We are given that a tire manufacturer warranties its tires to last at least 20,000 miles or "you get a new set of tires." In its past experience, a set of these tires last on average 26,000 miles with S.D. 5,000 miles. Assume that the wear is normally distributed.

<em>Let X = wearing of tires</em>

So, X ~ N( $\mu=26,000,\sigma^{2}=5,000^{2}$ )

Now, the z score probability distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = average lasting of tires = 26,000 miles

$\sigma$ = standard deviation = 5,000 miles

So, probability that a set of tires wears out before 20,000 miles is given by = P(X < 20,000 miles)

P(X < 20,000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{20,000-26,000}{5,000}$ ) = P(Z < -1.2) = 1 - P(Z $\leq$ 1.2)

= 1 - 0.88493 = 0.1151

Therefore, probability that a set of tires wears out before 20,000 miles is 0.1151.

4 0

3 years ago

Use distributive property to create an equivalent expression to 7x + 56

OverLord2011 [107]

You will have to factor this equation. This means you have to find the lowest number that goes into both 7 and 56 (in other terms, gcf(7, 56)). You will put that as the leading coefficient in this factored term. gcf(7,56) = 7 so 7 will be leading coefficient. It will look somewhat like this 7(a + b) (a and b are just random variables representing what will go inside the parentheses). a and b can be determined by dividing each term in this expression by 7. Hence,
7x + 56 = 7(x + 8).This can obviously also be proven by the distributive property.

7 0

2 years ago

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