Answer:
(a)


(b)
B. The sample is too small to make judgments about skewness or symmetry.
Step-by-step explanation:
Given:


Solving (a):
First, calculate the difference between the recorded TBBMC for both operators:

The last row which represents the difference between 1 and 2 is calculated using absolute values. So, no negative entry is recorded.
The mean is then calculated as:




Next, calculate the standard deviation (s).
This is calculated using:

So, we have



Solving (b):
Of the given options (A - E), option B is correct because the sample is actually too small
The difference between Pre-Image and Image is given as follows:
- Image is the shape AFTER a transformation is the picture of the transformation.
- A transformation's preimage is the shape BEFORE the change.
<h3>How do you define relationships between Image and Preimage?</h3>
Usually, the difference between image and pre-image is the way or method of transformation.
<h3>What is transformation in math?</h3>
A transformation is a broad phrase covering four distinct methods of changing the shape and/or location of a point, line, or geometric figure.
The Pre-Image is the original shape of the item, and the Image during the transformation is the final shape and location of the object.
The types of transformation in math are;
- translation
- rotation
- reflection, and
- dilation.
Learn more about pre-image:
brainly.com/question/8405245
#SPJ1
Answer:
Roman
Step-by-step explanation:
An easy way to solve this is looking at whether the line has a positive or negative slope. We read graphs from left to right so <u>if the line is increasing from left to right, it has a positive slope while a line decreasing from left to right has a negative slope</u>. Since the line shown is decreasing from left to right, it will have a negative slope and Roman is the only one out of the two that shows a negative slope. To check to see if he's correct, we can check the y-intercept. The y-intercept of the line is 8 and Roman also has 8 written in the expression. Hence, meaning Roman is correct.
Best of Luck!
4 for part A
17 for part B