Answer: Choice B) I and II only
The mean and median are considered measures of center as they represent the average of a data set. The average basically being a point that collectively speaks for all of the data. For example, if you have a group of basketball players whose heights range from 5'11" to 6'9", then there is no single height to report; however, we can compute the average to get a basic idea of the single height. The interquartile range (IQR) is a measure of variability or spread of the data. The higher the IQR, the more spread out the data is. Recall that 50% of the data is represented by the IQR and that IQR = Q3 - Q1 where Q1 and Q3 are the first and third quartiles respectively. So because the IQR is a measure of spread, it is not considered a center point.
a. Let 
be a random variable representing the weight of a ball bearing selected at random. We're told that 
, so
where 
. This probability is approximately
b. Let 
be a random variable representing the weight of the 
-th ball that is selected, and let 
be the mean of these 4 weights,
The sum of normally distributed random variables is a random variable that also follows a normal distribution,
so that
Then
c. Same as (b).
RT = 8 , RS = 3 AND ST = 5.4
RV = 4 , RU = 1.5 AND UV = 2.7
∴ RT/RV = 8/4 = 2
AND RS/RU = 3/1.5 = 2
AND ST/UV = 5.4/2.7 = 2
∴Δ RTS IS SIMILAR TO ΔRVU
SO, the best statement is:
Each pair of sides corresponds with a common ratio of 2.
Let x + 1.5 be length of segment above left hand
Let 2x be length of segment below left hand
15 = x + 1.5 + 2x
15 = 3x + 1.5
15 - 1.5 = 3x
13.5 = 3x
13.5/3= 3x/3
4.5 = x
x = 4.5
This is the length above the left hand.
2x must be then 4.5*2 = 9 foot
This is the length below the left hand.
Checking
4.5 + 9 + 1.5 = 15 foot
Adding length below left hand & length of left hand to right hand:
9 + 1.5 = 10.5
The right hand is hence 10.5 foot far up the pole.
2/4= 10/20
1 4/10=14/10 14/10=28/20
10/20×28/20=7/10
I hope this helped!!!:)
