Suppose abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min. determine the travel time x such that 22.96% of the 60 days have a travel time that is at least x.

The travel time is at least 28 minute.

Step-by-step explanation:

Given : Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min.

To determine : The travel time x such that 22.96% of the 60 days have a travel time that is at least x.

Solution :

Mean, μ=35.6 min

Standard deviation, σ=10.3 min

we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.

Using z-table, the z-score that will give us 0.2296 is -0.74 (Shown in attached graph)

The formula of z-score is given by:

Approximately the value of x is 28.

The travel time is at least 28 minute.

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation:

ok so the answer to your question is simple as long as you look at the different dimensions that contain the exemplary letters that combined in the right order give you words that eventually within the 5th dimension will conform and give you the correct answer to your question. hope i , stop looking at this easily, look at it with a spectral view.

step-by-step explanation:

told you