We have k coins. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0, 1]. We assume that conditioned on Q -q, all coin tosses are independent. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, fori- 1,2,..., k. (Tt includes the toss that results in the first Heads.) You may find the following integral useful: For any non-negative integers k and m k!m! e* (1-q)" dq = (k + m + 1)! 1. Find the PMF of T1.(Express your answer in terms of t using standard notation.) For t = 1, 2, . . .' PT (t) 2. Find the least mean squares (LMS) estimate of Q based on the observed value, t, of Ti.(Express your answer in terms of t using standard notation.) 3. We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate of Q given the number of tosses needed, Tİ-ti , . . . , T = tk, for each coin. Choose the correct expression for q k -1 k +1 O none of the above

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Complex mathematical expressions are in the question

Complete the standard form of the equation that represents the quadratic relationship displayed above, where a, b, and c are rational numbers.

Cone w has a radius of 8 cm and a height of 5 cm. square pyramid x has the same base area and height as cone w. paul and manuel disagree on how the volumes of cone w and square pyramid x are related. examine their arguments. which statement explains whose argument is correct and why? paul manuel the volume of square pyramid x is equal to the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is three times the volume of cone w. this can be proven by finding the base area and volume of cone w, along with the volume of square pyramid x. the base area of cone w is π(r2) = π(82) = 200.96 cm2. the volume of cone w is one third(area of base)(h) = one third(200.96)(5) = 334.93 cm3. the volume of square pyramid x is (area of base)(h) = (200.96)(5) = 1,004.8 cm3. paul's argument is correct; manuel used the incorrect formula to find the volume of square pyramid x. paul's argument is correct; manuel used the incorrect base area to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect formula to find the volume of square pyramid x. manuel's argument is correct; paul used the incorrect base area to find the volume of square pyramid x.

Solve for x and y a x= 13.3 y= 16.7 b x= 23.3 y= 12.5 c x= 7.5 y= 16.7 d x=7.5 y= 12.5