Answer:
Kindly check explanation
Step-by-step explanation:
Given the data:
Origibal data set
8, 11, 15, 22, 6
Rearranging The value :
x : 6, 8, 11, 15,22
Mean = Σx / n
n = sample size =5
Mean = (6+8+11+15+22) = 62
Mean = 62 / 5 = 12.4
Median = 1/2(n+1)th term
Median = 1/2(6)th term = 6/2 = 3rd term = 11
A.)
X = 4, 5, 15, 23, 19
MEAN = (4 +5 + 15 + 23 + 19) / 5 = 13.2
B.)
X = 2, 6, 5, 23, 34
Mean = (2+6+5+23+24) / 5
Mean = 60 / 2 = 12
C.)
X = 3, 7, 12, 21, 13
X = 3, 7, 12, 13, 21
Mean = (3+7+12+13+21) /5 = 11.2
Median = 1/2(6)th term = 6/2 = 3rd term = 12
According to Google, yes.... any details, I have no idea, but I hope I helped in any way, haha
Good luck~~
Answer:

Step-by-step explanation:

Substitute the value of 

Answer:
(Sorry this is a late response) 15.38=101.508 or 101.5
Step-by-step explanation:
You would divide 3 and 5 or 5 and 3 because 3 and 5 equals the fraction 3/5. The dividend is 3 and the 5 is the divisor. The quotient of this is 0.6. So the next step would be to add the 6 from 6 3/5 so this is the whole number so add it in front of 0.6 so it would turn into 6.6. The last step is to multiply the weight of the orange by the mg of vitamin C/oz that is 15.38 so multiply 6.6 by 15.38 this would equal 101.508 so you would just round this to a better answer and that would make it 101.5.
Tell me if I am wrong.. thanks
Answer:
Step-by-step explanation:
Having the information on how many events there are and how many people in each event there would help me personally solve this
what i can tell you is its a probability thing a tree diagram is starting with something, like flipping a coin, and creating a branch for heads and tails, 0.5 for each branch. like the attachment I have on here. there's only 2 probable results from a coin, but if I have 5 events with 50 competitors I've created a lot more probable outcomes, it also depends on the events, if one of my competitors in 6'9" and ones 5'2" and the event is a dunk contest it would be slightly unfair and the probability of the person who is 5'2" changing your tree diagram :)