answer:We calculate the probability that the first, second, and third guests each receive one of each type of roll, ensuring the fourth guest's set is automatically correct. For the first guest: - Probability = frac{text{ways to choose 1 nut, 1 cheese, 1 fruit, 1 veggie}}{text{ways to choose any 4 rolls out of 16}} - frac{4 cdot 4 cdot 4 cdot 4}{binom{16}{4}} = frac{256}{1820} = frac{64}{455} For the second guest (12 rolls left, 3 of each type): - frac{3 cdot 3 cdot 3 cdot 3}{binom{12}{4}} = frac{81}{495} = frac{27}{165} For the third guest (8 rolls left, 2 of each type): - frac{2 cdot 2 cdot 2 cdot 2}{binom{8}{4}} = frac{16}{70} = frac{8}{35} The overall probability: - frac{64}{455} cdot frac{27}{165} cdot frac{8}{35} = frac{13824}{2538375} This reduces to: - Prime factor the numerator and denominator and simplify the fraction. Calculating and simplifying, we find: - frac{13824}{2538375} simplifies to frac{16}{2925}, so m=16 and n=2925, resulting in m+n=boxed{2941}.

answer:# Solution: Part 1: Finding the value of m Given that f(x) = frac{3x+m}{{x}^{2}+1} is an odd function, we can use the property of odd functions, which states that f(-x) = -f(x). For an odd function, when x=0, f(0) should be equal to 0 if m is not part of the odd function characteristic. Thus, we evaluate f(0): f(0) = frac{3cdot0+m}{0^2+1} = frac{m}{1} = m Since f(x) is odd, f(0) must be 0, leading to: m = 0 Therefore, when m=0, f(x) = frac{3x}{{x}^{2}+1}, and for all x in mathbb{R}, f(-x) = frac{-3x}{{x}^{2}+1} = -f(x), confirming that f(x) is indeed an odd function. Thus, we have: boxed{m = 0} Part 2: Monotonicity of the function f(x) To determine the monotonicity of f(x) = frac{3x}{{x}^{2}+1} on the intervals [0,1] and [1,+infty), we consider the difference f(x_1) - f(x_2) for any x_1, x_2 in [0,+infty) with x_1 < x_2: f(x_1) - f(x_2) = frac{3x_1}{{x_1}^2+1} - frac{3x_2}{{x_2}^2+1} = frac{3(x_1-x_2)(1-x_1x_2)}{({x_1}^2+1)({x_2}^2+1)} - **On [1,+infty):** For 1 leq x_1 < x_2, we have x_1 - x_2 < 0 and x_1x_2 > 1, which implies 1 - x_1x_2 < 0. Since the denominator ({x_1}^2+1)({x_2}^2+1) > 0, it follows that f(x_1) - f(x_2) > 0, meaning f(x_1) > f(x_2). Therefore, f(x) is monotonically decreasing on [1,+infty). - **On [0,1]:** For 0 leq x_1 < x_2 leq 1, we have x_1 - x_2 < 0 and x_1x_2 < 1, which implies 1 - x_1x_2 > 0. Since the denominator ({x_1}^2+1)({x_2}^2+1) > 0, it follows that f(x_1) - f(x_2) < 0, meaning f(x_1) < f(x_2). Therefore, f(x) is monotonically increasing on [0,1]. Thus, we conclude that f(x) is monotonically increasing on [0,1] and monotonically decreasing on [1,+infty). The proof is as follows: boxed{text{Monotonically increasing on } [0,1] text{ and monotonically decreasing on } [1,+infty)} Part 3: Range of the real number n Given that for any x_1, x_2 in [0,+infty), f(x_1) - f(x_2) + n leq 0 always holds, we consider the maximum and minimum values of f(x): From Part 2, we know that f(x)_{max} = f(1) = frac{3}{2} and f(x)_{min} = f(0) = 0 (since f(x) > 0 for x > 0). Therefore, we have: -n geq f(x)_{max} - f(x)_{min} = frac{3}{2} - 0 = frac{3}{2} This implies that the range of n is: boxed{n leq -frac{3}{2}} Hence, the range of the real number n is (-infty, -frac{3}{2}].

answer:(1) Since f'(x) = frac{2}{x} - 6x - 11, we can compute the derivative at x=1: [ f'(1) = frac{2}{1} - 6(1) - 11 = -15. ] Also, f(1) = 2ln(1) - 3(1)^2 - 11(1) = -14. Therefore, the equation of the tangent line to the curve y=f(x) at the point (1, f(1)) is: [ y - (-14) = -15(x - 1). ] Simplifying this, we get: [ y = -15x + 1. ] Hence, the equation of the tangent line is boxed{y = -15x + 1}. (2) Let's consider g(x) = f(x) - (a-3)x^2 - (2a-13)x - 1. Simplifying, we get: [ g(x) = 2ln x - ax^2 + (2-2a)x - 1. ] The derivative of g(x) is: [ g'(x) = frac{2}{x} - 2ax + 2 - 2a = frac{-2ax^2 + (2-2a)x + 2}{x}. ] When a leq 0, since x > 0, we have g'(x) > 0, which means g(x) is increasing on the interval (0, +infty). Furthermore, evaluating g(x) at x=1, we get: [ g(1) = -a + 2 - 2a - 1 = 1 - 3a > 0. ] So the inequality f(x) leq (a-3)x^2 + (2a-13)x + 1 does not always hold true. When a > 0, we have: [ g'(x) = frac{-2a(x - frac{1}{a})(x + 1)}{x}. ] Setting g'(x) = 0, we find x = frac{1}{a}. Thus, g'(x) > 0 for x in (0, frac{1}{a}) and g'(x) < 0 for x in (frac{1}{a}, +infty). This means g(x) is increasing on the interval (0, frac{1}{a}) and decreasing on (frac{1}{a}, +infty). Consequently, the maximum value of g(x) is at x = frac{1}{a}: [ gleft(frac{1}{a}right) = 2ln frac{1}{a} + frac{1}{a} - 3. ] Simplifying this, we get: [ gleft(frac{1}{a}right) = frac{1}{a} - 2ln a - 3 leq 0. ] Let h(a) = frac{1}{a} - 2ln a - 3. The function h(a) is decreasing on the interval (0, +infty). We know that h(1) = -2 < 0. Therefore, for a geq 1, h(a) < 0. As a result, the smallest integer value for a is boxed{a = 1}.

answer:First, let's calculate the total number of steps Chanhee took during the 13 minutes of walking. If Chanhee walks at a pace of 90 steps per minute, then in 13 minutes she would have taken: 90 steps/minute * 13 minutes = 1170 steps Now, since Chanhee goes 0.45 meters for one step, to find the total distance she walked, we multiply the number of steps by the distance per step: 1170 steps * 0.45 meters/step = 526.5 meters Therefore, Chanhee walked a total of boxed{526.5} meters.