( PLASE HELP ME) Solve the following equation.
2 answers:
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Part A:
Given
defined by 
but
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
defined by
Note that in
, -1 = 1 and f(0) = 0 and f(1) = -1 = 1, so we can also use the formular 
and
Therefore, the function is a homomorphism.
Part C:
Given
, defined by 
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
, defined by 
but
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
, 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 
in 
.
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.
Given
but
Since, f(xy) ≠ f(x)f(y)
Therefore, the function is not a homomorphism.
Part B:
Given
Note that in
and
Therefore, the function is a homomorphism.
Part C:
Given
Since, f(x+y) ≠ f(x) + f(y), therefore, the function is not a homomorphism.
Part D:
Given
but
Since, h(ab) ≠ h(a)h(b), therefore, the funtion is not a homomorphism.
Part E:
Given
Then, for any
and
Therefore, the function is a homomorphism.
Answer:
The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 8 - 1 = 7
95% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of . So we have T = 2.3246
The margin of error is:
In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 0.32 - 0.0502 = 0.2698 mg/kg
The upper end of the interval is the sample mean added to M. So it is 0.32 + 0.0502 = 0.3702 mg/kg
The 95% confidence interval for the concentration in whitefish found in Yellowknife Bay is (0.2698 mg/kg, 0.3702 mg/kg).