answer:On the first two days, William rode his horse for a maximum of 6 hours per day, so that's 6 hours x 2 days = 12 hours. On the next two days, he rode his horse for only 1.5 hours per day, so that's 1.5 hours x 2 days = 3 hours. On the remaining two days, he rode his horse for half of the maximum time, so that's 6 hours / 2 = 3 hours per day. Therefore, that's 3 hours x 2 days = 6 hours. So in total, William rode his horse for 12 hours + 3 hours + 6 hours = 21 hours. 21 The answer is: 21

answer:There are 52 cards in the deck, so the probability that the first card is a Jack is frac{4}{52}. After the first card is dealt, there are 51 cards left in the deck, so the probability that the second card is a Queen is frac{4}{51}. After the first two cards are dealt, there are 50 cards left in the deck, so the probability that the third card is a King is frac{4}{50}. Therefore, the probability that all three cards are Jack, Queen, and King in order is frac{4}{52} cdot frac{4}{51} cdot frac{4}{50} = frac{64}{132600}. Simplifying the fraction, we get frac{8}{16575}. Therefore, the probability that the first card is a Jack, the second card is a Queen, and the third card is a King is boxed{frac{8}{16575}}.The answer is: frac{8}{16575}

answer:We want to find the value of log_8 2. This means we want to find the exponent to which we must raise 8 to get 2. In other words, we want to find x such that 8^x = 2. Rewriting 8 as 2^3, we have (2^3)^x = 2. Using the property (a^m)^n = a^{mn}, we can simplify this to 2^{3x} = 2. Since 2^1 = 2, we can equate the exponents to get 3x = 1. Dividing both sides by 3, we find x = frac{1}{3}. Therefore, log_8 2 = boxed{frac{1}{3}}. The answer is: frac{1}{3}

answer:If Boast Pool has 75 fish, then Onum Lake has 75 + 25 = 100 fish. Riddle Pond has half as many fish as Onum Lake, so it has 100 / 2 = 50 fish. The total number of fish in all three bodies of water is 75 + 100 + 50 = 225 fish. To find the average number of fish, we divide the total number of fish by the number of bodies of water, which is 3. So, the average number of fish in all three bodies of water is 225 / 3 = 75 fish. 75 The answer is: 75