The y-intercept of the line is (0 -6). The given function has no vertical and horizontal asymptotes
<h3>y-intercept and asymptotes of an equation</h3>
A linear equation is an equation that has a leading degree of 1. Given the equation
g(x) = 4x - 6
The y-intercept occurs at a point where x = 0
g(0) = 4(0) - 6
g(0) = -6
Hence the y-intercept of the line is (0 -6)
The given function has no vertical and horizontal asymptotes
Learn more on intercept here: brainly.com/question/18831322
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Answer:
2x2+2+2+2=10 2x2x2x2+2divied by 2 +0=9
Step-by-step explanation:
there you go!
Answer:
![Range=14](https://tex.z-dn.net/?f=Range%3D14)
![\sigma^2 =32.4](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D32.4)
![\sigma = 5 .7](https://tex.z-dn.net/?f=%5Csigma%20%3D%205%20.7)
The standard deviation will remain unchanged.
Step-by-step explanation:
Given
![Data: 136, 129, 141, 139, 138, 127](https://tex.z-dn.net/?f=Data%3A%20136%2C%20129%2C%20141%2C%20139%2C%20138%2C%20127)
Solving (a): The range
This is calculated as:
![Range = Highest - Least](https://tex.z-dn.net/?f=Range%20%3D%20Highest%20-%20Least)
Where:
![Highest = 141; Least = 127](https://tex.z-dn.net/?f=Highest%20%3D%20141%3B%20Least%20%3D%20127)
So:
![Range=141-127](https://tex.z-dn.net/?f=Range%3D141-127)
![Range=14](https://tex.z-dn.net/?f=Range%3D14)
Solving (b): The variance
First, we calculate the mean
![\bar x = \frac{1}{n} \sum x](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%20%5Csum%20x)
![\bar x = \frac{1}{6} (136+ 129+ 141+ 139+ 138+ 127)](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%28136%2B%20129%2B%20141%2B%20139%2B%20138%2B%20127%29)
![\bar x = \frac{1}{6} *810](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20%5Cfrac%7B1%7D%7B6%7D%20%2A810)
![\bar x = 135](https://tex.z-dn.net/?f=%5Cbar%20x%20%3D%20135)
The variance is calculated as:
![\sigma^2 =\frac{1}{n-1}\sum(x - \bar x)^2](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%28x%20-%20%5Cbar%20x%29%5E2)
So, we have:
![\sigma^2 =\frac{1}{6-1}*[(136 - 135)^2 +(129 - 135)^2 +(141 - 135)^2 +(139 - 135)^2 +(138 - 135)^2 +(127 - 135)^2]](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7B6-1%7D%2A%5B%28136%20-%20135%29%5E2%20%2B%28129%20-%20135%29%5E2%20%2B%28141%20-%20135%29%5E2%20%2B%28139%20-%20135%29%5E2%20%2B%28138%20-%20135%29%5E2%20%2B%28127%20-%20135%29%5E2%5D)
![\sigma^2 =\frac{1}{5}*[162]](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D%5Cfrac%7B1%7D%7B5%7D%2A%5B162%5D)
![\sigma^2 =32.4](https://tex.z-dn.net/?f=%5Csigma%5E2%20%3D32.4)
Solving (c): The standard deviation
This is calculated as:
![\sigma = \sqrt {\sigma^2 }](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B%5Csigma%5E2%20%7D)
![\sigma = \sqrt {32.4}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%20%7B32.4%7D)
--- approximately
Solving (d): With the stated condition, the standard deviation will remain unchanged.
The number -4 is a whole number