D.
BECAUSE I HAVE TO SI I CAN YOUR LIPAT KA MAN
Answer:
probative value
Explanation:
The Fourth Amendment to the United States Constitution provides legal protection against search and seizure without good reason and without a search and seizure warrant. As we can see from the question above, Mick Stoner was charged with marijuana, which was taken from a legal dispensary in Denver, seized without a search warrant, so Mick Stoner's defense attorney disputes the evidential value of the evidence, claiming that the truck's initial search violated the Fourth Amendment.
To dispute the probative value of the evidence means that the evidence gathered, in this case, is not guaranteed, by itself, to support a condemnatory sentence, thus requiring the repetition in court of some of the evidence produced.
Answer:
ARE THESE FREE POINTS!?!?!?!
Explanation:
<h2>PLEASE MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!</h2>
Answer:
Concentric circles are circles with a common center. The region between two concentric circles of different radii is called an annulus. Any two circles can be made concentric by inversion by picking the inversion center as one of the limiting points.
1. Picking any two points on the outer circle and connecting them gives 1/3.
2. Picking any random point on a diagonal and then picking the chord that perpendicularly bisects it gives 1/2.
3. Picking any point on the large circle, drawing a line to the center, and then drawing the perpendicularly bisected chord gives 1/4.
So some care is obviously needed in specifying what is meant by "random" in this problem.
Given an arbitrary chord BB^' to the larger of two concentric circles centered on O, the distance between inner and outer intersections is equal on both sides (AB=A^'B^'). To prove this, take the perpendicular to BB^' passing through O and crossing at P. By symmetry, it must be true that PA and PA^' are equal. Similarly, PB and PB^' must be equal. Therefore, PB-PA=AB equals PB^'-PA^'=A^'B^'. Incidentally, this is also true for homeoids, but the proof is nontrivial.
