We are drawing a single card from a standard 52- deck. find the following probability. p(two | no face card)

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

See attachment below and find the equivalent of tan(∠qsr)

Suppose you are choosing a 6-digit personal access code. this code is made up of 4 digits chosen from 1 to 9, followed by 2 letters chosen from a to z. any of these digits or letters can be repeated. find the total number of personal access codes that can be formed. 492,804 341,172 39,917,124 4,435,236