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 such that 17.36% of the 60 days have a travel time that is at least . you may find this standard normal distribution table or this list of software manuals useful. give your answer rounded to two decimal places.

the answer is 45.28 minutes

Step-by-step explanation:

Let X be the travel time in minutes

X is normally distributed with μ= 36.6 and σ=10.3 i.e. X~N (35.6, 10.3)

The travel time x such that 17.36% of the 60 days have a travel time that is at least x is given by,

P(X>x)= 0.1736

P(X<x)= 0.8264

P(Z<x−35.6/10.3)= 0.8264

P(Z<0.940)=0.8264

x−35.6/10.3= 0.940

x=0.940* 10.3+35.6=45.28

Therefore, x is equal to 45.28 minutes

x+y = 29, 5x + 2y = 100

step-by-step explanation:

i'm guessing it was a multiple first option was the correct answer