Use the information given in the diagram to prove that m∠JGI = One-half(b – a), where a and b represent the degree measures of arcs FH and JI. A circle is shown. Secants G J and G I intersect at point G outside of the circle. Secant G J intersects the circle at point F. Secant G I intersects the circle at point H. The measure of arc F H is a. The measure of arc J I is b. A dotted line is drawn from point J to point H. Angles JHI and GJH are inscribed angles. We have that m∠JHI = One-half b and m∠GJH = One-halfa by the . Angle JHI is an exterior angle of triangle . Because the measure of an exterior angle is equal to the sum of the measures of the remote interior angles, m∠JHI = m∠JGI + m∠GJH. By the , One-halfb = m∠JGI + One-halfa. Using the subtraction property, m∠JGI = One-halfb – One-halfa. Therefore, m∠JGI = One-half(b – a) by the distributive property.

Amachine that produces a special type of transistor (a component of computers) has a 2% defective rate. the production is considered a random process where each transistor is independent of the others. (a) what is the probability that the 10th transistor produced is the first with a defect? (b) what is the probability that the machine produces no defective transistors in a batch of 100? (c) on average, how many transistors would you expect to be produced before the first with a defect? what is the standard deviation? (d) another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. on average how many transistors would you expect to be produced with this machine before the first with a defect? what is the standard deviation? (e) based on your answers to parts (c) and (d), how does increasing the probability of an event a↵ect the mean and standard deviation of the wait time until success?