The tread life of a particular brand of tire is normally distributed with mean 60,000 miles and standard deviation 3800 miles. Suppose 35 tires are randomly selected for a quality assurance test. Find the probability that the mean tread life from this sample of 35 tires is greater than 59,000 miles. You may use your calculator, but show what you entered to find your answer. Round decimals to the nearest ten-thousandth (four decimal places).

P [ x > 59000} = 0,6057

Step-by-step explanation:

We assume Normal Distribution

P [ x > 59000} = (x - μ₀ ) /σ/√n

P [ x > 59000} =  (59000 - 60000)/ 3800

P [ x > 59000} = - 1000/3800/√35

P [ x > 59000} = - 1000*5,916 /3800

P [ x > 59000} =  - 5916/3800

P [ x > 59000} = - 1,55

We look for p value for that z score n z-table and find

P [ x > 59000} = 0,6057

if this is really you s much. i only have 3 points left and i have 2 questions to ask

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation: