C as you know what you’re going through to your point of what time you’re leaving the school you
Answer:

Step-by-step explanation:
We are given the following in the question:
Let
be the proportion of the internet sales and
be the proportion of the store sale.
Hypothesis:
We have to conduct a hypothesis to check that the Internet sales are more than 10 percent higher than store sales.
Thus, we can design the null and alternative hypothesis as:

Alternate Hypothesis:
The alternate hypothesis states that the proportion of the internet sales is greater than the proportion of store sales by 10 percent.
Answer:
A
Step-by-step explanation:
<u>Slope- intercept </u><u>form</u>
y= mx +c, where m is the slope and c is the y-intercept.
The given equation is in the slope-intercept form and hence we can easily identify its slope by looking at the coefficient of x.
y= 2x +7
slope= 2
Parallel lines have the same slope, thus the slope of the parallel line is also 2.
Substitute m= 2 into the general equation:
y= 2x +c
Substitute a pair of coordinates into the equation to find the value of c:
When x= 3, y= 11,
11= 2(3) +c
11= 6 +c
c= 11 -6
c= 5
Therefore the equation of the line is y= 2x +5.
Step-by-step explanation:

#Hopeitshelp
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}