💡First, determine the rate at which the cows produce milk. If ( x ) cows produce ( y ) gallons of milk in ( z ) days, then the total milk production per day for all ( x ) cows is ( frac{y}{z} ) gallons. Therefore, the milk production per cow per day is ( frac{y}{zx} ) gallons.Next, account for the daily waste of 10%. This means that only 90% of the milk is effectively collected each day. Thus, the effective milk production per cow per day becomes ( 0.9 times frac{y}{zx} ) gallons.Now, scale this production to ( w ) cows and ( v ) days. If one cow produces ( 0.9 times frac{y}{zx} ) gallons of milk per day, then ( w ) cows will produce ( w times 0.9 times frac{y}{zx} = 0.9 times frac{wy}{zx} ) gallons of milk per day. Over ( v ) days, the total milk production will be ( 0.9 times frac{wy}{zx} times v = 0.9 times frac{wyv}{zx} ) gallons.Therefore, the total amount of milk produced by ( w ) cows in ( v ) days, after accounting for the daily waste, is ( 0.9 times frac{wyv}{zx} ) gallons.

💡Okay, so I have this problem about an ellipse and a line passing through its left focus. I need to find the value of m and then determine the maximum and minimum values of the product |FA|·|FB| where A and B are the intersection points of the line and the ellipse. Hmm, let me break this down step by step.First, part (I): Determine the value of m.The ellipse is given parametrically as:x = 2 cos φy = √3 sin φI remember that parametric equations of an ellipse can be converted to the standard form. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis and b is the semi-minor axis. From the parametric equations, I can see that when φ = 0, x = 2 and y = 0, so that's the rightmost point. When φ = π/2, x = 0 and y = √3, so that's the topmost point. Therefore, a = 2 and b = √3.Now, the foci of an ellipse are located at (±c, 0) where c² = a² - b². Let me compute c:c² = a² - b² = 2² - (√3)² = 4 - 3 = 1So, c = 1. Therefore, the foci are at (-1, 0) and (1, 0). The left focus F is at (-1, 0).Now, the line l is given parametrically as:x = m + t cos αy = t sin αThis is a straight line with direction determined by α, and parameter t. Since the line passes through the left focus F, which is (-1, 0), we can substitute F into the parametric equations to find m.So, substituting x = -1 and y = 0 into the parametric equations:-1 = m + t cos α0 = t sin αFrom the second equation, 0 = t sin α. So, either t = 0 or sin α = 0.If sin α = 0, then the line is horizontal. But regardless, since the line passes through F, let's consider t = 0.If t = 0, then from the first equation:-1 = m + 0 => m = -1So, m must be -1. That seems straightforward.Wait, but does this hold for any t? Let me think. If the line passes through F, then there must exist some t such that x = -1 and y = 0. So, substituting y = 0 gives t sin α = 0. So, either t = 0 or sin α = 0.If sin α ≠ 0, then t must be 0. Plugging t = 0 into x gives x = m. Since x must be -1, m = -1.If sin α = 0, then the line is horizontal, and y = 0 for all t. So, in that case, the line is the x-axis. Since it passes through F, which is (-1, 0), m must still be -1 because when t = 0, x = m, which is -1.So, regardless of α, m must be -1. Okay, so part (I) is solved, m = -1.Now, part (II): Find the maximum and minimum values of |FA|·|FB| where A and B are the intersection points of line l and ellipse C.First, let me write down the equations again.Ellipse C:x = 2 cos φy = √3 sin φWhich, as I found earlier, is equivalent to (x²)/4 + (y²)/3 = 1.Line l:x = m + t cos α = -1 + t cos αy = t sin αSo, substituting m = -1, the parametric equations of the line are:x = -1 + t cos αy = t sin αI need to find the points where this line intersects the ellipse. So, substitute x and y from the line into the ellipse equation.Let me do that.Substitute x = -1 + t cos α and y = t sin α into (x²)/4 + (y²)/3 = 1.So,[(-1 + t cos α)²]/4 + [(t sin α)²]/3 = 1Let me expand this.First, expand (-1 + t cos α)²:= (1 - 2 t cos α + t² cos² α)So,[1 - 2 t cos α + t² cos² α]/4 + [t² sin² α]/3 = 1Multiply through:(1/4) - (2 t cos α)/4 + (t² cos² α)/4 + (t² sin² α)/3 = 1Simplify each term:1/4 - (t cos α)/2 + (t² cos² α)/4 + (t² sin² α)/3 = 1Now, let's collect like terms.First, the constant term: 1/4Linear term: - (t cos α)/2Quadratic terms: (t² cos² α)/4 + (t² sin² α)/3So, let's write the equation as:[ (cos² α)/4 + (sin² α)/3 ] t² - (cos α)/2 t + 1/4 - 1 = 0Simplify the constants:1/4 - 1 = -3/4So, the quadratic equation becomes:[ (cos² α)/4 + (sin² α)/3 ] t² - (cos α)/2 t - 3/4 = 0Let me denote the coefficients:A = (cos² α)/4 + (sin² α)/3B = - (cos α)/2C = -3/4So, the quadratic is A t² + B t + C = 0We can write this as:[ (cos² α)/4 + (sin² α)/3 ] t² - (cos α)/2 t - 3/4 = 0Now, to find the points of intersection, we need to solve for t. Let me denote t1 and t2 as the roots of this quadratic equation.From quadratic equation theory, we know that:t1 + t2 = -B/A = [ (cos α)/2 ] / [ (cos² α)/4 + (sin² α)/3 ]andt1 t2 = C/A = ( -3/4 ) / [ (cos² α)/4 + (sin² α)/3 ]But in our case, we are interested in |FA|·|FB|. Since F is the point (-1, 0), which is on the line l, the parameter t corresponds to the distance from F to any point on the line. So, when t = 0, we are at F. Therefore, the parameters t1 and t2 correspond to the distances from F to points A and B.Wait, but actually, in parametric equations, t is a parameter, not necessarily the distance. So, the actual distance |FA| would be the Euclidean distance between F and A, which can be expressed in terms of t.But since F is on the line l, and the parametric equations are given with t as a parameter, the distance from F to a point on the line is |t| times the direction vector's magnitude. However, in this case, the direction vector is (cos α, sin α), which is a unit vector because cos² α + sin² α = 1. Therefore, the parameter t corresponds to the actual distance from F to the point on the line.Wait, let me verify that. The parametric equations are:x = -1 + t cos αy = 0 + t sin αSo, the direction vector is (cos α, sin α), which is a unit vector because sqrt( (cos α)^2 + (sin α)^2 ) = 1. Therefore, t is indeed the signed distance from F to the point (x, y). So, |FA| is |t1| and |FB| is |t2|. Therefore, |FA|·|FB| = |t1 t2|.But in the quadratic equation, t1 t2 = C/A = (-3/4)/A. So, |t1 t2| = |C/A| = | (-3/4)/A | = (3/4)/A because A is positive.So, |FA|·|FB| = (3/4)/ABut A = (cos² α)/4 + (sin² α)/3So, let me write that:|FA|·|FB| = (3/4) / [ (cos² α)/4 + (sin² α)/3 ]Let me simplify the denominator:(cos² α)/4 + (sin² α)/3 = (3 cos² α + 4 sin² α)/12Therefore,|FA|·|FB| = (3/4) / [ (3 cos² α + 4 sin² α)/12 ] = (3/4) * (12)/(3 cos² α + 4 sin² α) = (9)/(3 cos² α + 4 sin² α)So, |FA|·|FB| = 9 / (3 cos² α + 4 sin² α)Now, I need to find the maximum and minimum values of this expression as α varies.Let me denote D = 3 cos² α + 4 sin² αSo, |FA|·|FB| = 9 / DTherefore, to find the extrema of 9/D, I can find the extrema of D and then take reciprocals.So, let's analyze D = 3 cos² α + 4 sin² αI can write this as:D = 3 cos² α + 4 sin² α = 3 (cos² α + sin² α) + sin² α = 3 + sin² αBecause cos² α + sin² α = 1, so 3 cos² α + 4 sin² α = 3 (1 - sin² α) + 4 sin² α = 3 - 3 sin² α + 4 sin² α = 3 + sin² αSo, D = 3 + sin² αTherefore, |FA|·|FB| = 9 / (3 + sin² α)Now, sin² α varies between 0 and 1. Therefore, D varies between 3 and 4.Hence, 9/D varies between 9/4 and 9/3, which is 9/4 and 3.Therefore, the maximum value of |FA|·|FB| is 3, and the minimum value is 9/4.Wait, let me double-check that.When sin² α = 0, D = 3, so |FA|·|FB| = 9/3 = 3.When sin² α = 1, D = 4, so |FA|·|FB| = 9/4.Yes, that seems correct.So, the maximum value is 3, and the minimum value is 9/4.Therefore, the answer to part (II) is that |FA|·|FB| has a maximum value of 3 and a minimum value of 9/4.Let me just recap to make sure I didn't make any mistakes.1. Converted the parametric ellipse to standard form, found a, b, c, and the foci.2. Substituted the focus into the line equation to find m = -1.3. Substituted the parametric line into the ellipse equation, expanded, and formed a quadratic in t.4. Recognized that |FA|·|FB| is |t1 t2|, which from quadratic equation is |C/A|.5. Simplified the expression and found that it depends on sin² α.6. Analyzed the range of sin² α to find the extrema.Everything seems to check out. I think I'm confident with this solution.

💡First, I need to determine the total number of teachers at the university. There are 120 full professors, 100 associate professors, 80 lecturers, and 60 teaching assistants, which adds up to 360 teachers in total.Next, I know that 16 lecturers were selected in the sample. Since the sampling is stratified, the proportion of lecturers in the sample should be the same as their proportion in the entire population. The proportion of lecturers in the total population is 80 out of 360. To find the total sample size ( n ), I set up the equation:[frac{16}{n} = frac{80}{360}]Solving for ( n ), I cross-multiply:[16 times 360 = 80 times n][5760 = 80n][n = frac{5760}{80} = 72]Therefore, the total sample size ( n ) is 72.

💡First, I need to determine how many segments of 2/5 of a mile are in the total trip distance of 3.6 miles. To do this, I divide 3.6 miles by 2/5 miles per segment.Next, I calculate the total cost for the distance traveled by multiplying the number of segments by the charge per segment, which is 0.35.Finally, I subtract the total cost of the distance from the overall trip charge to find the initial fee.

💡Okay, so I have this geometry problem here. Let me try to visualize it first. There's a triangle ABC, with points D on AC and F on AB. Lines AD and CF intersect at point P. I'm given the ratios AP:PD = 3:2 and FP:PC = 3:4. I need to find the ratio AF/FB.Hmm, ratios in triangles... Maybe I can use mass point geometry or coordinate geometry. I remember mass point is useful for such ratio problems. Let me try that.First, let's recall mass point. It assigns masses to the vertices such that the masses are inversely proportional to the lengths. So, if AP:PD is 3:2, then the mass at A should be 2 and at D should be 3? Wait, no, actually, mass points are inversely proportional, so if AP:PD is 3:2, then mass at A is 2 and mass at D is 3. That way, the mass at P would be 2 + 3 = 5.Similarly, for FP:PC = 3:4, so the mass at F is 4 and at C is 3, making the mass at P equal to 4 + 3 = 7.Wait, but P has two different masses from the two different cevians. That can't be right. Maybe I need to adjust the masses so that the mass at P is consistent.So, from AD, mass at P is 5, and from CF, mass at P is 7. To make them consistent, I can scale the masses. The least common multiple of 5 and 7 is 35. So, I can multiply the masses from AD by 7 and from CF by 5.So, from AD: mass at A is 2*7=14, mass at D is 3*7=21, mass at P is 5*7=35.From CF: mass at F is 4*5=20, mass at C is 3*5=15, mass at P is 7*5=35.Okay, now mass at P is consistent as 35.Now, mass at D is 21, which is on AC. So, masses at A and C must add up to 21. Wait, mass at A is 14, so mass at C should be 21 - 14 = 7? Wait, no, that's not right. Mass points on a line add up at the point. So, if D is on AC, the masses at A and C should be proportional to the lengths CD:DA.Wait, I think I need to clarify. The mass at A is 14, mass at C is 15 (from CF). Wait, but D is on AC, so masses at A and C should be such that mass at A / mass at C = CD / DA.But wait, mass at A is 14, mass at C is 15, so CD/DA = 14/15? Hmm, but from AD, the ratio AP:PD is 3:2, which relates to the masses. Maybe I'm mixing something up.Alternatively, maybe I should use coordinate geometry. Let me assign coordinates to the triangle. Let me place point A at (0,0), B at (b,0), and C at (c,d). Then, points D and F can be expressed in terms of these coordinates.Point D is on AC. Let me parameterize AC. If I let D divide AC in the ratio k:1, so coordinates of D would be ((k*c)/(k+1), (k*d)/(k+1)). Similarly, point F is on AB. Let me say F divides AB in the ratio m:1, so coordinates of F would be ((m*b)/(m+1), 0).Now, lines AD and CF intersect at P. Let me find the coordinates of P by solving the equations of AD and CF.Equation of AD: It goes from A(0,0) to D((k*c)/(k+1), (k*d)/(k+1)). So, parametric equations can be written as x = t*(k*c)/(k+1), y = t*(k*d)/(k+1), where t is a parameter.Equation of CF: It goes from C(c,d) to F((m*b)/(m+1), 0). The parametric equations can be written as x = c - s*(c - (m*b)/(m+1)), y = d - s*d, where s is another parameter.At the intersection point P, these coordinates must be equal. So,t*(k*c)/(k+1) = c - s*(c - (m*b)/(m+1))andt*(k*d)/(k+1) = d - s*d.From the second equation, let's solve for s:t*(k)/(k+1) = 1 - sSo, s = 1 - t*(k)/(k+1)Now, plug this into the first equation:t*(k*c)/(k+1) = c - [1 - t*(k)/(k+1)]*(c - (m*b)/(m+1))Let me simplify the right side:c - (c - (m*b)/(m+1)) + [1 - t*(k)/(k+1)]*(m*b)/(m+1)Wait, that might not be the best approach. Let me instead expand the right side:c - s*(c - (m*b)/(m+1)) = c - [1 - t*(k)/(k+1)]*(c - (m*b)/(m+1))= c - (c - (m*b)/(m+1)) + t*(k)/(k+1)*(c - (m*b)/(m+1))= (m*b)/(m+1) + t*(k)/(k+1)*(c - (m*b)/(m+1))So, now, left side is t*(k*c)/(k+1), right side is (m*b)/(m+1) + t*(k)/(k+1)*(c - (m*b)/(m+1))Let me write this as:t*(k*c)/(k+1) = (m*b)/(m+1) + t*(k*c)/(k+1) - t*(k*m*b)/( (k+1)(m+1) )Subtract t*(k*c)/(k+1) from both sides:0 = (m*b)/(m+1) - t*(k*m*b)/( (k+1)(m+1) )So,(m*b)/(m+1) = t*(k*m*b)/( (k+1)(m+1) )Divide both sides by (m*b)/(m+1):1 = t*k/(k+1)So,t = (k+1)/kWait, but t is a parameter in the parametric equation of AD. If t = (k+1)/k, then that would mean P is beyond D on AD, but according to the ratio AP:PD = 3:2, P should be between A and D. So, t should be less than 1. Hmm, that suggests something is wrong.Wait, maybe my parametrization is off. Let me check.In the parametric equation for AD, I used t from 0 to 1 to go from A to D. Similarly, for CF, s from 0 to 1 goes from C to F. So, if t is (k+1)/k, which is greater than 1, that would mean P is beyond D on AD, which contradicts the given ratio AP:PD = 3:2, meaning P is between A and D.So, perhaps I made a mistake in the algebra. Let me go back.From the second equation:t*(k*d)/(k+1) = d - s*dDivide both sides by d:t*k/(k+1) = 1 - sSo, s = 1 - t*k/(k+1)Now, plug this into the first equation:t*(k*c)/(k+1) = c - [1 - t*k/(k+1)]*(c - (m*b)/(m+1))Let me expand the right side:c - (c - (m*b)/(m+1)) + t*k/(k+1)*(c - (m*b)/(m+1))= (m*b)/(m+1) + t*k/(k+1)*(c - (m*b)/(m+1))So, the equation becomes:t*(k*c)/(k+1) = (m*b)/(m+1) + t*k/(k+1)*(c - (m*b)/(m+1))Let me subtract t*k/(k+1)*(c - (m*b)/(m+1)) from both sides:t*(k*c)/(k+1) - t*k/(k+1)*(c - (m*b)/(m+1)) = (m*b)/(m+1)Factor out t*k/(k+1):t*k/(k+1)*(c - (c - (m*b)/(m+1))) = (m*b)/(m+1)Simplify inside the parentheses:c - c + (m*b)/(m+1) = (m*b)/(m+1)So,t*k/(k+1)*(m*b)/(m+1) = (m*b)/(m+1)Divide both sides by (m*b)/(m+1):t*k/(k+1) = 1So,t = (k+1)/kAgain, same result. Hmm, this suggests that t = (k+1)/k, which is greater than 1, implying P is beyond D on AD, but according to the problem, AP:PD = 3:2, so P is between A and D, so t should be 3/(3+2) = 3/5.Wait, maybe my parametrization is incorrect. Let me try a different approach. Instead of parameterizing AD and CF, maybe I can use the given ratios to express the coordinates of P in two different ways and then equate them.Let me express P in terms of AD first. Since AP:PD = 3:2, P divides AD in the ratio 3:2. So, using section formula, coordinates of P would be:P = ( (2*A + 3*D) ) / (3 + 2) = (2*A + 3*D)/5Similarly, since FP:PC = 3:4, P divides CF in the ratio 3:4. So, coordinates of P would be:P = ( (4*C + 3*F) ) / (3 + 4) = (4*C + 3*F)/7So, now I have two expressions for P:(2*A + 3*D)/5 = (4*C + 3*F)/7Let me write this equation:7*(2*A + 3*D) = 5*(4*C + 3*F)14*A + 21*D = 20*C + 15*FNow, I need to express D and F in terms of A, B, and C.Point D is on AC, so let me express D as a weighted average of A and C. Let me say D divides AC in the ratio m:1, so D = (m*C + A)/(m + 1)Similarly, point F is on AB, so F divides AB in the ratio n:1, so F = (n*B + A)/(n + 1)Now, substitute D and F into the equation:14*A + 21*( (m*C + A)/(m + 1) ) = 20*C + 15*( (n*B + A)/(n + 1) )Let me multiply through by (m + 1)(n + 1) to eliminate denominators:14*A*(m + 1)(n + 1) + 21*(m*C + A)*(n + 1) = 20*C*(m + 1)(n + 1) + 15*(n*B + A)*(m + 1)Now, expand each term:Left side:14*A*(m + 1)(n + 1) = 14A*(mn + m + n + 1)21*(m*C + A)*(n + 1) = 21*(m*C*(n + 1) + A*(n + 1)) = 21m*C*(n + 1) + 21A*(n + 1)Right side:20*C*(m + 1)(n + 1) = 20C*(mn + m + n + 1)15*(n*B + A)*(m + 1) = 15*(n*B*(m + 1) + A*(m + 1)) = 15n*B*(m + 1) + 15A*(m + 1)Now, let's collect like terms.Left side:14A*(mn + m + n + 1) + 21m*C*(n + 1) + 21A*(n + 1)Right side:20C*(mn + m + n + 1) + 15n*B*(m + 1) + 15A*(m + 1)Now, let's bring all terms to the left side:14A*(mn + m + n + 1) + 21m*C*(n + 1) + 21A*(n + 1) - 20C*(mn + m + n + 1) - 15n*B*(m + 1) - 15A*(m + 1) = 0Now, let's expand each term:14A*mn + 14A*m + 14A*n + 14A + 21m*C*n + 21m*C + 21A*n + 21A - 20C*mn - 20C*m - 20C*n - 20C - 15n*B*m - 15n*B - 15A*m - 15A = 0Now, let's group like terms:Terms with A*mn: 14A*mnTerms with A*m: 14A*m - 15A*m = -A*mTerms with A*n: 14A*n + 21A*n = 35A*nTerms with A: 14A + 21A - 15A = 20ATerms with C*mn: 21m*C*n - 20C*mn = (21m - 20)C*mnTerms with C*m: 21m*C - 20C*m = (21m - 20m)C = m*CTerms with C*n: 21m*C*n - 20C*n = (21m - 20)C*nWait, I think I might have messed up the grouping. Let me try again.Looking back:14A*mn + 14A*m + 14A*n + 14A + 21m*C*n + 21m*C + 21A*n + 21A - 20C*mn - 20C*m - 20C*n - 20C - 15n*B*m - 15n*B - 15A*m - 15ANow, group by variables:A terms:14A*mn + (14A*m - 15A*m) + (14A*n + 21A*n) + (14A + 21A - 15A)= 14A*mn - A*m + 35A*n + 20AC terms:21m*C*n + 21m*C - 20C*mn - 20C*m - 20C*n - 20C= (21m - 20)C*mn + (21m - 20m)C + (-20C*n) - 20C= (21m - 20)C*mn + m*C - 20C*n - 20CB terms:-15n*B*m - 15n*B= -15n*m*B - 15n*BSo, putting it all together:14A*mn - A*m + 35A*n + 20A + (21m - 20)C*mn + m*C - 20C*n - 20C -15n*m*B - 15n*B = 0This seems complicated. Maybe I should consider that A, B, C are position vectors, so the coefficients of A, B, C must each be zero for the equation to hold.So, let's set the coefficients of A, B, and C to zero.Coefficient of A:14*mn - m + 35*n + 20 = 0Coefficient of B:-15n*m -15n = 0Coefficient of C:(21m - 20)mn + m - 20n - 20 = 0So, now we have three equations:1) 14mn - m + 35n + 20 = 02) -15mn -15n = 03) (21m - 20)mn + m - 20n - 20 = 0Let me start with equation 2:-15mn -15n = 0Factor out -15n:-15n(m + 1) = 0So, either n = 0 or m = -1But n = 0 would mean F coincides with A, which isn't possible since F is on AB. Similarly, m = -1 would mean D is beyond A on AC, which isn't the case. So, this suggests that my approach might be flawed or I made a mistake in the algebra.Wait, maybe I made a mistake in expanding the terms. Let me double-check the expansion.Looking back at the equation after moving all terms to the left:14A*(mn + m + n + 1) + 21m*C*(n + 1) + 21A*(n + 1) - 20C*(mn + m + n + 1) - 15n*B*(m + 1) - 15A*(m + 1) = 0Wait, perhaps I made a mistake in expanding 21m*C*(n + 1). It should be 21m*C*n + 21m*C, not 21m*C*n + 21m*C*n. Wait, no, 21m*C*(n + 1) is 21m*C*n + 21m*C.Similarly, 21A*(n + 1) is 21A*n + 21A.Similarly, -20C*(mn + m + n + 1) is -20C*mn -20C*m -20C*n -20C.-15n*B*(m + 1) is -15n*B*m -15n*B.-15A*(m + 1) is -15A*m -15A.So, the expansion seems correct.So, back to equation 2: -15n(m + 1) = 0. Since n ≠ 0 and m ≠ -1, this suggests that there's no solution unless I made a wrong assumption.Wait, maybe my initial assumption about the parametrization of D and F is incorrect. Let me try a different approach.Instead of using mass point, maybe I can use Menelaus' theorem or Ceva's theorem.Wait, Ceva's theorem relates the ratios of the cevians. In triangle ABC, if AD, BE, and CF are concurrent, then (AF/FB)*(BD/DC)*(CE/EA) = 1.But in this case, we only have two cevians, AD and CF, intersecting at P. So, maybe I can introduce another cevian to apply Ceva.Alternatively, maybe I can use Menelaus' theorem on triangle ADC with transversal CF.Wait, Menelaus' theorem states that for a triangle, if a line crosses the sides (or their extensions), the product of the segment ratios is equal to 1.Let me try applying Menelaus' theorem to triangle ADC with transversal CF.Wait, triangle ADC: points are A, D, C. The transversal CF intersects AD at P, DC at some point, and AC at F? Wait, no, CF starts at C and goes to F on AB, which is not on AC. Hmm, maybe not the best approach.Alternatively, apply Menelaus to triangle ACF with transversal PD.Wait, not sure. Maybe it's better to use coordinate geometry again but with a different setup.Let me assign coordinates differently. Let me set A at (0,0), B at (1,0), and C at (0,1). So, triangle ABC is a right triangle for simplicity.Then, point D is on AC. Since AC is from (0,0) to (0,1), let me say D is at (0, d), where 0 < d < 1.Point F is on AB, which is from (0,0) to (1,0). Let me say F is at (f, 0), where 0 < f < 1.Now, lines AD and CF intersect at P.Equation of AD: It's the line from (0,0) to (0,d), which is the vertical line x=0. Wait, but that's just the y-axis. But CF is from C(0,1) to F(f,0). Let me find the equation of CF.The slope of CF is (0 - 1)/(f - 0) = -1/f. So, equation is y - 1 = (-1/f)(x - 0), which simplifies to y = (-1/f)x + 1.Intersection point P is where x=0 and y = (-1/f)(0) + 1 = 1. So, P is at (0,1). But that's point C. But according to the problem, P is the intersection of AD and CF, which in this coordinate system is point C. But in the problem, P is inside the triangle, so this suggests that my coordinate choice might not be suitable because in this case, AD is along AC, making P coincide with C, which contradicts the given ratios.So, maybe I should choose a different coordinate system where AD is not along AC. Let me instead place A at (0,0), B at (1,0), and C at (0,1). Then, point D is on AC, which is from (0,0) to (0,1). Let me say D is at (0, d). Point F is on AB, which is from (0,0) to (1,0). Let me say F is at (f, 0).Now, line AD is from (0,0) to (0,d), which is the vertical line x=0. Line CF is from (0,1) to (f,0). The equation of CF is y = (-1/f)x + 1.Intersection point P is where x=0, so y = 1. So, P is at (0,1), which is point C. But in the problem, P is inside the triangle, so this setup doesn't work because AD is along AC, making P coincide with C. Therefore, I need to choose a different coordinate system where AD is not along AC.Let me instead place A at (0,0), B at (1,0), and C at (0,1). Then, point D is on AC, which is from (0,0) to (0,1). Let me say D is at (0, d). Point F is on AB, which is from (0,0) to (1,0). Let me say F is at (f, 0).Now, line AD is from (0,0) to (0,d), which is the vertical line x=0. Line CF is from (0,1) to (f,0). The equation of CF is y = (-1/f)x + 1.Intersection point P is where x=0, so y = 1. So, P is at (0,1), which is point C. Again, same problem. So, maybe I need to place D not on AC but somewhere else? Wait, no, D is on AC by the problem statement.Wait, perhaps I should not place C at (0,1). Let me instead place C at (c,0), making ABC a triangle with A at (0,0), B at (1,0), and C at (c,0). Wait, but then AC is from (0,0) to (c,0), which is along the x-axis, same as AB. That would make ABC a degenerate triangle. Not good.Alternatively, place C at (0,1), A at (0,0), B at (1,0), and D on AC at (0,d). Then, line AD is x=0, and CF is from (0,1) to (f,0). Their intersection is at (0,1), which is C. So, again, P is C, which contradicts the given ratios.Hmm, maybe I need to choose a different coordinate system where AD is not along AC. Let me try placing A at (0,0), B at (1,0), and C at (0,1). Then, point D is on AC, which is from (0,0) to (0,1). Let me say D is at (0, d). Point F is on AB, which is from (0,0) to (1,0). Let me say F is at (f, 0).Now, line AD is from (0,0) to (0,d), which is x=0. Line CF is from (0,1) to (f,0). Equation of CF: y = (-1/f)x + 1.Intersection P is at x=0, y=1, which is point C. Again, same issue.Wait, maybe I'm misunderstanding the problem. It says D lies on AC and F lies on AB. So, AD is from A to D on AC, and CF is from C to F on AB. Their intersection is P inside the triangle.But in my coordinate system, AD is along AC, making P coincide with C. So, perhaps I need to adjust the coordinate system so that AD is not along AC.Wait, maybe I should place C not on the y-axis. Let me try placing A at (0,0), B at (1,0), and C at (0,1). Then, AC is from (0,0) to (0,1). Point D is on AC at (0,d). Point F is on AB at (f,0). Line AD is x=0, line CF is from (0,1) to (f,0). Their intersection is at (0,1), which is C. So, again, P is C.This suggests that in this coordinate system, P is always C, which contradicts the given ratios. Therefore, maybe I need to choose a different coordinate system where AD is not along AC.Wait, perhaps I should place C at (1,1), so that AC is from (0,0) to (1,1). Then, point D can be somewhere along AC, say at (t,t). Point F is on AB, which is from (0,0) to (1,0). Let me say F is at (f,0).Now, line AD is from (0,0) to (t,t). Its parametric equation is x = t*s, y = t*s, where s ranges from 0 to 1.Line CF is from (1,1) to (f,0). The slope is (0 - 1)/(f - 1) = -1/(f - 1). So, equation is y - 1 = (-1/(f - 1))(x - 1).Simplify: y = (-1/(f - 1))x + (1/(f - 1)) + 1 = (-1/(f - 1))x + (1 + (f - 1))/(f - 1) = (-1/(f - 1))x + f/(f - 1)Now, find intersection P of AD and CF.From AD: x = t*s, y = t*sFrom CF: y = (-1/(f - 1))x + f/(f - 1)Set equal:t*s = (-1/(f - 1))*(t*s) + f/(f - 1)Multiply both sides by (f - 1):t*s*(f - 1) = -t*s + fBring all terms to left:t*s*(f - 1) + t*s - f = 0Factor t*s:t*s*(f - 1 + 1) - f = 0Simplify:t*s*f - f = 0Factor f:f*(t*s - 1) = 0Since f ≠ 0 (as F is on AB, not at A), we have t*s - 1 = 0 => s = 1/tBut s must be ≤ 1 since it's along AD from A to D. So, 1/t ≤ 1 => t ≥ 1. But D is on AC from A(0,0) to C(1,1), so t must be ≤ 1. Therefore, t = 1, which makes D coincide with C. But then P would be at C, which again contradicts the given ratios.Hmm, this is getting frustrating. Maybe I need to use mass point correctly.Let me try mass point again. From AP:PD = 3:2, so mass at A is 2, mass at D is 3, mass at P is 5.From FP:PC = 3:4, so mass at F is 4, mass at C is 3, mass at P is 7.To make mass at P consistent, scale the masses so that mass at P is LCM of 5 and 7, which is 35.So, scale the first ratio by 7: mass at A = 2*7=14, mass at D=3*7=21, mass at P=35.Scale the second ratio by 5: mass at F=4*5=20, mass at C=3*5=15, mass at P=35.Now, mass at D is 21, which is on AC. So, masses at A and C must add to 21. Mass at A is 14, so mass at C should be 21 - 14 = 7. But from the second scaling, mass at C is 15. Contradiction.Wait, that suggests that my mass point approach is inconsistent. Maybe I need to adjust the masses differently.Alternatively, maybe I should use the area method. Let me consider the areas created by the cevians.But perhaps it's better to use vectors. Let me denote vectors with position vectors from A as the origin.Let me denote vector A as origin, so A = (0,0). Let vector B = b, vector C = c.Point D is on AC, so vector D = k*c, where 0 < k < 1.Point F is on AB, so vector F = m*b, where 0 < m < 1.Point P is the intersection of AD and CF.Parametrize AD: P = t*D = t*k*c, where t is a parameter.Parametrize CF: P = c + s*(F - c) = c + s*(m*b - c) = (1 - s)*c + s*m*b.Set equal:t*k*c = (1 - s)*c + s*m*bSince vectors are linearly independent (as A, B, C are non-collinear), coefficients must match.So,For c component: t*k = 1 - sFor b component: 0 = s*m => s = 0 or m = 0But m ≠ 0 (since F is on AB, not at A), so s = 0.But s = 0 gives P = c, which is point C, contradicting the given ratios. So, this suggests that my parametrization is incorrect.Wait, maybe I should parametrize AD differently. Let me write AD as A + t*(D - A) = t*D, since A is origin.Similarly, CF can be parametrized as C + s*(F - C) = c + s*(m*b - c).Set equal:t*k*c = c + s*(m*b - c)Rearrange:t*k*c - c = s*(m*b - c)Factor c:c*(t*k - 1) = s*m*b - s*cBring all terms to left:c*(t*k - 1 + s) - s*m*b = 0Since b and c are linearly independent, their coefficients must be zero:For c: t*k - 1 + s = 0For b: -s*m = 0From b component: s*m = 0. Since m ≠ 0, s = 0.Then from c component: t*k - 1 + 0 = 0 => t*k = 1 => t = 1/kBut t must be ≤ 1 since P is on AD between A and D. So, 1/k ≤ 1 => k ≥ 1. But D is on AC, so k ≤ 1. Therefore, k = 1, which makes D = C, and P = C, which contradicts the given ratios.This suggests that my approach is flawed. Maybe I need to use a different method.Wait, perhaps I should use the concept of similar triangles or coordinate geometry with a different setup.Let me try coordinate geometry again, but this time place A at (0,0), B at (1,0), and C at (0,1). Then, AC is from (0,0) to (0,1), and AB is from (0,0) to (1,0).Point D is on AC at (0,d), and point F is on AB at (f,0).Line AD is x=0, line CF is from (0,1) to (f,0). Equation of CF: y = (-1/f)x + 1.Intersection P is at x=0, y=1, which is point C. Again, same problem.Wait, maybe I need to place C not on the y-axis. Let me place C at (c,0), making ABC a degenerate triangle. No, that's not good.Alternatively, place C at (0,1), A at (0,0), B at (1,0), and D on AC at (0,d). Then, line AD is x=0, line CF is from (0,1) to (f,0). Their intersection is at (0,1), which is C. So, P is C, which contradicts the given ratios.This suggests that in any coordinate system where AC is vertical, AD is along AC, making P coincide with C. Therefore, I need to choose a different coordinate system where AD is not along AC.Let me try placing A at (0,0), B at (1,0), and C at (1,1). So, AC is from (0,0) to (1,1). Point D is on AC at (t,t), where 0 < t < 1. Point F is on AB at (f,0), where 0 < f < 1.Line AD is from (0,0) to (t,t). Its parametric equation is x = t*s, y = t*s, where s ∈ [0,1].Line CF is from (1,1) to (f,0). The slope is (0 - 1)/(f - 1) = -1/(f - 1). Equation: y - 1 = (-1/(f - 1))(x - 1).Simplify: y = (-1/(f - 1))x + (1/(f - 1)) + 1 = (-1/(f - 1))x + (1 + f - 1)/(f - 1) = (-1/(f - 1))x + f/(f - 1)Now, find intersection P of AD and CF.From AD: x = t*s, y = t*sFrom CF: y = (-1/(f - 1))x + f/(f - 1)Set equal:t*s = (-1/(f - 1))*(t*s) + f/(f - 1)Multiply both sides by (f - 1):t*s*(f - 1) = -t*s + fBring all terms to left:t*s*(f - 1) + t*s - f = 0Factor t*s:t*s*(f - 1 + 1) - f = 0Simplify:t*s*f - f = 0Factor f:f*(t*s - 1) = 0Since f ≠ 0, we have t*s - 1 = 0 => s = 1/tBut s must be ≤ 1 since it's along AD from A to D. So, 1/t ≤ 1 => t ≥ 1. But D is on AC from A(0,0) to C(1,1), so t ≤ 1. Therefore, t = 1, which makes D coincide with C, and P would be at C, which contradicts the given ratios.This is the same issue as before. It seems that in any coordinate system where AC is a straight line from A, AD ends up coinciding with AC, making P coincide with C, which is not the case in the problem.Therefore, maybe I need to approach this problem differently, perhaps using mass point correctly.Let me try mass point again, carefully.Given AP:PD = 3:2, so mass at A is 2, mass at D is 3, mass at P is 5.Given FP:PC = 3:4, so mass at F is 4, mass at C is 3, mass at P is 7.To make mass at P consistent, scale the masses so that mass at P is 35 (LCM of 5 and 7).So, scale the first ratio by 7: mass at A = 2*7=14, mass at D=3*7=21, mass at P=35.Scale the second ratio by 5: mass at F=4*5=20, mass at C=3*5=15, mass at P=35.Now, mass at D is 21, which is on AC. So, masses at A and C must add to 21. Mass at A is 14, so mass at C should be 21 - 14 = 7. But from the second scaling, mass at C is 15. Contradiction.This suggests that my mass point approach is inconsistent. Maybe I need to adjust the masses differently.Wait, perhaps I need to consider that mass at D is 21, which is on AC, so masses at A and C must be in the ratio CD:DA. Since mass at A is 14 and mass at C is 15, the ratio CD:DA = 14:15.So, CD/DA = 14/15 => DA/CD = 15/14.But DA + CD = AC, so DA = (15/29)*AC and CD = (14/29)*AC.Similarly, mass at F is 20, which is on AB. Masses at A and B must add to 20. Mass at A is 14, so mass at B is 20 - 14 = 6.Therefore, the ratio AF/FB = mass at B / mass at A = 6/14 = 3/7.Wait, that seems promising. So, AF/FB = 3/7.Let me verify this.From mass point, masses at A=14, B=6, C=15.Mass at D=21 (on AC), mass at F=20 (on AB).Mass at P=35.So, AF/FB = mass at B / mass at A = 6/14 = 3/7.Yes, that makes sense.So, the answer is 3/7.

💡Alright, so I have this probability problem here, and I need to figure out which pair of events are complementary. Let me read it again carefully."In a bag, there are 3 white balls and 4 black balls. Three balls are drawn randomly from the bag. Which of the following pairs of events are complementary events? ( )"Then there are four options: A, B, C, and D. Each option pairs two events, and I need to determine which pair is complementary.First, I need to recall what complementary events are. From what I remember, complementary events are two events where one event is the complement of the other. That means if one event happens, the other cannot happen, and vice versa. Also, the two events together cover all possible outcomes. So, they are mutually exclusive and exhaustive.In other words, for two events E and F to be complementary, E must occur if and only if F does not occur, and F must occur if and only if E does not occur. Additionally, there are no other possibilities outside of E and F.So, with that in mind, let's look at each option one by one.**Option A: Exactly one white ball and all white balls**Okay, so the first event is drawing exactly one white ball, and the second event is drawing all white balls. Let's think about whether these can happen at the same time.If I draw exactly one white ball, that means I have one white and two black balls. On the other hand, drawing all white balls means I have three white balls. These two events cannot happen at the same time because you can't have both exactly one white and all white balls in the same draw. So, they are mutually exclusive.But are they complementary? That means, does every possible outcome fall into either exactly one white ball or all white balls? Let's see.When drawing three balls, the possible number of white balls could be 0, 1, 2, or 3. So, if I have exactly one white ball, that's one possibility, and all white balls is another. But what about the cases where I have two white balls or no white balls? Those aren't covered by either of these events. So, these two events are not exhaustive because there are other possible outcomes. Therefore, they are not complementary events.So, Option A is not the correct answer.**Option B: At least one white ball and all black balls**Alright, the first event is drawing at least one white ball, and the second event is drawing all black balls.Let's see if these can happen at the same time. If I draw at least one white ball, that means I have one or more white balls. Drawing all black balls means I have zero white balls. So, these two events cannot happen together; they are mutually exclusive.Now, are they exhaustive? That is, does every possible outcome fall into either at least one white ball or all black balls?When drawing three balls, the number of white balls can be 0, 1, 2, or 3. If I have at least one white ball, that covers the cases where I have 1, 2, or 3 white balls. The only case not covered is when I have 0 white balls, which is exactly the second event: all black balls.So, together, these two events cover all possible outcomes. They are mutually exclusive and exhaustive, which means they are complementary events.So, Option B seems to be a strong candidate for the correct answer.But let's check the other options to be thorough.**Option C: At least one white ball and at least two white balls**First event: at least one white ball. Second event: at least two white balls.Are these mutually exclusive? Let's see. If I have at least two white balls, that means I have two or three white balls. But the first event includes having one, two, or three white balls. So, if I have two or three white balls, both events occur simultaneously. Therefore, these events are not mutually exclusive.Since they can happen at the same time, they cannot be complementary. Complementary events must be mutually exclusive.So, Option C is not the correct answer.**Option D: At least one white ball and at least one black ball**First event: at least one white ball. Second event: at least one black ball.Are these mutually exclusive? Let's think. If I draw three balls, it's possible to have one white and two black balls. In this case, both events occur: there is at least one white ball and at least one black ball. Therefore, these events can happen together, meaning they are not mutually exclusive.Since they are not mutually exclusive, they cannot be complementary events.So, Option D is also not the correct answer.After reviewing all the options, only Option B consists of two events that are mutually exclusive and exhaustive, making them complementary events.