Answer:
There are asymptotes at x = three-halves and x = negative one-third.
Step-by-step explanation:
f(x) = (x + 1)/ (6x^2 - 7x - 3)
= (x + 1)( / (6x^2 + 2x - 9x - 3)
= (x + 1) / (2x(3x + 1) - 3(3x + 1))
= (x + 1) / (2x - 3)(3x + 1)
Now x = 3/2 and x = -1/3 both make te denominator zero so these are both asymptotes.
The answer would be letter B
Answer:
The sample consisting of 64 data values would give a greater precision.
Step-by-step explanation:
The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).
That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.
Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.
The two sample sizes are:
<em>n</em>₁ = 25
<em>n</em>₂ = 64
The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.
Width for n = 25:
Width for n = 64:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B64%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Thus, the sample consisting of 64 data values would give a greater precision
What is the question?
I'm assuming it is to find the length and width.
+_= plus or minus
(X+36)
____________
| |
(X) | |
|____________|
X^2+36X-2040<0
X<-36+_(36^2-4*-2040)^(1/2)
-----------------------------------
2
X<-18+_2((591)^(1/2))
This is probably not what you wanted, sorry