^3) or 8.Lastly, the term (0.25^{frac{1}{2}}) is the square root of 0.25. Since (0.25) is (frac{1}{4}), the square root is (frac{1}{2}).Now, I combine all these simplified terms:[frac{10}{4} - 1 + 8 + frac{1}{2}]Converting (frac{10}{4}) to a decimal gives 2.5. Adding and subtracting the terms:[2.5 - 1 + 8 + 0.5 = 10]

💡First, I need to find the radius of the cone. Since the diameter is 12 cm, the radius is half of that, which is 6 cm.Next, I'll use the formula for the volume of a cone, which is ( V = frac{1}{3} pi r^2 h ). Plugging in the values, I get:( V = frac{1}{3} pi (6)^2 (8) )Calculating the squared radius:( 6^2 = 36 )Then, multiply by the height:( 36 times 8 = 288 )Finally, divide by 3:( frac{288}{3} = 96 )So, the volume of the cone is ( 96pi ) cubic centimeters.

💡Okay, so I have this integral to solve: the integral from 0 to 3 of the square root of (9 minus x squared) dx. Hmm, let me think about how to approach this. I remember that integrals involving square roots of quadratic expressions often relate to geometric shapes, maybe circles or something like that. First, let me write down the integral to make it clearer:∫₀³ √(9 - x²) dxAlright, so the function inside the integral is √(9 - x²). I recall that the equation y = √(9 - x²) represents the upper half of a circle with radius 3 centered at the origin. Because if I square both sides, I get y² = 9 - x², which rearranges to x² + y² = 9. That's definitely the equation of a circle with radius 3.So, the integral from 0 to 3 of √(9 - x²) dx is essentially finding the area under the curve of this upper semicircle from x = 0 to x = 3. Since the circle is symmetric, the area from 0 to 3 would be a quarter of the entire circle's area. Wait, let me make sure about that. The entire circle has an area of πr², which in this case is π*(3)² = 9π. If I take the upper half, that's a semicircle, so its area is (1/2)*9π = (9π)/2. But since we're integrating from 0 to 3, which is only the right half of the semicircle, that should be a quarter of the entire circle, right? So, a quarter of 9π is (9π)/4.But hold on, let me double-check that reasoning. If I consider the entire circle, it's symmetric in all four quadrants. So, the area in the first quadrant (where both x and y are positive) is indeed a quarter of the circle. Since we're integrating from 0 to 3, which is along the x-axis in the first quadrant, the area under the curve should be that quarter-circle area.Alternatively, I could approach this integral using substitution. Maybe a trigonometric substitution since we have a square root of a quadratic. Let's see, if I let x = 3 sin θ, then dx = 3 cos θ dθ. Let me try substituting that in.So, substituting, when x = 0, sin θ = 0, so θ = 0. When x = 3, sin θ = 1, so θ = π/2. So, the limits of integration change from 0 to π/2.Now, substituting into the integral:∫₀³ √(9 - x²) dx = ∫₀^(π/2) √(9 - (3 sin θ)²) * 3 cos θ dθSimplify inside the square root:√(9 - 9 sin² θ) = √(9(1 - sin² θ)) = √(9 cos² θ) = 3 cos θSo, the integral becomes:∫₀^(π/2) 3 cos θ * 3 cos θ dθ = ∫₀^(π/2) 9 cos² θ dθHmm, okay, so now I have 9 times the integral of cos squared theta from 0 to pi over 2. I remember that the integral of cos squared can be simplified using a power-reduction formula. The formula is:cos² θ = (1 + cos 2θ)/2So, substituting that in:9 ∫₀^(π/2) (1 + cos 2θ)/2 dθ = (9/2) ∫₀^(π/2) (1 + cos 2θ) dθNow, split the integral:(9/2) [ ∫₀^(π/2) 1 dθ + ∫₀^(π/2) cos 2θ dθ ]Compute each integral separately.First integral: ∫₀^(π/2) 1 dθ = [θ]₀^(π/2) = π/2 - 0 = π/2Second integral: ∫₀^(π/2) cos 2θ dθ. Let me make a substitution here. Let u = 2θ, so du = 2 dθ, which means dθ = du/2. When θ = 0, u = 0. When θ = π/2, u = π.So, the integral becomes:∫₀^π cos u * (du/2) = (1/2) ∫₀^π cos u du = (1/2) [sin u]₀^π = (1/2)(sin π - sin 0) = (1/2)(0 - 0) = 0So, the second integral is 0.Putting it all together:(9/2) [ π/2 + 0 ] = (9/2)(π/2) = 9π/4Okay, so that confirms it. Whether I think about it geometrically or compute it using substitution, I end up with the same result: 9π/4.I guess another way to think about it is recognizing that the integral of √(a² - x²) from 0 to a is always (π a²)/4, which is the area of a quarter-circle. So, in this case, a is 3, so it's (π * 3²)/4 = 9π/4.Just to make sure I didn't make any mistakes in the substitution method, let me go through the steps again quickly.1. Let x = 3 sin θ, so dx = 3 cos θ dθ.2. When x = 0, θ = 0; when x = 3, θ = π/2.3. Substitute into the integral: √(9 - 9 sin² θ) = 3 cos θ.4. The integral becomes 9 ∫ cos² θ dθ from 0 to π/2.5. Use the power-reduction formula: cos² θ = (1 + cos 2θ)/2.6. Integrate term by term: (9/2)(θ + (sin 2θ)/2) evaluated from 0 to π/2.7. At π/2: θ = π/2, sin 2θ = sin π = 0.8. At 0: θ = 0, sin 0 = 0.9. So, the result is (9/2)(π/2) = 9π/4.Yep, that all checks out. I think I'm confident that the answer is 9π/4.**Final Answer**The value of the integral is boxed{dfrac{9pi}{4}}.