Answer:
x < 3
Step-by-step explanation:
I think it’s y = 10x + 10
Answer: 0.1824
Step-by-step explanation:
Given : The mileage per day is distributed normally with
Mean : ![\mu=110\text{ miles per day}](https://tex.z-dn.net/?f=%5Cmu%3D110%5Ctext%7B%20miles%20per%20day%7D)
Standard deviation : ![\sigma=38\text{ miles per day}](https://tex.z-dn.net/?f=%5Csigma%3D38%5Ctext%7B%20miles%20per%20day%7D)
Let X be the random variable that represents the distance traveled by truck in one day .
Now, calculate the z-score :-
![z=\dfrac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
For x= 132 miles per day.
![z=\dfrac{132-110}{38}\approx0.58](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B132-110%7D%7B38%7D%5Capprox0.58)
For x= 159 miles per day.
![z=\dfrac{159-110}{38}\approx1.29](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B159-110%7D%7B38%7D%5Capprox1.29)
Now by using standard normal distribution table, the probability that a truck drives between 132 and 159 miles in a day will be :-
![P(132](https://tex.z-dn.net/?f=P%28132%3Cx%3C159%29%3DP%280.58%3Cz%3C1.29%29%5C%5C%5C%5C%3DP%28z%3C1.29%29-P%280.58%29%5C%5C%5C%5C%3D%200.9014747-0.7190426%3D0.1824321%5Capprox0.1824)
Hence, the probability that a truck drives between 132 and 159 miles in a day =0.1824
The answer is 11/4 or 2 3/4