Answer:
Maybe because the situation changed
Step-by-step explanation:
Answer:
≈ -5.1857
≈ -5.4857
≈ 3.7262
Step-by-step explanation:
Rewrite the equation system as:



Now, write the system in its augmented matrix form:
![\left[\begin{array}{cccc}6&8&0&-75\\-3&6&6&5\\2&-9&0&39\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D6%268%260%26-75%5C%5C-3%266%266%265%5C%5C2%26-9%260%2639%5Cend%7Barray%7D%5Cright%5D)
applying row reduction process to its associated augmented matrix:
Swap R1 and R3, and then Swap R1 and R2:
![\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\6&8&0&-75\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C2%26-9%260%2639%5C%5C6%268%260%26-75%5Cend%7Barray%7D%5Cright%5D)
R3+2R1
![\left[\begin{array}{cccc}-3&6&6&5\\2&-9&0&39\\0&20&12&-65\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C2%26-9%260%2639%5C%5C0%2620%2612%26-65%5Cend%7Barray%7D%5Cright%5D)
3R2+2R1
![\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&20&12&-65\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C0%26-15%2612%26127%5C%5C0%2620%2612%26-65%5Cend%7Barray%7D%5Cright%5D)
15R3+20R2
![\left[\begin{array}{cccc}-3&6&6&5\\0&-15&12&127\\0&0&420&1565\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-3%266%266%265%5C%5C0%26-15%2612%26127%5C%5C0%260%26420%261565%5Cend%7Barray%7D%5Cright%5D)
Now we have a simplified system:


From (3):
(4)
Replacing (4) in (2)
(5)
Finally replacing (5) and (4) in (1)

Answer:
7 6/18
Step-by-step explanation:
Since A (3,4) A’(9,12)
You can see 3*3=9
Therefor it was enlarged by 3times
Scale factor of 3
Answer:
The requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.
Step-by-step explanation:
A normal-distribution is an accurate symmetric-distribution of experimental data-values.
If we create a histogram on data-values that are normally distributed, the figure of columns form a symmetrical bell shape.
If X
N (µ, σ²), then
, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z
N (0, 1).
The distribution of these z-variates is known as the standard normal distribution.
Thus, the requirements that are necessary for a normal probability distribution to be a standard normal probability distribution are <em>µ</em> = 0 and <em>σ</em> = 1.