Answer:where’s the gobba goop
Step-by-step explanation:
Answer:
A. The risk of type l error increases and becomes too high.
Step-by-step explanation:
The overall level of significance increases as the number of t- tests ( used for comparing two, three or four means separately ) increases. This in turn increases the risk of type I error.
We might be tempted to apply the two sample t- test to all possible pairwise comparisons of means. This type of running t- test to several means comparison has the risk of increasing overall level of significance
5×.98 is 4.90 then add 19.29 and add 14.25. this equals $38.44 hope this helps.
Answer:
![A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-4%262%5C%5C2%266%26-6%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given
Required
Find the standard matrix
The standard matrix (A) is given by
Where
![T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=T%28x%29%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
becomes
![Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]](https://tex.z-dn.net/?f=Ax%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx_1%26x_2%26x_3%5C%5C-%26%26x_n%5Cend%7Barray%7D%5Cright%5D)
The x on both sides cancel out; and, we're left with:
![A = [T(e_1)\ T(e_2)\ T(e_3)]](https://tex.z-dn.net/?f=A%20%3D%20%5BT%28e_1%29%5C%20T%28e_2%29%5C%20T%28e_3%29%5D)
Recall that:
In matrix:
is represented as: ![\left[\begin{array}{c}a\\b\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Da%5C%5Cb%5Cend%7Barray%7D%5Cright%5D)
So:
![T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]](https://tex.z-dn.net/?f=T%28e_1%29%20%3D%20%281%2C2%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D1%5C%5C2%5Cend%7Barray%7D%5Cright%5D)
![T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_2%29%20%3D%20%28-4%2C6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-4%5C%5C6%5Cend%7Barray%7D%5Cright%5D)
![T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]](https://tex.z-dn.net/?f=T%28e_3%29%20%3D%20%282%2C-6%29%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D2%5C%5C-6%5Cend%7Barray%7D%5Cright%5D)
Substitute the above expressions in
Hence, the standard of the matrix A is:
3276800/8 = 409600 (how many bacteria in an hour)
Then just multiply that number by the amount of hours you want to solve for
9 hours = 3686400
10 Hours = 4096000
