8736
x 6.
6x6....6x3.....6x7.....6x8
it would equal=52,416. hope this helped...
Answer: 4442000000
Step-by-step explanation:
Answer:
C
Step-by-step explanation:
To simplify this equation,
-x+10<2
-10 from both sides
-x<-8
Take out the negative
x<8
This means x is less than 8
For ≤ and ≥ , use a closed dot to indicate the number itself is part of the solution.
For < and >, use an open circle to indicate the number itself is not part of the solution.
In this case you use a open circle and becasue x is less than 8, the arrow pointing left is correct.
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}
The only way for two integers to have an odd product is each integer is odd.
For example, 1*3=3 (odd), 1 and 3 are both odd.
Or,
5*11=55, all of 5,11,55 are odd.
The sum of two odd integers is always even, so the condition of even sum is automatically satisfied when the product is odd.
Out of the four integers, there are only two odd numbers, so choose the pair to be these two odd numbers and you'd get the right answer.
