Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with

$\mu=42,100\text{ miles}$

$\sigma=2,510\text{ miles}$

Let x be the random variable that represents the lifetime of the tires .

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x= 44,500 miles

$z=\dfrac{44500-42100}{2510}\approx0.96$

For x= 48,000 miles

$z=\dfrac{48000-42100}{2510}\approx2.35$

Using the standard normal distribution table , we have

The p-value : $P(44500$

$P(z$

Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles = 0.1591

3 0

3 years ago

Zared plays basketball on his high school team. One of the things he needs to practice is his free throws. On his first shot, th

Marat540 [252]

Answer:

lol brainliest

Step-by-step explanation:

8 0

2 years ago

to cover a rectangular region of her yard penny needs at least 170.5 square feet of sod. the length of the region is 15.5 feet.

pshichka [43]

To cover at least 170.5 sq.ft. the width must be at least 11 feet
W=170.5sq.ft / 15.5 ft = 11ft

7 0

2 years ago