answer:1. **Calculate Rachel's running details:** - Rachel completes a lap every 75 seconds. - In 15 minutes (900 seconds), Rachel completes (frac{900}{75} = 12) laps. 2. **Determine Rachel's position for the photo:** - Rachel runs one-fifth of a lap in (frac{1}{5} times 75 = 15) seconds. - To be in the one-fifth of the track centered on the starting line, Rachel must be within ±15 seconds of the starting line when the photo is taken. 3. **Calculate Robert's running details:** - Robert completes a lap every 70 seconds. - In 15 minutes (900 seconds), Robert completes (frac{900}{70} approx 12.86) laps, nearly completing his 13th lap. 4. **Determine Robert's position for the photo:** - Robert runs one-fifth of a lap in (frac{1}{5} times 70 = 14) seconds. - To be in the one-fifth of the track centered on the starting line, Robert must be within ±14 seconds of the starting line. 5. **Calculate the overlap time when both are in the picture:** - The time interval during which the photo is taken is from 900 to 960 seconds. - Both Rachel and Robert need to be within ±15 and ±14 seconds of the starting line, respectively. - The overlap for both being in the picture is determined by these ±15 and ±14-second intervals around the starting line. 6. **Calculate the probability:** - Both are in the picture if they are within the smaller of the two intervals, ±14 seconds, around the starting line. - The probability they are both in the picture is (frac{2 times 14}{60} = frac{28}{60} = frac{7}{15}). Conclusion: The probability that both Rachel and Robert are in the picture is frac{7{15}}. The final answer is boxed{C}

answer:Since the denominator of a fraction cannot be zero, and x + 2 is in the denominator, we have x + 2 neq 0. Solving this, we get x neq -2. Therefore, the range of the independent variable is boxed{x neq -2}.

answer:1. **Initial Setup:** - We have a regular 2n-gon with empty cups placed at its vertices. - Petya and Vasya take turns pouring tea into the cups, starting with Petya. - On each turn, a player can either pour tea into one empty cup or into two cups that are symmetric about the center of the 2n-gon, provided both cups are empty. - The player unable to make a move loses the game. 2. **Case of Odd n:** - Assume n is odd. - Petya can secure a win with the following strategy: - On the first move, Petya pours tea into two diametrically opposite cups. - On each subsequent move, Petya pours tea into the cups that are symmetric relative to this diameter, corresponding to the cups Vasya filled on the previous turn. - Before Vasya's turn, any pair of symmetric cups is simultaneously either both empty or both filled. - If Vasya fills two diametrically opposite cups, Petya responds by filling the other symmetric pair of diametrically opposite cups. - Since n is odd, no diameter is symmetric to itself, ensuring that there always exists at least one pair of diametrically opposite cups which remain empty for Petya to fill in response to Vasya’s move. 3. **Conclusion for Odd n:** - Therefore, when n is odd, Petya can always replicate Vasya's moves symmetrically and ensure that he will never be left without a move, ensuring that Petya wins independent of Vasya's actions. 4. **Case of Even n:** - Now, assume n is even, i.e., n = 2m. - We can divide the set of cups into m quartets, each forming the vertices of a square. - To avoid losing, Vasya should follow this strategy: - If Petya fills two cups, they must be in the vertices of one square. - Vasya then pours tea into the two remaining vertices of the same square. - If Petya fills only one cup, Vasya fills the cup in the adjacent vertex of the same square. - Following this strategy, Vasya can always ensure that he has a response move by filling the remaining positions in the vertices of these squares. 5. **Conclusion for Even n:** - Therefore, when n is even, Vasya can always find a response to Petya's move and force a win. # Conclusion: - Petya wins if and only if n is odd, ensuring that Petya has a winning strategy independent of Vasya's actions. [ boxed{text{Petya wins if } n text{ is odd.}} ]

answer:To find the average percentage decrease, we first calculate the total percentage decrease, which is frac{2000 - 1280}{2000} times 100% = 36%. Since the price was reduced twice, the average percentage decrease per reduction is frac{36%}{2} = 18%. Therefore, the correct answer is boxed{text{C: 18%}}.