answer:Solution: ① The regression line hat y=b hat x+a always passes through the sample center point ( overset{ .}{x}, overset{ .}{y}), which is correct according to the definition of the regression line equation; ② "x=6" can lead to "x²-5x-6=0", but not necessarily the other way around, so it should be a sufficient but not necessary condition, hence incorrect; ③ The negation of "There exists x_0 in mathbb{R}, such that x_0^2+2x_0+3<0" is "For all x in mathbb{R}, x^2+2x+3 geq 0", hence incorrect; ④ If "Proposition p lor q" is true, then p, q at least one is true, then neg p, neg q at least one is false, hence "Proposition neg p land neg q" is false, hence incorrect. Therefore, the answer is boxed{text{B}}. ① Judged based on the definition of the regression line; ② Judged based on the concept; ③ The negation of an existential proposition involves changing "exists" to "for all" and negating the conclusion; ④ Concluding that at least one of p, q is true leads to at least one of neg p, neg q being false, leading to the conclusion. This question examines the concepts of regression line equations, necessary and sufficient conditions, negation of propositions, and compound propositions. It is a basic question type that should be mastered.

answer:(Ⅰ) Given the length of the major axis of the ellipse is a = 2, and the semi-focal distance c = sqrt{a^2 - b^2}, from the eccentricity e = frac{c}{a} = frac{sqrt{4 - b^2}}{2} = frac{sqrt{3}}{2} we get b^2 = 1. Therefore, the top vertex of the ellipse is (0, 1), which is the focus of the parabola. Thus, p = 2 and the equation of the parabola is x^2 = 4y. (Ⅱ) As line l has a slope that exists and is not zero, we can assume that the equation of line l is y = k(x + 1). Let E(x_1, y_1) and F(x_2, y_2) be the points of intersection. Since y = frac{x^2}{4}, we have y' = frac{x}{2}. Therefore, the slopes of the tangents l_1 and l_2 are frac{x_1}{2} and frac{x_2}{2} respectively. When l_1 is perpendicular to l_2, we have frac{x_1}{2} cdot frac{x_2}{2} = -1, leading to x_1 cdot x_2 = -4. Substituting y = k(x + 1) into the equation x^2 = 4y, we get x^2 = 4k(x + 1), which simplifies to x^2 - 4kx - 4k = 0. The discriminant must be positive, so Delta = (-4k)^2 - 4 cdot (-4k) > 0, which yields k < -1 or k > 0. Additionally, since x_1 cdot x_2 = -4, we have -4k = -4, therefore k = 1. Thus, the equation of line l is boxed{x - y + 1 = 0}.

answer:Solution: (1) Since begin{cases} f(0)=b=6 f(1)=a+b+1=5 end{cases} Rightarrow begin{cases} a=-2 b=6 end{cases} Rightarrow f(x)=x^{2}-2x+6; (2) Since f(x)=x^{2}-2x+6=(x-1)^{2}+5, and x in [-2,2], the parabola opens upwards, and its axis of symmetry is x=1, therefore When x=1, the minimum value of f(x) is boxed{5}, and when x=-2, the maximum value of f(x) is boxed{14}.

answer:- **Starting Point**: [ frac{n+500}{50} = lfloor sqrt{n} rfloor ] Simplifying the left side: [ frac{n+500}{50} = frac{n}{50} + 10 ] This expression must equal lfloor sqrt{n} rfloor, and frac{n}{50} + 10 must be an integer, hence n must satisfy n equiv 0 pmod{50}. - **Exploring Values**: If n = 50k for some integer k, then: [ frac{50k + 500}{50} = k + 10 = lfloor sqrt{50k} rfloor ] For lfloor sqrt{50k} rfloor = k + 10, we need to find k such that: [ sqrt{50k} approx k + 10 ] Squaring both sides for approximation: [ 50k approx (k + 10)^2 = k^2 + 20k + 100 ] [ 0 approx k^2 - 30k + 100 ] Solving this quadratic equation: [ k = frac{30 pm sqrt{900 - 400}}{2} = frac{30 pm sqrt{500}}{2} ] [ k approx frac{30 pm 22.36}{2} ] [ k_1 approx 26.18, quad k_2 approx 3.82 ] Since k must be an integer, test k = 26 and k = 4, corresponding: [ n = 50 times 26 = 1300, quad n = 50 times 4 = 200 ] For n = 1300: [ frac{1300 + 500}{50} = 36, quad lfloor sqrt{1300} rfloor = 36 ] For n = 200: [ frac{200 + 500}{50} = 14, quad lfloor sqrt{200} rfloor = 14 ] - **Conclusion with Boxed Answer**: The values of n that satisfy the equation are 200 and 1300. Thus, there are boxed{2} such integers.