Answer:
D. Always about the parameter only
Step-by-step explanation: A Null Hypothesis is a hypothesis used in statistics to show that there is no variation among the variables of interest. A Null Hypothesis also can be said to be giving a proposal that there is not significant difference between the observed data.
An alternative hypothesis is a hypothesis in statistics given contrary to the Null Hypothesis,a Null hypothesis can also be seen in statistics as stating that something is happening among the data.
Answer:
C. x<27 and x>-3
Step-by-step explanation:
[x-12]<15
-15<x-12<15
-15+12<×<15+12
-3<×<27
x>-3 and x<27
Following transformations on Triangle ABC will result in the Triangle A'B'C'
a) Reflection the triangle across x-axis
b) Shift towards Right by 2 units
c) Shift upwards by 6 units
In Triangle ABC, the coordinates of the vertices are:
A (1,9)
B (3, 12)
C (4, 4)
In Triangle A'B'C, the coordinates of the vertices are:
A' (3, -3)
B' (5, -6)
C' (6, 2)
First consider point A of Triangle ABC.
Coordinate of A are (1, 9). If we reflect it across x-axis the coordinate of new point will be (1, -9). Moving it 2 units to right will result in the point (3, -9). Moving it 6 units up will result in the point (3,-3) which are the coordinates of point A'.
Coordinates of B are (3,12). Reflecting it across x-axis, we get the new point (3, -12). Moving 2 units towards right, the point is translated to (5, -12). Moving 6 units up we get the point (5, -6), which are the coordinate of B'.
The same way C is translated to C'.
Thus the set of transformations applied on ABC to get A'B'C' are:
a) Reflection the triangle across x-axis
b) Shift towards Right by 2 units
c) Shift upwards by 6 units
Answer:
a) -4
b) 1
c) 1
Step-by-step explanation:
a) The matrix A is given by:
![A=\left[\begin{array}{ccc}-3&0&1\\2&-4&2\\-3&-2&1\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3%260%261%5C%5C2%26-4%262%5C%5C-3%26-2%261%5Cend%7Barray%7D%5Cright%5D)
to find the eigenvalues of the matrix you use the following:
where lambda are the eigenvalues and I is the identity matrix. By replacing you obtain:
![A-\lambda I=\left[\begin{array}{ccc}-3-\lambda&0&1\\2&-4-\lambda&2\\-3&-2&1-\lambda\end{array}\right]](https://tex.z-dn.net/?f=A-%5Clambda%20I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-3-%5Clambda%260%261%5C%5C2%26-4-%5Clambda%262%5C%5C-3%26-2%261-%5Clambda%5Cend%7Barray%7D%5Cright%5D)
and by taking the determinant:
![[(-3-\lambda)(-4-\lambda)(1-\lambda)+(0)(2)(-3)+(2)(-2)(1)]-[(1)(-4-\lambda)(-3)+(0)(2)(1-\lambda)+(2)(-2)(-3-\lambda)]=0\\\\-\lambda^3-6\lambda^2-12\lambda-16=0](https://tex.z-dn.net/?f=%5B%28-3-%5Clambda%29%28-4-%5Clambda%29%281-%5Clambda%29%2B%280%29%282%29%28-3%29%2B%282%29%28-2%29%281%29%5D-%5B%281%29%28-4-%5Clambda%29%28-3%29%2B%280%29%282%29%281-%5Clambda%29%2B%282%29%28-2%29%28-3-%5Clambda%29%5D%3D0%5C%5C%5C%5C-%5Clambda%5E3-6%5Clambda%5E2-12%5Clambda-16%3D0)
and the roots of this polynomial is:
hence, the real eigenvalue of the matrix A is -4.
b) The multiplicity of the eigenvalue is 1.
c) The dimension of the eigenspace is 1 (because the multiplicity determines the dimension of the eigenspace)
