answer:To find the expected value of 1/X, we need to compute the integral of (1/x) times the probability density function (pdf) of X, over the entire range of X. In this case, the pdf of X is given by: f(x) = { 2x, for 0 ≤ x ≤ 1 0, otherwise } The expected value of 1/X is therefore: E(1/X) = ∫[0,1] (1/x) * 2x dx To evaluate this integral, we can simplify the integrand to: E(1/X) = ∫[0,1] 2 dx However, this simplification is incorrect, as the original integral had (1/x) as a factor, which is not defined at x=0. The correct approach is to recognize that the integral ∫[0,1] (1/x) * 2x dx is an improper integral, since the integrand is not defined at x=0. To evaluate this integral, we can start by rewriting it as: E(1/X) = ∫[a,1] (1/x) * 2x dx where a is a small positive number. Evaluating this integral, we get: E(1/X) = ∫[a,1] 2 dx/x = 2 ∫[a,1] dx/x = 2 [ln(x)] from a to 1 = 2 [ln(1) - ln(a)] = -2 ln(a) Now, we take the limit as a approaches 0 from the right: E(1/X) = lim (a→0+) -2 ln(a) However, this limit does not exist, as ln(a) approaches -∞ as a approaches 0. This indicates that the expected value of 1/X does not exist, or is infinite. In other words, the expected value of 1/X is not well-defined for this particular distribution, due to the singularity at x=0. This is a common issue when dealing with inverse random variables, and it highlights the importance of carefully considering the properties of the distribution before attempting to compute expectations.

answer:To solve this problem, let's consider the snail's progress each day. The snail climbs 3 feet during the day and slips back 2 feet at night, so it effectively moves 1 foot up the well each day. However, on the 18th day, the snail climbs 3 feet and reaches the top of the well, which is 20 feet high. Since it's already at the top, it won't slip back at night, so we don't need to subtract the 2 feet it would normally slip back. Therefore, the snail takes 18 days to reach the top of the well.

answer:To find out how many days it will take the snail to reach the top of the well, let's think about the progress it makes each day. The snail effectively moves 1 foot up the well each day, as it climbs 3 feet and slips back 2 feet. Since the well is 20 feet deep, if the snail moved 1 foot up every day without any exceptions, it would take 20 days to reach the top. However, on the final day, the snail will climb 3 feet and reach the top of the well without slipping back. In the first 17 days, the snail effectively moves 17 feet (1 foot per day). On the 18th day, when it climbs 3 feet, it will reach 20 feet (17 feet from the previous days + 3 feet on the final day). Therefore, it takes the snail 18 days to reach the top of the well, since it climbs 3 feet on the final day and doesn't slip back, allowing it to cover the remaining distance to the top.

answer:To find out how many ways the manager can choose 5 employees out of 20 to award the bonuses to, we need to consider that the order in which the employees are chosen doesn't matter. This is a combination problem, because we are selecting a group of people without regard to their order. The number of ways to choose a group of 5 employees out of 20 can be calculated using the combination formula: 20 choose 5. This is often written as C(20, 5) or 20C5 and can be calculated as 20! / (5! * (20-5)!). Breaking this down further, we divide the factorial of 20 (the total number of employees) by the factorial of 5 (the number of employees to choose) and the factorial of 15 (the remaining number of employees). Simplifying this expression, C(20, 5) equals 20! / (5! * 15!), which equals (20 * 19 * 18 * 17 * 16) / (5 * 4 * 3 * 2 * 1). After calculating the result, the manager can choose 5 employees out of 20 to award the bonuses to in 15,504 ways.