Part A:
Given
![f:Z \rightarrow Z,](https://tex.z-dn.net/?f=f%3AZ%20%5Crightarrow%20Z%2C%20)
defined by
![f(x)=-x](https://tex.z-dn.net/?f=f%28x%29%3D-x)
![f(x+y)=-(x+y)=-x-y \\ \\ f(x)+f(y)=-x+(-y)=-x-y](https://tex.z-dn.net/?f=f%28x%2By%29%3D-%28x%2By%29%3D-x-y%20%5C%5C%20%20%5C%5C%20f%28x%29%2Bf%28y%29%3D-x%2B%28-y%29%3D-x-y)
but
![f(xy)=-xy \\ \\ f(x)\cdot f(y)=-x\cdot-y=xy](https://tex.z-dn.net/?f=f%28xy%29%3D-xy%20%5C%5C%20%20%5C%5C%20f%28x%29%5Ccdot%20f%28y%29%3D-x%5Ccdot-y%3Dxy)
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
![f:Z_2 \rightarrow Z_2,](https://tex.z-dn.net/?f=f%3AZ_2%20%5Crightarrow%20Z_2%2C%20)
defined by
![f(x)=-x](https://tex.z-dn.net/?f=f%28x%29%3D-x)
Note that in
![Z_2](https://tex.z-dn.net/?f=Z_2)
, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular
![f(x)=x](https://tex.z-dn.net/?f=f%28x%29%3Dx)
![f(x+y)=x+y \\ \\ f(x)+f(y)=x+y](https://tex.z-dn.net/?f=f%28x%2By%29%3Dx%2By%20%5C%5C%20%20%5C%5C%20f%28x%29%2Bf%28y%29%3Dx%2By)
and
![f(xy)=xy \\ \\ f(x)\cdot f(y)=xy](https://tex.z-dn.net/?f=f%28xy%29%3Dxy%20%5C%5C%20%20%5C%5C%20f%28x%29%5Ccdot%20f%28y%29%3Dxy)
Therefore, the function is a homomorphism.
Part C:
Given
![g:Q\rightarrow Q](https://tex.z-dn.net/?f=g%3AQ%5Crightarrow%20Q)
, defined by
![g(x)= \frac{1}{x^2+1}](https://tex.z-dn.net/?f=g%28x%29%3D%20%5Cfrac%7B1%7D%7Bx%5E2%2B1%7D%20)
![g(x+y)= \frac{1}{(x+y)^2+1} = \frac{1}{x^2+2xy+y^2+1} \\ \\ g(x)+g(y)= \frac{1}{x^2+1} + \frac{1}{y^2+1} = \frac{y^2+1+x^2+1}{(x^2+1)(y^2+1)} = \frac{x^2+y^2+2}{x^2y^2+x^2+y^2+1}](https://tex.z-dn.net/?f=g%28x%2By%29%3D%20%5Cfrac%7B1%7D%7B%28x%2By%29%5E2%2B1%7D%20%3D%20%5Cfrac%7B1%7D%7Bx%5E2%2B2xy%2By%5E2%2B1%7D%20%20%5C%5C%20%20%5C%5C%20g%28x%29%2Bg%28y%29%3D%20%5Cfrac%7B1%7D%7Bx%5E2%2B1%7D%20%2B%20%5Cfrac%7B1%7D%7By%5E2%2B1%7D%20%3D%20%5Cfrac%7By%5E2%2B1%2Bx%5E2%2B1%7D%7B%28x%5E2%2B1%29%28y%5E2%2B1%29%7D%20%3D%20%5Cfrac%7Bx%5E2%2By%5E2%2B2%7D%7Bx%5E2y%5E2%2Bx%5E2%2By%5E2%2B1%7D%20)
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
![h:R\rightarrow M(R)](https://tex.z-dn.net/?f=h%3AR%5Crightarrow%20M%28R%29)
, defined by
![h(a)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)](https://tex.z-dn.net/?f=h%28a%29%3D%20%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-a%260%5C%5Ca%260%5Cend%7Barray%7D%5Cright%29%20)
![h(a+b)= \left(\begin{array}{cc}-(a+b)&0\\a+b&0\end{array}\right)= \left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right) \\ \\ h(a)+h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)+ \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)=\left(\begin{array}{cc}-a-b&0\\a+b&0\end{array}\right)](https://tex.z-dn.net/?f=h%28a%2Bb%29%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-%28a%2Bb%29%260%5C%5Ca%2Bb%260%5Cend%7Barray%7D%5Cright%29%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-a-b%260%5C%5Ca%2Bb%260%5Cend%7Barray%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20h%28a%29%2Bh%28b%29%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-a%260%5C%5Ca%260%5Cend%7Barray%7D%5Cright%29%2B%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-b%260%5C%5Cb%260%5Cend%7Barray%7D%5Cright%29%3D%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-a-b%260%5C%5Ca%2Bb%260%5Cend%7Barray%7D%5Cright%29)
but
![h(ab)= \left(\begin{array}{cc}-ab&0\\ab&0\end{array}\right) \\ \\ h(a)\cdot h(b)= \left(\begin{array}{cc}-a&0\\a&0\end{array}\right)\cdot \left(\begin{array}{cc}-b&0\\b&0\end{array}\right)= \left(\begin{array}{cc}ab&0\\-ab&0\end{array}\right)](https://tex.z-dn.net/?f=h%28ab%29%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-ab%260%5C%5Cab%260%5Cend%7Barray%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20h%28a%29%5Ccdot%20h%28b%29%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-a%260%5C%5Ca%260%5Cend%7Barray%7D%5Cright%29%5Ccdot%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7D-b%260%5C%5Cb%260%5Cend%7Barray%7D%5Cright%29%3D%20%5Cleft%28%5Cbegin%7Barray%7D%7Bcc%7Dab%260%5C%5C-ab%260%5Cend%7Barray%7D%5Cright%29)
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
![f:Z_{12}\rightarrow Z_4](https://tex.z-dn.net/?f=f%3AZ_%7B12%7D%5Crightarrow%20Z_4)
, defined by
![\left([x_{12}]\right)=[x_4]](https://tex.z-dn.net/?f=%5Cleft%28%5Bx_%7B12%7D%5D%5Cright%29%3D%5Bx_4%5D)
, where
![[u_n]](https://tex.z-dn.net/?f=%5Bu_n%5D)
denotes the lass of the integer
![u](https://tex.z-dn.net/?f=u)
in
![Z_n](https://tex.z-dn.net/?f=Z_n)
.
Then, for any
![[a_{12}],[b_{12}]\in Z_{12}](https://tex.z-dn.net/?f=%5Ba_%7B12%7D%5D%2C%5Bb_%7B12%7D%5D%5Cin%20Z_%7B12%7D)
, we have
![f\left([a_{12}]+[b_{12}]\right)=f\left([a+b]_{12}\right) \\ \\ =[a+b]_4=[a]_4+[b]_4=f\left([a]_{12}\right)+f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%2B%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Ba%2Bb%5D_%7B12%7D%5Cright%29%20%5C%5C%20%20%5C%5C%20%3D%5Ba%2Bb%5D_4%3D%5Ba%5D_4%2B%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29%2Bf%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
and
![f\left([a_{12}][b_{12}]\right)=f\left([ab]_{12}\right) \\ \\ =[ab]_4=[a]_4[b]_4=f\left([a]_{12}\right)f\left([b]_{12}\right)](https://tex.z-dn.net/?f=f%5Cleft%28%5Ba_%7B12%7D%5D%5Bb_%7B12%7D%5D%5Cright%29%3Df%5Cleft%28%5Bab%5D_%7B12%7D%5Cright%29%20%5C%5C%20%5C%5C%20%3D%5Bab%5D_4%3D%5Ba%5D_4%5Bb%5D_4%3Df%5Cleft%28%5Ba%5D_%7B12%7D%5Cright%29f%5Cleft%28%5Bb%5D_%7B12%7D%5Cright%29)
Therefore, the function is a homomorphism.
Answer: B.792
Step-by-step explanation:
Answer:
The correct answer is "$159 and $175".
Step-by-step explanation:
The give values are:
Mean,
= $167
Standard deviation,
= $40
Number of commuters,
n = 100
Now,
⇒ ![SE=\frac{\sigma}{\sqrt{n} }](https://tex.z-dn.net/?f=SE%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%20%7D)
On putting the given values, we get
⇒ ![=\frac{40}{\sqrt{100} }](https://tex.z-dn.net/?f=%3D%5Cfrac%7B40%7D%7B%5Csqrt%7B100%7D%20%7D)
⇒ ![=\frac{40}{10}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B40%7D%7B10%7D)
⇒ ![=4](https://tex.z-dn.net/?f=%3D4)
By using the 2 SE rule of thumb, we get
= ![$(167 - 2\times 4)](https://tex.z-dn.net/?f=%24%28167%20-%202%5Ctimes%204%29)
= ![167-8](https://tex.z-dn.net/?f=167-8)
=
($)
Or,
= ![(167 + 2\times 4)](https://tex.z-dn.net/?f=%28167%20%2B%202%5Ctimes%204%29)
= ![167+8](https://tex.z-dn.net/?f=167%2B8)
=
($)
i.e,
$159 and $175
Answer:
I wont be able to graph it for you, best way is to use desmos or wolfram alpha. This is very easy but since i cant draw it would be difficult to explain. Sorry.
Step-by-step explanation:
Answer:
45 sq. units
Step-by-step explanation:
We can draw a segment from F to S, splitting this figure into a rectangle and a triangle.
The area of a rectangle is given by the formula A=lw, where l is the length and w is the width.
The width of this figure is the distance of FW; this is 2. The length of this figure is the distance of WC; this is 9. This makes the area of the rectangle A=2(9) = 18 sq. units.
The area of a triangle is given by the formula A=1/2(b)(h), where b is the base and h is the height.
The base of the triangle is given by the distance of FS; this is 9. The height of the triangle is given from N to segment FS; this is 6. This makes the area of the triangle A=1/2(6)(9) = 27 sq. units.
This makes the total area 18+27 = 45 sq. units.