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Answer:
(d) see below
Step-by-step explanation:
Since each grid square is 1/4 unit. the coordinates in terms of grid squares are ...
A(0.50, -1.25) ⇒ (0.50/.25, -1.25/.25) = (2, -5) . . . grid squares
Point A is located 2 squares right and 5 squares down on the grid. This matches choice D.
Answer:
(-1, 4)
Step-by-step explanation:
-5+3/2 = -1
6+ 2/2 = 4
(-1, 4)
Yes, they certainly do exist.
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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}
Answer:
Step-by-step explanation:
We can write the given expression as :

As given, f = ∇f.
∇f =
+
+
We can write the partial derivative with respect to x, y and z.
___(Equation 1)
______(Equation 2)
______(Equation 3)
Take equation 2 and integrate with respect to y,

----------Equation 4
Derivate both sides w.r.t x , we get :

or

integrate
a(x,z)=b(z)
put in equation 4 ,
we get :

take derivative wrt z

we can take here:

integrate:

b(z) = C
The function can be written as :
from equation 4 :
Where C is a constant.