I’m pretty sure the answer would be Y=42°
Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations

The augmented matrix is
![\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C1%261%261%26k%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
The reduction of this matrix to row-echelon form is outlined below.

![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)

![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%263%26-2%26k%5E2-4%5Cend%7Barray%7D%5Cright%5D)

![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%260%260%26k%5E2-k-2%5Cend%7Barray%7D%5Cright%5D)
The last row determines, if there are solutions or not. To be consistent, we must have k such that


Case k = −1:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-1-2%5C%5C0%260%260%26%28-1%29%5E2-%28-1%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-3%5C%5C0%260%260%26-2%5Cend%7Barray%7D%5Cright%5D)
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%262-2%5C%5C0%260%260%26%282%29%5E2-%282%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D)
This gives the infinite many solution.
M5 is the same as m1 so it's 75°
Answer:
a)= 2
b) 6.324
c) P= 0.1217
Step-by-step explanation:
a) The mean of the sampling distribution of X`1- X`2 denoted by ux`-x` = u1-u2 is equal to the difference between population means i.e = 2 ( given in the question)
b) The standard deviation of the sampling distribution of X`1- X`2 ( standard error of X`1- X`2) denoted by σ_X`1- X`2 is given by
σ_X`1- X`2 = √σ²/n1 +σ²/n2
Var ( X`1- X`2) = Var X`1 + Var X`2 = σ²/n1 +σ²/n2
so
σ_X`1- X`2 =√20 +20 = 6.324
if the populations are normal the sampling distribution X`1- X`2 , regardless of sample sizes , will be normal with mean u1-u2 and variance σ²/n1 +σ²/n2.
Where as Z is normally distributed with mean zero and unit variance.
If we take X`1- X`2= 0 and u1-u2= 2 and standard deviation of the sampling distribution = 6.324 then
Z= 0-2/ 6.342= -0.31625
P(-0.31625<z<0)= 0.1217
The probability would be 0.1217
Answer:
33.29
Step-by-step explanation:
A=πr^2
A=π2^2=12.57
12.57/2=6.29
7-4=3
3x2=6
6/2=3
4x6=24
6.29+3+24=33.29