2/3 x 3/4 =6/12=1/2
(Multiply numerators, multiply denominators, simplify)
Answer:
= -3i + (3/4+2i) - (9/3+3i)
= (3/4 - 9/3) + (-3 + 2 -3)i
= -9/4 -4i
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

Answer:
Hi your question is incomplete attached below is the complete question
answer : p ( X < 725 ) = 0.0116
Step-by-step explanation:
Given data:
Average life of bulbs (μ ) = 750 hours
standard deviation (б ) = 55 hours
n ( sample size ) = 25
X = 725
<u>Probability that the mean life of a random sample of 25 bulbs will be less than725 hours </u>
p ( X < 725 ) = p (( X - μ )/ б √n < 725 - 750 / 55√25 )
= P ( Z > - 2.27 )
Hence P ( X < 725 ) = 0.0116 ( using Z-table )
Answer:

Step-by-step explanation:
We are given the two functions:

And we want to find:

So, by substitution:

Simplify. Distribute:

Rearrange:

Combine like terms. Hence:
