and
so

and you rationalize denominator by multiplying numerator and denominatr by
so that gives--

Your answer is 
Answer:
Step-by-step explanation:
Given data:
SS={0,1,2,3,4}
Let probability of moving to the right be = P
Then probability of moving to the left is =1-P
The transition probability matrix is:
![\left[\begin{array}{ccccc}1&P&0&0&0\\1-P&1&P&0&0\\0&1-P&1&P&0\\0&0&1-P&1&P\\0&0&0&1-P&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccccc%7D1%26P%260%260%260%5C%5C1-P%261%26P%260%260%5C%5C0%261-P%261%26P%260%5C%5C0%260%261-P%261%26P%5C%5C0%260%260%261-P%261%5Cend%7Barray%7D%5Cright%5D)
Calculating the limiting probabilities:
π0=π0+Pπ1 eq(1)
π1=(1-P)π0+π1+Pπ2 eq(2)
π2=(1-P)π1+π2+Pπ3 eq(3)
π3=(1-P)π2+π3+Pπ4 eq(4)
π4=(1-P)π3+π4 eq(5)
π0+π1+π2+π3+π4=1
π0-π0-Pπ1=0
→π1 = 0
substituting value of π1 in eq(2)
(1-P)π0+Pπ2=0
from
π2=(1-P)π1+π2+Pπ3
we get
(1-P)π1+Pπ3 = 0
from
π3=(1-P)π2+π3+Pπ4
we get
(1-P)π2+Pπ4 =0
from π4=(1-P)π3+π4
→π3=0
substituting values of π1 and π3 in eq(3)
→π2=0
Now
π0+π1+π2+π3+π4=0
π0+π4=1
π0=0.5
π4=0.5
So limiting probabilities are {0.5,0,0,0,0.5}
S+a=150
3s+8a=1000
a=-s+150
3s-8s+1200=1000
-5s=-200
s=40
a=110
40 student and 110 adult tickets were sold
Given:
m/n-3(m+n)
m=-10 and n=-2
Solving:
-10/-2-3(-10-2) Since m=-10 and n=-2, we substitute them in for the values since they have an equal relationship.
Now, follow Orders of Operations: PEMDAS
-10/-2-3(-12) Simplify parentheses (P)
There are no Exponents (E)
Multiplication (M) and Division (D) are =, so we do whatever operation comes from the left to right.
-10/-2=5 Division comes first on the left.
therefore:
5-3(-12)
-3(12)=-36, Multiplication is last on the right.
therefore:
5-36
5-36=-31 After M/D, we do Addition and Subtraction, which are also =, so we do them left to right.
Therefore -10/-2-3(-10-2)=-31