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

ANSWER ASAP ILL MARK BRAINLIEST!!!!

Korolek [52]

Answer:

$2 {x}^{2} - 6x - 1 = 0 \\ a = 2 \: \: \: b = - 6 \: \: \: c = - 1 \\ \\ x = \frac{6\binom{ + }{ - } \ \sqrt{ {6}^{2} - 4(2)( - 1)} }{2 \times 2} \\ \\ x = \frac{6 \binom{ + }{ - } \sqrt{44} }{4} \\ x = \frac{6 + \sqrt{44} }{4} \\ \\ or x = \frac{6 - \sqrt{44} }{4}$

I hope that is useful for you :)

8 0

2 years ago

Can someone please help me with this problem?

Bumek [7]

Answer:

f'(1) = 2

Step-by-step explanation:

f(x) = 2x^2 -2x +3

Take the derivative

f'(x) = 2 * 2x - 2 * 1

f'(x) = 4x -2

f'(1) = 4(1) -2

=4-2

4 0

3 years ago