According to a survey, 63\%63%63, percent of the Scottish population has visited woodland in the previous year. Skeptical about this claim, you decide to take a simple random sample of 650650650 people in this population, and you ask them if they had visited woodland in the previous year. You find that 60\%60%60, percent of the sample replied "yes" to your question.
Assuming that the original survey's 63\%63%63, percent claim is correct, what is the approximate probability that less than 60\%60%60, percent of the sample would report that they had visited woodland in the previous year?

5.25 ounces

step-by-step explanation:

divided 3/4 and 6

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