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

Solve the system by subtitution y = 3x + 18 y = 5x

elena-14-01-66 [18.8K]

Answer:

x=9

Step-by-step explanation:

y=3x+18=5x

thus 3x+18=5x

subtract 3x from both sides 18=2x

simplify for x x=9

3 0

3 years ago

A number is cut in half

cupoosta [38]

Answer:

Step-by-step explanation:

7 0

2 years ago

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