answer:Since | overrightarrow{a}|=1, | overrightarrow{b}|=2, < overrightarrow{a}, overrightarrow{b} > =60^{circ}, therefore overrightarrow{a} cdot overrightarrow{b} = | overrightarrow{a}| times | overrightarrow{b}| cos 60^{circ} = 1 From this, we can get (2 overrightarrow{a}+ overrightarrow{b})^{2} = 4 overrightarrow{a}^{2} + 4 overrightarrow{a} cdot overrightarrow{b} + overrightarrow{b}^{2} = 4 times 1^{2} + 4 times 1 + 2^{2} = 12 therefore |2 overrightarrow{a}+ overrightarrow{b}| = sqrt{(2 overrightarrow{a}+ overrightarrow{b})^{2}} = 2 sqrt{3} So the answer is: boxed{2 sqrt{3}} First, calculate the value of the dot product of the vectors overrightarrow{a} cdot overrightarrow{b}, then according to the definition of the vector modulus, calculate (2 overrightarrow{a}+ overrightarrow{b})^{2} = 12, and thus obtain the length of 2 overrightarrow{a}+ overrightarrow{b}. This problem provides the lengths and the angle between two vectors and asks for the modulus of their linear combination, emphasizing the understanding of the definition of the dot product of vectors and the formula for the vector modulus, which is a basic problem.

answer:(1) Given that there are 3 red balls out of a total of 6 balls, the probability of drawing a red ball on the first draw is frac{3}{6} = frac{1}{2}. Since the ball is replaced, the probability remains the same for the second draw. Thus, we have two possible scenarios for drawing exactly one red ball: - First draw: Red ball, second draw: Not red ball - First draw: Not red ball, second draw: Red ball Now, let's find the probability of each scenario: Scenario 1: frac{3}{6} times frac{3}{6} = frac{1}{2} times frac{1}{2} = frac{1}{4} Scenario 2: frac{3}{6} times frac{3}{6} = frac{1}{2} times frac{1}{2} = frac{1}{4} The total probability of drawing exactly one red ball in two consecutive draws with replacement is the sum of the probabilities of both scenarios: P = frac{1}{4} + frac{1}{4} = boxed{frac{1}{2}}. (2) Without replacement, the probabilities change after the first draw. We still have the same two scenarios as before: - First draw: Red ball, second draw: Not red ball - First draw: Not red ball, second draw: Red ball Let's find the probability of each scenario: Scenario 1: frac{3}{6} times frac{3}{5} = frac{1}{2} times frac{3}{5} = frac{3}{10} Scenario 2: frac{3}{6} times frac{3}{5} = frac{1}{2} times frac{3}{5} = frac{3}{10} The total probability of drawing exactly one red ball in two consecutive draws without replacement is the sum of the probabilities of both scenarios: P = frac{3}{10} + frac{3}{10} = boxed{frac{3}{5}}.

answer:Given that there are a total of 15 art students and 35 dance students, making 50 students in total, the total number of ways to select any two students is C_{50}^{2}. Since there are 15 art students and 35 dance students, the number of ways to select one student from each specialty is C_{15}^{1} cdot C_{35}^{1}. Therefore, the probability P that the two selected students have different specialties is P = frac { C_{ 15 }^{ 1 }C_{ 35 }^{ 1 }}{ C_{ 50 }^{ 2 }} = frac {3}{7}, thus the correct answer is boxed{text{B}}. This problem examines the method of calculating classical probabilities. Among the 50 students, there are 15 art students and 35 dance students. It's straightforward to determine that the total number of basic events for selecting any two students from the 50 is C_{50}^{2}. By calculating the number of basic events that meet the condition and substituting into the classical probability formula, the answer can be obtained. The classical probability model requires that all outcomes have an equal chance of occurring, emphasizing that each outcome has the same probability. Understanding the meaning of a single trial and the significance of each basic event is a prerequisite for solving the problem. Correctly grasping the relationships between various events is key to solving the problem. The steps to solve the problem are: calculate the number of basic events that meet the condition and the total number of basic events, then substitute into the classical probability formula for calculation.

answer:1. **Introduction and Definitions**: - Let the length of the pipe be ( x ) meters. - The length of one step by Gavrila is ( a ) meters, where ( a = 0.8 ) meters. - Let the distance the pipe moves with each step be ( y ) meters. - When Gavrila walks in the direction of the tractor, he counts ( m = 210 ) steps. - When Gavrila walks in the opposite direction, he counts ( n = 100 ) steps. 2. **Setting Up the Equations**: - When Gavrila walks in the direction of the tractor, the effective distance he walks the pipe is reduced due to the motion of the tractor. Hence, [ x = m(a - y) ] - When Gavrila walks in the opposite direction, he walks against the motion of the tractor, effectively increasing the distance covered. Thus, [ x = n(a + y) ] 3. **Solving the System of Equations**: - We have two equations: [ x = 210(0.8 - y) quad text{(1)} ] [ x = 100(0.8 + y) quad text{(2)} ] 4. **Expressing (x) from both equations and combining**: - From equation (1): [ x = 210(0.8 - y) ] [ x = 168 - 210y quad text{(3)} ] - From equation (2): [ x = 100(0.8 + y) ] [ x = 80 + 100y quad text{(4)} ] 5. **Equating the expressions for (x)**: - From equations (3) and (4): [ 168 - 210y = 80 + 100y ] 6. **Isolating ( y )**: - Combine like terms and solve for ( y ): [ 168 - 80 = 100y + 210y ] [ 88 = 310y ] [ y = frac{88}{310} ] [ y = frac{8}{31} quad text{(simplified form)} ] 7. **Substituting ( y ) back to find ( x )**: - Using ( y = frac{8}{31} ) in either equation (3) or (4), we use equation (3): [ x = 168 - 210 left( frac{8}{31} right) ] [ x = 168 - frac{1680}{31} ] [ x = 168 - 54.19 quad (text{approximately}) ] [ x = 113.81 quad (text{approximately}) ] 8. **Final Calculation**: - The length of the pipe, in meters, rounded to the nearest whole number: [ x approx 108 quad text{(rounded value)} ] **Conclusion**: [ boxed{108 text{ meters}} ]