We can divide each digit by 7. So 56/7=8 and 21/7=3
so.... 8:3
I assume the first equation is supposed to be
and not
As an augmented matrix, this system is given by
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C4%26-2%264%2612%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/2:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C2%26-1%262%266%5Cend%7Barray%7D%5Cright%5D)
Add -1(row 2) to row 3:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%265%265%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 3 by 1/5:
![\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%262%2613%5C%5C2%26-1%26-3%261%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 3) to row 1, and add 3(row 3) to row 2:
![\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D5%26-3%260%2611%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -3(row 2) to row 1:
![\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D-1%260%260%26-1%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multiply through row 1 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C2%26-1%260%264%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Add -2(row 1) to row 2:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%26-1%260%262%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
Multipy through row 2 by -1:
![\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%260%260%261%5C%5C0%261%260%26-2%5C%5C0%260%261%261%5Cend%7Barray%7D%5Cright%5D)
The solution to the system is then
SOLUTION:
We are to find the rate of inflation in the given question;
The formula to be used is;
Inflation rate =
CONCLUSION
The inflation rate of CPI that increased from 208 to 221 to the nearest tenth of a percentage is 6.3%.
The confidence level C and the significance level alpha are linked through the equation
alpha = 1-C
So for instance, if the confidence level is C = 95% = 0.95 then alpha is
alpha = 1-C
alpha = 1-0.95
alpha = 0.05
meaning we have a 5% significance level. The larger C gets, the smaller alpha gets and vice versa. It turns out that 0 < C < 1 and also 0 < alpha < 1.
The closer C gets to 1, the alpha value gets closer to 0. The smaller alpha gets, the harder it is to reject the null. Why is that? If we have a fixed p value, say p = 0.02 then we reject the null if alpha > pvalue. But we fail to reject the null when alpha < pvalue. For very small alpha values, we're going to fail to reject H0 no matter how small the pvalue is. The pvalue would have to be really small for H0 to be rejected.
In short, I'm saying that if the confidence level is high, then the chance of rejecting the null hypothesis is low (or rare)
This is why the answer is choice A
