answer:- The event of getting three tails and two heads in any order can be broken down into different specific orders: - The probability of a specific sequence (e.g., TTTHH) occurring where there are three tails and two heads is left(frac{1}{2}right)^5 = frac{1}{32} because each coin toss is an independent event. - The number of favorable sequences is the number of ways to choose 3 tails out of 5 tosses, which can be calculated using combinations (denoted as binom{n}{k}). The number of sequences where exactly 3 coins show tails and 2 coins show heads is binom{5}{3}. - Compute binom{5}{3} = frac{5 times 4 times 3}{3 times 2 times 1} = 10. - Multiply the single outcome probability by the number of favorable sequences: 10 times frac{1}{32} = frac{10}{32} = frac{5}{16}. boxed{frac{5}{16}}

answer:Let's denote the number of students in the Period 2 gym class as x. According to the problem, the Period 1 gym class has 5 fewer than twice as many students as in the Period 2 gym class. This relationship can be expressed as: [2x - 5 = 11] To find the value of x, we solve the equation step by step: [ begin{align*} 2x - 5 &= 11 2x &= 11 + 5 2x &= 16 x &= frac{16}{2} x &= 8 end{align*} ] Therefore, there are boxed{8} students in the Period 2 gym class.

answer:To analyze the problem systematically, let's consider each strategy player A can adopt and evaluate the outcomes: # Strategy 1: Choosing Containers A and B - **Volume of A and B combined**: The volume of A is a^2 cdot a = a^3, and the volume of B is a^2 cdot b = a^2b. So, the total volume for A and B is a^3 + a^2b. - **Volume of C and D combined**: Similarly, the volume of C is b^2 cdot a = ab^2, and the volume of D is b^2 cdot b = b^3. Thus, the total volume for C and D is ab^2 + b^3. - **Comparing Volumes**: The difference in volumes is (a^3 + a^2b) - (ab^2 + b^3) = a^2(a+b) - b^2(a+b) = (a+b)^2(a-b). Since (a+b)^2 > 0 but the sign of (a-b) is uncertain, this strategy does not guarantee a win. # Strategy 2: Choosing Containers A and C - **Volume of A and C combined**: The volume for A and C is a^3 + b^2a. - **Volume of B and D combined**: The volume for B and D is a^2b + b^3. - **Comparing Volumes**: The difference in volumes is (a^3 + b^2a) - (a^2b + b^3) = a(a^2 + b^2) - b(a^2 + b^2) = (a^2 + b^2)(a-b). Since a^2 + b^2 > 0 but the sign of (a-b) is uncertain, this strategy also does not guarantee a win. # Strategy 3: Choosing Containers A and D - **Volume of A and D combined**: The volume for A and D is a^3 + b^3. - **Volume of B and C combined**: The volume for B and C is a^2b + ab^2. - **Comparing Volumes**: The difference in volumes is (a^3 + b^3) - (a^2b + ab^2) = (a+b)(a^2 - ab + b^2) - ab(a+b) = (a+b)(a-b)^2. Since a neq b and both a, b > 0, we have (a+b)(a-b)^2 > 0, indicating a^3 + b^3 > a^2b + ab^2. Therefore, the only winning strategy for player A is to choose containers A and D first, as this is the only combination that guarantees a higher total volume of water than the remaining two containers, B and C. Thus, the number of winning strategies for player A is boxed{1}.

answer:To find all the real number sequences {b_n}_{n geq 1} and {c_n}_{n geq 1} that satisfy the given conditions, we will proceed step-by-step. 1. **Initial Conditions and Roots**: - Given that for any positive integer n, b_n leq c_n. - For any positive integer n, b_{n+1} and c_{n+1} are the two roots of the quadratic equation x^2 + b_n x + c_n = 0. 2. **Quadratic Equation Roots**: - The roots of the quadratic equation x^2 + b_n x + c_n = 0 are given by: [ x = frac{-b_n pm sqrt{b_n^2 - 4c_n}}{2} ] - Let b_{n+1} and c_{n+1} be the roots, then: [ b_{n+1} = frac{-b_n + sqrt{b_n^2 - 4c_n}}{2}, quad c_{n+1} = frac{-b_n - sqrt{b_n^2 - 4c_n}}{2} ] 3. **Case Analysis**: - Suppose b_1 neq 0 and c_1 neq 0. We need to consider the signs of b_n and c_n. 4. **Sign Analysis**: - If b_n and c_n are both positive or both negative, we need to check the discriminant b_n^2 - 4c_n to ensure the roots are real. - If b_n and c_n are of opposite signs, we need to ensure b_n leq c_n holds. 5. **Contradiction**: - For every case, there exists an n such that b_n, c_n < 0. Then: [ b_{n+1}^2 - 4c_{n+1} = left(frac{-b_n - sqrt{b_n^2 - 4c_n}}{2}right)^2 - 4 left(frac{-b_n + sqrt{b_n^2 - 4c_n}}{2}right) ] Simplifying this expression: [ b_{n+1}^2 - 4c_{n+1} = left(frac{b_n^2}{2} + 2b_n - c_nright) - left(frac{-b_n}{2} + 2right) sqrt{b_n^2 - 4c_n} ] This expression simplifies to: [ b_{n+1}^2 - 4c_{n+1} < left(frac{b_n^2}{2} + b_nright) - left(frac{-b_n}{2} + 2right)(-b_n) = 3b_n < 0 ] Hence, x^2 + b_{n+1}x + c_{n+1} = 0 does not have a real root, which is a contradiction. 6. **Special Cases**: - If b_1 = 0, then c_1 geq 0 but at the same time c_1 leq 0 for x^2 + b_1 x + c_1 = 0 to have a real root. Hence, c_1 = 0. - If c_1 = 0, then b_2 = 0, c_2 = -b_1. By a similar argument as above, we get c_2 = 0. Hence, b_1 = 0. Therefore, the only sequences that satisfy the given conditions are b_n equiv 0 and c_n equiv 0. The final answer is boxed{b_n equiv 0, ; c_n equiv 0}.