answer:To find the median: 1. The results are already arranged in increasing order: 2, 6, 8, 9, 13, 16, 16, 17, 20, 20. 2. With 10 data points, the median is the average of the 5th and 6th terms: - 5th term: 13 - 6th term: 16 - Median: frac{13 + 16}{2} = 14.5. [ boxed{14.5} ]

answer:# Problem: 求所有的函数 f: mathbb{R} rightarrow mathbb{R}, 满足 [ f(x y-1)+f(x) f(y)=2 x y-1. ] 1. **Substitute ( y = 0 ) into the given functional equation:** [ f(x cdot 0 - 1) + f(x) f(0) = 2 cdot x cdot 0 - 1. ] Simplifying the left side: [ f(-1) + f(x) f(0) = -1. ] This equation holds for all ( x in mathbb{R} ). Let's denote this as (1). 2. **Analyze the case where ( f(0) neq 0 ):** If ( f(0) neq 0 ), then the equation ( f(-1) + f(x) f(0) = -1 ) means that ( f(x) ) must be a constant function. Let ( f(x) = k ) for some constant ( k ). Substituting ( f(x) = k ) into the original functional equation: [ f(x y - 1) + f(x) f(y) = 2 x y - 1 ] becomes: [ k + k cdot k = 2 x y - 1. ] Simplifying this, we get: [ k + k^2 = 2 x y - 1. ] This equation must hold for all values of ( x ) and ( y ). However, it leads to a contradiction since the left side is a constant and the right side is dependent on ( x ) and ( y ). Therefore, there is no such constant ( k ) that can satisfy the given functional equation. Thus, we conclude that ( f(0) = 0 ). 3. **Substitute ( x = 1 ) and ( y = 1 ) into the original functional equation:** [ f(1 cdot 1 - 1) + f(1) f(1) = 2 cdot 1 cdot 1 - 1. ] Simplifying: [ f(0) + f(1)^2 = 1. ] Since ( f(0) = 0 ), the equation reduces to: [ f(1)^2 = 1. ] Therefore, ( f(1) = 1 ) or ( f(1) = -1 ). 4. **Case 1: Assume ( f(1) = 1 ):** We will try ( f(x) = x ) as a potential solution. Substitute ( f(x) = x ) into the original equation: [ f(xy - 1) + f(x) f(y) = 2xy - 1, ] gives: [ xy - 1 + x cdot y = 2xy - 1. ] This holds true because: [ xy - 1 + xy = 2xy - 1. ] Thus, ( f(x) = x ) is a solution. 5. **Case 2: Assume ( f(1) = -1 ):** We will check ( f(x) = -x ) as a potential solution. Substitute ( f(x) = -x ) into the original equation: [ f(xy - 1) + f(x) f(y) = 2xy - 1, ] gives: [ -(xy - 1) + (-x)(-y) = 2xy - 1. ] This simplifies to: [ -xy + 1 + xy = 2xy - 1. ] Since: [ 1 neq 2xy - 1, ] we find that ( f(x) = -x ) does not satisfy the original equation. 6. **Consider other forms:** Let us consider ( f(x) = -x^2 ) and substitute it into the original equation: [ f(xy - 1) + f(x) f(y) = 2xy - 1, ] gives: [ -(xy - 1)^2 + (-x^2)(-y^2) = 2xy - 1. ] This simplifies to: [ -(xy - 1)^2 + x^2 y^2 = 2xy - 1. ] Since: [ (xy - 1)^2 - x^2 y^2 = -2xy + 1, ] we find that: [ xy^2 - 2x y + 1 = -2xy + 1, ] which does not hold in general. Therefore, ( f(x) = -x^2 ) is not a solution. # Conclusion: The only solution to the functional equation is: [ boxed{f(x) = x}. ]

answer:To find out how deep the lost ship is below the sea level, we need to multiply the rate at which the scuba diver descends by the time it takes him to reach the ship. The diver descends at a rate of 30 feet per minute and it takes him 80 minutes to reach the ship. Depth = Rate × Time Depth = 30 feet/minute × 80 minutes Depth = 2400 feet Therefore, the lost ship is boxed{2400} feet below the sea level.

answer:Let M(x,y). The slope of AM, k_1 = frac{y}{x+5}, The slope of BM, k_2 = frac{y}{x-5}, Given that k_1 k_2 = frac{y}{x+5} times frac{y}{x-5} = -2 quad (x neq pm 5), Simplify and rearrange to get 2x^2 + y^2 = 50 quad (x neq pm 5), Thus, frac{x^2}{25} + frac{y^2}{50} = 1 quad (x neq pm 5), boxed{M text{ lies on the trajectory of an ellipse with the origin as the center and foci on the } y text{-axis (excluding the two endpoints of the major axis).}}