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

3w + 7=8 need help with this

LiRa [457]

Hope this helps you out 3w+7=8

The answer is w= 1/3

Step by step:

3w+7=8
Step 1: is subtract is 7 from both sides.
3w+7-7=8-7
3w=1

Step 2: Divide both sides by 3.
3w/3 = 1/3

Your solution is:

w=1/3

6 0

3 years ago

If I roll a 6 sided dice and then flip a coin,

lina2011 [118]

Answer:

1/12

Step-by-step explanation:

5 0

3 years ago

Total surface area of this shape

skelet666 [1.2K]

Answer:

6 * 10 * 6

Step-by-step explanation:

6 0

4 years ago