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.
27 inches ......... 3/8
x inches .............8/8
<h3>x = (27×8/8)/(3/8) = 27×8/3 = 216/3 = 72 inches (father)</h3>
Answer:
6.6 inches
Step-by-step explanation:
4.1 times 3 = 12.3
2.2 times 3 = 6.6
I think this is right... sorry if not
Answer:
Step-by-step explanation:
Using either the critical value rule or the p-value rule, a conclusion can be drawn at a level of significance (alpha)
The null hypothesis: u = hypothesized mean
Alternative hypothesis: u > u0 or u < u0 for a one tailed test
Alternative hypothesis for a two tailed test: u =/ u0
To draw a conclusion by failing to reject the null hypothesis as stated then: using critical value
Observed z score > critical z score for both the one and two tailed test.
Or using p value:
P-value > alpha for a one tailed test
P-value > alpha/2 for a two tailed test
Thus, if a one-sided null hypothesis for a single mean cannot be rejected at a given significance level, then the corresponding two-sided null hypothesis will also not be rejected at the same significance level.
