Consider the following observations on shear strength of a joint bonded in a particular manner: 30.0 4.4 33.1 66.7 81.5 22.2 40.4 16.4 73.7 36.6 109.9 a. Determine the value of the sample mean.
b. Determine the value of the sample median.
c. Why is it so different from the mean?
d. Determine the Lower Quartile (LQ).
e. Determine the Upper Quartile (UQ).
f. Calculate a trimmed mean by deleting the smallest and largest observations. What is the corresponding trimming percentage?

i think the answer is .36

step-by-step explanation:

what i did was only 2.88 divided by 8

the direct variation says that:

then the equation is of the form is given by:

            where k is the constant variation.

from the given table:

consider any value of x and m(x)=y

x = 2 and m(x)=y = 3

substitute these value in [1]; we have;

divide both sides by 2 we get;

⇒

check:

put x = -1 and m(x) = -1.5

substitute in the equation:

-1.5 = -1.5                   true

therefore, the function m(x) represents a direct variation

graph of this function as shown below:

answer:  

step-by-step explanation:

odd numbers are only divisible by 1 and itself.

even numbers are divisible by more than 1 and itself.

for example 6 is even because it is divisible by more than 1 and its self. 1 x 6= 6 ; 2 x 3= 6

and 7 is odd because only 7 x 1 equals seveb

inverse property is the answer

step-by-step explanation: