Determine the area under the standard normal curve that lies to the right of the z-score 0.16 and to the left of the z-score 0.26. z−0.2−0.10.00.10.20.30.40.50.000.42 070.46020.50000.53980.57930.61790.6 5540.69150.010.41680.45620.50400.54 380.58320.62170.65910.69500.020.412 90.45220.50800.54780.58710.62550.66 280.69850.030.40900.44830.51200.551 70.59100.62930.66640.70190.040.4052 0.44430.51600.55570.59480.63310.670 00.70540.050.40130.44040.51990.5596 0.59870.63680.67360.70880.060.39740 .43640.52390.56360.60260.64060.6772 0.71230.070.39360.43250.52790.56750 .60640.64430.68080.71570.080.38970. 42860.53190.57140.61030.64800.68440 .71900.090.38590.42470.53590.57530. 61410.65170.68790.7224
Use the value(s) from the table above.

Provide your answer below:

i'm sorry i don't know

step-by-step explanation:

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