Answer:
-3 ![\frac{9}{11}](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B11%7D)
Step-by-step explanation:
It’s 569.8
.................
You can solve this problem through factoring.
First, you have the equation,
![h(x) = \frac{x^2-36}{x-6}](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%5Cfrac%7Bx%5E2-36%7D%7Bx-6%7D)
Then, you can factor the numerator.
![h(x) = \frac{(x+6)(x-6)}{x-6}](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%5Cfrac%7B%28x%2B6%29%28x-6%29%7D%7Bx-6%7D)
You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.
You are left with ![h(x) = x+6](https://tex.z-dn.net/?f=h%28x%29%20%3D%20x%2B6)
The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.
The answer would be 7.225.
Answer:
option f is right
Step-by-step explanation:
Given that data is collected to perform the following hypothesis test.
![H_0: \mu = 5.5\\H_a: \mu >5.5](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu%20%3D%205.5%5C%5CH_a%3A%20%5Cmu%20%3E5.5)
(right tailed test)
Sample mean = 5.4
p value = 0.1034
when p value = 0.1034 we normally accept null hypothesis. i.e chances of null hypothesis true is the probability of obtaining test results at least as extreme as the results actually observed during the test, assuming that the null hypothesis is correct
f) If the mean µ does not differ significantly from 5.5 (that is, if the null hypothesis is true), then the probability of obtaining a sample mean y as far or farther from 5.5 than 5.4 is .1034.
.