answer:The function f(x) is an exponential function with a negative exponent. This means that the function is always positive and decreasing. The range of the function is the set of all possible output values. Since the function is always positive, the minimum value of the range is 0. To find the maximum value of the range, we need to find the value of x that makes the exponent of the exponential function as small as possible. This occurs when x = 0. Therefore, the maximum value of the range is f(0) = e^{-frac{4}{3}} = frac{1}{e^{29/28}}. Thus, the range of the function is (0, frac{1}{e^{29/28}}]. The range of the function is (0, frac{1}{e^{29/28}}].

answer:To find f(g(x)), we substitute the entire function g(x) into f(x) wherever x appears: f(g(x)) = frac{1}{10(g(x)) + 17} + 13 = frac{1}{10left(dfrac{1}{9x-6}right) + 17} + 13 Next, we multiply the numerator and denominator of the main fraction by 9x-6: f(g(x))= frac{9x-6}{(9x-6)cdot 10left(dfrac{1}{9x-6}right) + 17} + 13 =frac{9x-6}{10left(1right) + 17(9x-6)} + 13 Simplifying further, we get: f(g(x))= frac{9x - 6}{153x-92} + frac{13(153x - 92)}{153-92} Combining the two fractions, we obtain: f(g(x))= frac{9x - 6 + 13(153x - 92)}{153x-92} Simplifying the numerator, we get: f(g(x))= frac{1998x - 1202}{153x-92} Therefore, f(g(x)) = dfrac{1998x-1202}{153x-92}.

answer:We can use conditional probability to break down the desired probability into smaller parts. First, we condition on the event that X_1 > alpha_k. This means that X_1 is not among the k smallest values in the sample. Given this condition, the remaining N-1 values are still independent and identically distributed with CDF F(x). Next, we consider the probability that X_{(1)}leq alpha_1, X_{(2)}<alpha_2, ...,X_{(k)}<alpha_{k} given that X_1 > alpha_k. This is the probability that the k smallest values in the sample are all less than alpha_k, which is equal to F(alpha_1)left[1-F(alpha_k)right]^{N-k} prod_{i=2}^{k}left[F(alpha_i)-F(alpha_{i-1})right] Finally, we multiply these two probabilities together to get the desired result: P(X_{(1)}leq alpha_1, X_{(2)}<alpha_2, ...,X_{(k)}<alpha_{k}, X_1>alpha_k) = frac{(N-1)!}{(N-k-1)!}F(alpha_1)left[1-F(alpha_k)right]^{N-k} prod_{i=2}^{k}left[F(alpha_i)-F(alpha_{i-1})right] The probability that X_{(1)}leq alpha_1, X_{(2)}<alpha_2, ...,X_{(k)}<alpha_{k}, X_1>alpha_k can be computed as follows: P(X_{(1)}leq alpha_1, X_{(2)}<alpha_2, ...,X_{(k)}<alpha_{k}, X_1>alpha_k) = frac{(N-1)!}{(N-k-1)!}F(alpha_1)left[1-F(alpha_k)right]^{N-k} prod_{i=2}^{k}left[F(alpha_i)-F(alpha_{i-1})right]

answer:Given Data: Mass of the first object: m₁ = 32.5 g Mass of the second object: m₂ = 10.5 g Initial velocity of the first object: u₁ = -18.5 cm/s (moving to the left) Initial velocity of the second object: u₂ = 14.5 cm/s (moving to the right) Let v₁ and v₂ be the final velocities of the objects after the collision. Using the principle of conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ Substituting the given values: (32.5 g)(-18.5 cm/s) + (10.5 g)(14.5 cm/s) = (32.5 g)v₁ + (10.5 g)v₂ Solving for v₂: v₂ = [(32.5 g)(-18.5 cm/s) + (10.5 g)(14.5 cm/s)] / (10.5 g) v₂ ≈ 9.95 cm/s (moving to the right) Using the equation for elastic collision: u₁ - u₂ = -(v₁ - v₂) Substituting the values: -18.5 cm/s - 14.5 cm/s = -(v₁ - 9.95 cm/s) v₁ ≈ -23.05 cm/s (moving to the left) Therefore, the final velocity of the first object is -23.05 cm/s (moving to the left) and the final velocity of the second object is 9.95 cm/s (moving to the right).