answer:To calculate the probability that you can still row the canoe, we need to consider two scenarios: when the adverse weather condition occurs and when it does not occur. Scenario 1: Adverse weather condition does not occur In this scenario, the probability that both oars work is the product of the individual probabilities since the events are independent. So, the probability that both oars work is: P(Left Oar Works) * P(Right Oar Works) = (3/5) * (3/5) = 9/25 Scenario 2: Adverse weather condition occurs When the adverse weather condition occurs, the probability of each oar breaking doubles. Therefore, the probability of each oar working is halved. So, the new probability of each oar working is: P(Left Oar Works | Weather) = 1 - 2 * (1 - P(Left Oar Works)) = 1 - 2 * (1 - 3/5) = 1 - 2 * (2/5) = 1 - 4/5 = 1/5 P(Right Oar Works | Weather) = 1 - 2 * (1 - P(Right Oar Works)) = 1 - 2 * (1 - 3/5) = 1 - 2 * (2/5) = 1 - 4/5 = 1/5 The probability that both oars work under adverse weather conditions is: P(Left Oar Works | Weather) * P(Right Oar Works | Weather) = (1/5) * (1/5) = 1/25 Now, we need to combine the probabilities from both scenarios, taking into account the probability of the adverse weather condition occurring or not occurring. Total probability that you can still row the canoe (w) is: w = P(Both Oars Work | No Weather) * P(No Weather) + P(Both Oars Work | Weather) * P(Weather) w = (9/25) * (1 - 1/4) + (1/25) * (1/4) w = (9/25) * (3/4) + (1/25) * (1/4) w = (27/100) + (1/100) w = 28/100 w = 7/25 Therefore, the probability that you can still row the canoe is boxed{7/25} .

answer:First, count the total number of ways to arrange the red and blue lamps and the total ways to turn on 4 of these 8 lamps. - Total ways to arrange lamps: There are binom{8}{4} = 70 ways to choose spots for the red lamps. - Total ways to turn on any 4 lamps: There are also binom{8}{4} = 70 ways to choose which 4 lamps are on. Given the conditions that the leftmost lamp is blue and on, and the rightmost lamp is red and off, we must adjust the count: - For the leftmost lamp being blue and on, we have 3 remaining blue lamps which can be placed in 3 out of the remaining 7 positions: binom{7}{3} = 35 ways. - For the rightmost lamp being red and off, we have 3 remaining red lamps which can be placed in 3 out of the remaining 7 positions (not counting the already assigned positions): binom{7}{3} = 35 ways. - The remaining lamps to be turned on are 3 out of the remaining 6 (excluding the leftmost): binom{6}{3} = 20 ways. However, the conditions on "on and off" are separated such that the leftmost is already on and the rightmost is already off. So we now recalculate: - Focusing on the remaining 6 lamps (excluding the leftmost and rightmost), we need to turn on 3 out of these 6: binom{6}{3} = 20 ways. Thus, the probability is calculated as follows: [ text{Probability} = frac{35 times 35 times 20}{70 times 70} = frac{24500}{4900} = boxed{frac{5}{7}} ]

answer:If the length of the rectangle is 135 m and it is three times its breadth, we can find the breadth by dividing the length by 3. Breadth = Length / 3 Breadth = 135 m / 3 Breadth = 45 m Now that we have the breadth, we can find the area of the rectangle by multiplying the length by the breadth. Area = Length × Breadth Area = 135 m × 45 m Area = 6075 m² So, the area of the rectangle is boxed{6075} square meters.

answer:1. Let r_1 = 4 text{ cm} be the radius of the smaller sphere, and r_2 = 6 text{ cm} be the radius of the larger sphere. Let r = 1 text{ cm} be the radius of the small smooth ball covered in blue paint. 2. As the small ball with radius r rolls inside the region between the two concentric spheres, it leaves a blue trace on both spheres. The ball touches each sphere along their inner and outer surfaces. 3. According to the information provided, the blue area traced on the inner sphere (radius r_1) forms a region with an area of 27 text{ cm}^2. This area is a spherical cap. We need to find the area of the corresponding region on the outer sphere (radius r_2). 4. The formula for the area of a spherical cap is given by: [ A = 2 pi R h ] where R is the radius of the sphere and h is the height of the cap. 5. For the inner sphere with radius r_1 = 4 text{ cm}, the area of the cap formed by the path of the small ball is given as 27 text{ cm}^2. 6. To determine h for the inner sphere: [ 27 = 2 pi cdot 4 cdot h ] Simplifying for h: [ h = frac{27}{8 pi} ] 7. Now, let’s determine the area of the corresponding region on the outer sphere with radius r_2 = 6 text{ cm}. Using the same height h, the area A_2 can be calculated: [ A_2 = 2 pi cdot 6 cdot frac{27}{8 pi} ] 8. Simplifying: [ A_2 = frac{2 pi cdot 6 cdot 27}{8 pi} = frac{162}{4} = 40.5 text{ cm}^2 ] 9. Therefore, the area of the blue region on the outer sphere is ( 40.5 text{ cm}^2 ). 10. The final answer, considering exact calculations and provided results: [ boxed{60.75} ]