Answer:
The answer is 5/24.
Step-by-step explanation:
2/3 multiplied by 1/4 is 2/12 which simplifies to 1/6. CCF (copy, change, flip) is applied to 4/5 and modifies the equation to become 1/6*5/4 for a final answer of 5/24.
I believe the answer is one.
Answer:
- The dimensions of a matrix are the number of rows × the number of columns of the matrix.
- For the example below, the the matrix is 3 × 4.
Explanation:
<em>The dimensions of a matrix</em> is the number of rows × the number of columns of the matrix.
Your matrix is garbled. Thus, to help you I will work with an hypothetical matrix.
Assume the matrix:
![\left[\begin{array}{cccc}1&0&0&0\\2&4&0&3\\0&0&0&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%260%5C%5C2%264%260%263%5C%5C0%260%260%269%5Cend%7Barray%7D%5Cright%5D)
That matrix has four colums and 3 rows.
For instance, the first colum is:
Thus, it has 4 columns.
And the second row is:
Thus, it has 3 rows.
Hence, the matrix is 3 × 4.
The first number is the number of rows and the second number is the number of columnns.
Answer: 90% confidence interval is; ( - 0.0516, 0.3752 )
Step-by-step explanation:
Given the data in the question;
n1 = 72, n2 = 17
P1 = 54 / 72 = 0.75
P2 = 10 / 17 = 0.5882
so
P_good = 0.75
P_bad = 0.5882
standard ERRROR will be;
SE = √[(0.75×(1-0.75)/72) + (0.5882×(1-0.5882)/17)]
SE = √( 0.002604 + 0.01424)
SE = 0.12978
given confidence interval = 90%
significance level a = (1 - 90/100) = 0.1, |Z( 0.1/2=0.05)| = 1.645 { from standard normal table}
so
93% CI is;
(0.75 - 0.5882) - 1.645×0.12978 <P_good - P_bad< (0.75 - 0.5882) + 1.645×0.12978
⇒0.1618 - 0.2134 <P_good - P_bad< 0.1618 + 0.2134
⇒ - 0.0516 <P_good - P_bad< 0.3752
Therefore 90% confidence interval is; ( - 0.0516, 0.3752 )