Answer: 0.9726
Step-by-step explanation:
Let x be the random variable that represents the distance the tires can run until they wear out.
Given : The top-selling Red and Voss tire is rated 50,000 miles, which means nothing. In fact, the distance the tires can run until they wear out is a normally distributed random variable with a 
67,000 miles and a 
5,200 miles.
Then , the probability that a tire wears out before 60,000 miles :
[using p-value table for z]
Hence, the probability that a tire wears out before 60,000 miles= 0.9726
Answer:
3/32
Step-by-step explanation:
One approach to doing this probelm is to evaluate all three functions F, G and H at x = 2:
F(2) = 2(2) - 1 = 3
G(2) = 3(2) + 2 = 8
H(2) = (2)^2 = 4
3/8
Then (F/G)(2) = (3/8), and (F/G/H) = -------- = 3/32
4
Look up what is 75% of 31,500
fg^3 / g^3f = 1
f^4 / f*f*f*f = f^4 / f^4 = 1
f^2g^3 / f^3g^2 = g/f
f*f/f^2 = f^2 / f^2 = 1
Answer:
f^2g^3 / f^3g^2
You could rewrite
as
and be tempted to cancel out the factors of
. But this cancellation is only valid when
.
When
, you end up with the indeterminate form
, which is why
is not a zero.
