D1,..,d9 = 0,0,2,2,2,3,4,6,8 //there are 9 values, in ascending order
Q2 (median) = d5 = 2 //value in the middle
Q1 = (d2+d3) / 2 = (0+2)/2 = 1
(Q1 is middle value of d1,d2,d3,d4, but there is no middle element among four elements, that is why arithmetic mean is taken)
Q3 = (d7+d8) / 2 = (4+6)/2 = 5
interquantlie range = IQR = Q3 - Q1 = 5 -1 = 4
answer: 4
Answer:
0
Step-by-step explanation:
given that we roll a fair die repeatedly until we see the number four appear and then we stop.
the number 4 can appear either in I throw, or II throw or .... indefinitely
So X = the no of throws can be from 1 to infinity
This is a discrete distribution countable.
Sample space= {1,2,.....}
b) Prob ( 4 never appears) = Prob (any other number appears in all throws)
= 
where n is the number of throws
As n tends to infinity, this becomes 0 because 5/6 is less than 1.
Hence this probability is approximately 0
Or definitely 4 will appear atleast once.
They will see each other every 30 days but won't every 15 days
Answer:
ounces of marigold seeds.
Step-by-step explanation:
We have been given that Jane uses 2/5 of an ounce of lupine seeds for every ounce of Marigold seeds. We are asked to find the ounces of Marigold seeds, which Jane will use, if she used one ounce of lupine seeds.
We will use proportions to solve our given problem.

Let us simplify our proportion as shown below:


Since Jane used 1 pounce of lupine seeds, so we will substitute this value in our proportion as:


Therefore, Jane will use
ounces of marigold seeds, if she used one ounce of lupine seeds.