answer：First, identify the total number of balls in the bag, which is 7.Next, determine the number of red balls, which is 2.The probability of drawing a red ball is the ratio of the number of red balls to the total number of balls.Therefore, the probability is 2 divided by 7, which simplifies to 2/7.

answer：Okay, so I have these two algebraic expressions to simplify, and I need to figure out how to do it step by step. Let me start with the first one: 5a² - 7 + 4a - 2a² - 9a + 3. Hmm, I remember that simplifying means combining like terms. Like terms are terms that have the same variable raised to the same power. So, I need to look for terms with a², terms with a, and constant numbers.First, let's identify the like terms in the first expression. I see 5a² and -2a². Those are like terms because they both have a squared. Then, I have 4a and -9a, which are like terms because they both have a to the first power. Finally, there are the constants: -7 and +3. Okay, so let's combine them one by one. For the a² terms: 5a² minus 2a². That should be 5 minus 2, which is 3, so 3a². Next, for the a terms: 4a minus 9a. That's 4 minus 9, which is -5, so -5a. Lastly, the constants: -7 plus 3. That's -4. Putting it all together, the simplified form should be 3a² - 5a - 4. I think that's it for the first one. Let me double-check to make sure I didn't miss any terms or make a calculation error. 5a² - 2a² is definitely 3a². 4a - 9a is -5a, and -7 + 3 is -4. Yep, that looks correct.Now, moving on to the second expression: (5x² - 6x) - 3(2x² - 3x). This one looks a bit trickier because there's a distributive property involved. I need to remember to distribute the -3 across both terms inside the parentheses. So, starting with the first part, (5x² - 6x), I can leave that as is for now. Then, I have to distribute the -3 to both 2x² and -3x inside the second parentheses. Let's do that step by step. First, -3 times 2x². That should be -6x². Then, -3 times -3x. Multiplying two negatives gives a positive, so that's +9x. So now, the expression becomes 5x² - 6x - 6x² + 9x. Now, I need to combine like terms again. Let's look for the x² terms and the x terms. The x² terms are 5x² and -6x². Combining those: 5x² minus 6x² is -1x², or just -x². Next, the x terms: -6x and +9x. Combining those: -6x plus 9x is 3x. So, putting it all together, the simplified form is -x² + 3x. Wait a second, let me make sure I did that correctly. Distribute the -3: -3*2x² is -6x², and -3*(-3x) is +9x. Then, combining 5x² - 6x² gives -x², and -6x + 9x gives 3x. Yeah, that seems right. I think I've got both expressions simplified correctly. The first one is 3a² - 5a - 4, and the second one is -x² + 3x. I should probably write them in the standard form, which usually means writing the terms in descending order of exponents. For the first expression, 3a² is already first, followed by -5a, and then the constant -4. For the second expression, -x² is first, then +3x. Just to be thorough, let me check if there are any other like terms or if I missed any signs. In the first expression, all the a² terms are combined, all the a terms are combined, and the constants are combined. No other like terms left. In the second expression, after distributing, I combined the x² terms and the x terms correctly. I think I'm confident with these answers. Simplifying expressions is all about carefully combining like terms and making sure the signs are correct, especially when distributing negative signs. It's easy to make a mistake with the signs, but as long as I take it step by step, I should be fine.

answer：First, identify the number of visitors from yesterday, which is 247 people.Next, calculate today's visitors by adding 131 more people to yesterday's count:247 + 131 = 378.Finally, add yesterday's and today's visitors to find the total number of people who entered the museum over the two days:247 + 378 = 625.

answer：Okay, so I have this vector problem here, and I need to figure out which expression can't be simplified to (overrightarrow{PQ}). The options are A, B, C, and D. Let me go through each one step by step.Starting with option A: (overrightarrow{AB} + (overrightarrow{PA} + overrightarrow{BQ})). Hmm, let's break this down. First, I know that vector addition is associative, so I can rearrange the terms. If I group (overrightarrow{PA}) and (overrightarrow{AB}) together, that should give me (overrightarrow{PB}), right? Because moving from P to A and then from A to B is the same as moving directly from P to B. So, (overrightarrow{PA} + overrightarrow{AB} = overrightarrow{PB}). Then, adding (overrightarrow{BQ}) to that, I get (overrightarrow{PB} + overrightarrow{BQ}). But (overrightarrow{PB} + overrightarrow{BQ}) is just (overrightarrow{PQ}), since you're moving from P to B and then B to Q, which is the same as P to Q. So, option A simplifies to (overrightarrow{PQ}). Got it.Moving on to option B: ((overrightarrow{AB} + overrightarrow{PC}) + (overrightarrow{BA} - overrightarrow{QC})). Let me simplify this. First, notice that (overrightarrow{AB} + overrightarrow{BA}) is zero because they are equal in magnitude but opposite in direction. So, those two cancel each other out. That leaves me with (overrightarrow{PC} - overrightarrow{QC}). Hmm, (overrightarrow{PC} - overrightarrow{QC}) can be rewritten as (overrightarrow{PC} + overrightarrow{CQ}) because subtracting (overrightarrow{QC}) is the same as adding (overrightarrow{CQ}). Now, (overrightarrow{PC} + overrightarrow{CQ}) is just (overrightarrow{PQ}), since you're moving from P to C and then C to Q, which is the same as P to Q. So, option B also simplifies to (overrightarrow{PQ}).Next, option C: (overrightarrow{QC} - overrightarrow{QP} + overrightarrow{CQ}). Let me see. First, (overrightarrow{QC}) and (overrightarrow{CQ}) are opposites, right? So, (overrightarrow{QC} + overrightarrow{CQ} = 0). That leaves me with (-overrightarrow{QP}). But (-overrightarrow{QP}) is the same as (overrightarrow{PQ}), because it's just the vector in the opposite direction. So, option C simplifies to (overrightarrow{PQ}) as well.Now, onto option D: (overrightarrow{PA} + overrightarrow{AB} - overrightarrow{BQ}). Let me try to simplify this. First, (overrightarrow{PA} + overrightarrow{AB}) is (overrightarrow{PB}), similar to option A. So, that gives me (overrightarrow{PB} - overrightarrow{BQ}). Hmm, (overrightarrow{PB} - overrightarrow{BQ}) doesn't seem to simplify directly to (overrightarrow{PQ}). Let me think about the geometry here. If I have (overrightarrow{PB}) and I subtract (overrightarrow{BQ}), what does that represent? It's like moving from P to B and then subtracting the vector from B to Q. That would be equivalent to moving from P to B and then from Q to B, which doesn't directly give me the vector from P to Q. Wait, maybe I can visualize this. If I have points P, A, B, and Q, and I'm trying to get from P to Q. In option D, I'm going from P to A, then A to B, which gets me to B, and then subtracting the vector from B to Q. So, instead of going from B to Q, I'm going from Q to B. That would place me at a point that's not Q but somewhere else. So, it doesn't seem like this simplifies to (overrightarrow{PQ}). Let me double-check. If I have (overrightarrow{PA} + overrightarrow{AB}), that's definitely (overrightarrow{PB}). Then, subtracting (overrightarrow{BQ}) is like adding (-overrightarrow{BQ}), which is (overrightarrow{QB}). So, (overrightarrow{PB} + overrightarrow{QB}). But (overrightarrow{PB} + overrightarrow{QB}) doesn't simplify to (overrightarrow{PQ}). Instead, it's like moving from P to B and then from Q to B, which doesn't result in a direct vector from P to Q. So, putting it all together, options A, B, and C all simplify to (overrightarrow{PQ}), but option D doesn't. Therefore, the answer must be D.

answer：Okay, so I have this problem about an ellipse, and I need to find its eccentricity. Let me try to understand the problem step by step.First, the ellipse is given by the equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a > b > 0). The foci of the ellipse are (F_1) and (F_2), with (F_1) being the left focus and (F_2) the right focus. There's a point (P) on the ellipse such that the line segment (PF_2) is perpendicular to the line segment (F_1F_2), and the angle (angle PF_1F_2) is (30^circ). I need to find the eccentricity of the ellipse.Alright, let me recall some properties of ellipses. The standard form is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a) is the semi-major axis and (b) is the semi-minor axis. The distance between the center and each focus is (c), where (c^2 = a^2 - b^2). The eccentricity (e) is given by (e = frac{c}{a}).So, my goal is to find (e = frac{c}{a}). To do that, I need to relate (c) and (a) somehow using the given conditions.Given that (PF_2 perp F_1F_2), this means that the line segment from (P) to (F_2) is perpendicular to the major axis (F_1F_2). So, if I imagine the ellipse, (F_1F_2) is the major axis, which is along the x-axis in this case since the equation is in terms of (x^2) and (y^2). Therefore, (PF_2) must be a vertical line segment from (F_2).Let me sketch this mentally. The major axis is horizontal, from (F_1) to (F_2). Point (P) is somewhere on the ellipse such that when you draw a line from (P) to (F_2), it is vertical. So, (P) must be directly above or below (F_2). Since the angle (angle PF_1F_2) is (30^circ), which is an acute angle, I think (P) is above (F_2), forming a triangle with (F_1) and (F_2).Let me denote the coordinates of the foci. Since the ellipse is centered at the origin (because the equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1)), the foci are at ((-c, 0)) and ((c, 0)). So, (F_1 = (-c, 0)) and (F_2 = (c, 0)).Point (P) is on the ellipse, and (PF_2) is vertical, so the x-coordinate of (P) is the same as that of (F_2), which is (c). Therefore, the coordinates of (P) are ((c, y)), where (y) is the y-coordinate we need to find.Since (P) lies on the ellipse, plugging into the ellipse equation:[frac{c^2}{a^2} + frac{y^2}{b^2} = 1]So,[frac{y^2}{b^2} = 1 - frac{c^2}{a^2}]But (c^2 = a^2 - b^2), so substituting:[frac{y^2}{b^2} = 1 - frac{a^2 - b^2}{a^2} = 1 - 1 + frac{b^2}{a^2} = frac{b^2}{a^2}]Therefore,[y^2 = frac{b^4}{a^2}]So,[y = frac{b^2}{a}]Thus, the coordinates of (P) are ((c, frac{b^2}{a})).Now, let's consider the triangle (PF_1F_2). We know that (angle PF_1F_2 = 30^circ). Let me visualize this triangle. Points (F_1), (F_2), and (P) form a triangle with (F_1) at ((-c, 0)), (F_2) at ((c, 0)), and (P) at ((c, frac{b^2}{a})).So, the sides of the triangle are:1. (F_1F_2): This is the distance between (F_1) and (F_2), which is (2c).2. (PF_2): This is the vertical distance from (P) to (F_2), which is (frac{b^2}{a}).3. (PF_1): This is the distance from (P) to (F_1). Let me compute this distance.Using the distance formula between (P(c, frac{b^2}{a})) and (F_1(-c, 0)):[PF_1 = sqrt{(c - (-c))^2 + left(frac{b^2}{a} - 0right)^2} = sqrt{(2c)^2 + left(frac{b^2}{a}right)^2} = sqrt{4c^2 + frac{b^4}{a^2}}]So, now we have the lengths of all sides of triangle (PF_1F_2):- (F_1F_2 = 2c)- (PF_2 = frac{b^2}{a})- (PF_1 = sqrt{4c^2 + frac{b^4}{a^2}})Given that (angle PF_1F_2 = 30^circ), we can use the Law of Sines or the Law of Cosines to relate the sides and the angle.Let me think about which one would be more straightforward. Since we know one angle and some sides, maybe the Law of Sines would work.In triangle (PF_1F_2), the angle at (F_1) is (30^circ). The sides opposite to the angles are:- Opposite to (angle PF_1F_2 = 30^circ) is (PF_2 = frac{b^2}{a})- Opposite to (angle PF_2F_1) is (PF_1 = sqrt{4c^2 + frac{b^4}{a^2}})- Opposite to (angle F_1PF_2) is (F_1F_2 = 2c)Wait, but I don't know the other angles. Maybe the Law of Sines is not directly helpful here. Alternatively, maybe using the Law of Cosines on the angle (30^circ) would be better.In triangle (PF_1F_2), applying the Law of Cosines at angle (F_1):[(PF_2)^2 = (PF_1)^2 + (F_1F_2)^2 - 2 cdot PF_1 cdot F_1F_2 cdot cos(30^circ)]Plugging in the known values:[left(frac{b^2}{a}right)^2 = left(sqrt{4c^2 + frac{b^4}{a^2}}right)^2 + (2c)^2 - 2 cdot sqrt{4c^2 + frac{b^4}{a^2}} cdot 2c cdot cos(30^circ)]Simplify each term:Left-hand side (LHS):[left(frac{b^2}{a}right)^2 = frac{b^4}{a^2}]Right-hand side (RHS):First term: (left(sqrt{4c^2 + frac{b^4}{a^2}}right)^2 = 4c^2 + frac{b^4}{a^2})Second term: ((2c)^2 = 4c^2)Third term: (2 cdot sqrt{4c^2 + frac{b^4}{a^2}} cdot 2c cdot cos(30^circ) = 4c cdot sqrt{4c^2 + frac{b^4}{a^2}} cdot frac{sqrt{3}}{2} = 2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}})Putting it all together:[frac{b^4}{a^2} = left(4c^2 + frac{b^4}{a^2}right) + 4c^2 - 2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}}]Simplify RHS:[4c^2 + frac{b^4}{a^2} + 4c^2 - 2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}} = 8c^2 + frac{b^4}{a^2} - 2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}}]So, equation becomes:[frac{b^4}{a^2} = 8c^2 + frac{b^4}{a^2} - 2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}}]Subtract (frac{b^4}{a^2}) from both sides:[0 = 8c^2 - 2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}}]Let me rearrange this:[2sqrt{3}c cdot sqrt{4c^2 + frac{b^4}{a^2}} = 8c^2]Divide both sides by (2c) (assuming (c neq 0), which it isn't because (a > b > 0)):[sqrt{3} cdot sqrt{4c^2 + frac{b^4}{a^2}} = 4c]Square both sides to eliminate the square roots:[3 cdot left(4c^2 + frac{b^4}{a^2}right) = 16c^2]Expand the left side:[12c^2 + frac{3b^4}{a^2} = 16c^2]Subtract (12c^2) from both sides:[frac{3b^4}{a^2} = 4c^2]Multiply both sides by (frac{a^2}{3}):[b^4 = frac{4c^2a^2}{3}]Take square roots on both sides:[b^2 = frac{2ca}{sqrt{3}}]But we know from the relationship in an ellipse that (c^2 = a^2 - b^2). Let me substitute (b^2) from the above equation into this.So,[c^2 = a^2 - frac{2ca}{sqrt{3}}]Hmm, this seems a bit complicated. Let me see if I can express (c) in terms of (a). Let me denote (e = frac{c}{a}), which is the eccentricity we need to find. Then, (c = ea), and (b^2 = a^2(1 - e^2)).Wait, but from earlier, we have (b^2 = frac{2ca}{sqrt{3}}). Substituting (c = ea):[b^2 = frac{2ea cdot a}{sqrt{3}} = frac{2e a^2}{sqrt{3}}]But we also have (b^2 = a^2(1 - e^2)). Therefore,[a^2(1 - e^2) = frac{2e a^2}{sqrt{3}}]Divide both sides by (a^2):[1 - e^2 = frac{2e}{sqrt{3}}]Rearrange the equation:[e^2 + frac{2e}{sqrt{3}} - 1 = 0]This is a quadratic equation in terms of (e). Let me write it as:[e^2 + frac{2}{sqrt{3}} e - 1 = 0]Let me solve for (e) using the quadratic formula. For an equation (ax^2 + bx + c = 0), the solutions are:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Here, (a = 1), (b = frac{2}{sqrt{3}}), and (c = -1). Plugging into the formula:[e = frac{ -frac{2}{sqrt{3}} pm sqrt{left(frac{2}{sqrt{3}}right)^2 - 4 cdot 1 cdot (-1)}}{2 cdot 1}]Compute discriminant:[left(frac{2}{sqrt{3}}right)^2 - 4(1)(-1) = frac{4}{3} + 4 = frac{4}{3} + frac{12}{3} = frac{16}{3}]So,[e = frac{ -frac{2}{sqrt{3}} pm sqrt{frac{16}{3}} }{2}]Simplify square root:[sqrt{frac{16}{3}} = frac{4}{sqrt{3}}]So,[e = frac{ -frac{2}{sqrt{3}} pm frac{4}{sqrt{3}} }{2}]This gives two solutions:1. (e = frac{ -frac{2}{sqrt{3}} + frac{4}{sqrt{3}} }{2} = frac{ frac{2}{sqrt{3}} }{2} = frac{1}{sqrt{3}} approx 0.577)2. (e = frac{ -frac{2}{sqrt{3}} - frac{4}{sqrt{3}} }{2} = frac{ -frac{6}{sqrt{3}} }{2} = -frac{3}{sqrt{3}} = -sqrt{3})Since eccentricity is always positive and less than 1 for an ellipse, we discard the negative solution. Therefore, (e = frac{1}{sqrt{3}}).But (frac{1}{sqrt{3}}) can be rationalized as (frac{sqrt{3}}{3}), which is approximately 0.577. Looking back at the options given:A: (frac{sqrt{3}}{6}) ≈ 0.288B: (frac{1}{3}) ≈ 0.333C: (frac{1}{2}) = 0.5D: (frac{sqrt{3}}{3}) ≈ 0.577So, the correct answer is D.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I found that (P) is at ((c, frac{b^2}{a})).2. Calculated the distance (PF_1) correctly.3. Applied the Law of Cosines correctly on the angle at (F_1).4. Solved the equation step by step, leading to a quadratic in (e).5. Solved the quadratic and found (e = frac{sqrt{3}}{3}).Everything seems to check out. So, I think the answer is indeed D.**Final Answer**boxed{D}

answer：First, I need to determine the rate of interest for the first year, which I'll denote as ( R% ).For the first year, the interest on Rs. 9000 at ( R% ) is calculated as:[text{Interest for the first year} = frac{9000 times R times 1}{100} = 90R]The amount after the first year is:[text{Amount after the first year} = 9000 + 90R]For the second year, the interest rate is 5%. The interest for the second year is:[text{Interest for the second year} = (9000 + 90R) times frac{5}{100} = (9000 + 90R) times 0.05]The total amount after two years is:[text{Total amount} = 9000 + 90R + (9000 + 90R) times 0.05]Given that the total amount after two years is Rs. 9828, I can set up the equation:[9000 + 90R + (9000 + 90R) times 0.05 = 9828]Expanding and simplifying the equation:[9000 + 90R + 450 + 4.5R = 9828]Combining like terms:[94.5R = 378]Solving for ( R ):[R = frac{378}{94.5} = 4]Therefore, the rate of interest for the first year is ( 4% ).