Reserve problems chapter 4 section 5 problem 2 a laptop company claims up to 10.1 hours of wireless web usage for its newest laptop battery life. however, reviews on this laptop shows many complaints about low battery life. a survey on battery life reported by customers shows that it follows a normal distribution with mean 9.5 hours and standard deviation 30 minutes. (a) what is the probability that the battery life is at least 10.1 hours? round your answer to four decimal places (e. g. 98.7654).

0.1151

Step-by-step explanation:

Given that a laptop company claims up to 10.1 hours of wireless web usage for its newest laptop battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 9.5 hours and standard deviation 30 minutes.

If X is the battery life then X is N(9.5, 0.5)

(we convert into one unit for calculations purpose here hours)

a) the probability that the battery life is at least 10.1 hours

=

we get the probability that the battery life is at least 10.1 hours is 0.1151

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )

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