Out of 2500 light bulbs that were tested, 135 were defective. Based on this information calculate the following probabilities. Enter your answers as a reduced fraction using / for the fraction bar.
P(defective) =
P(not defective) =

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v