Answer:
<em>c. ABBA counterbalancing
</em>
Explanation:
The student should not use the method because it is a progressive error management technique for each subject by introducing all <em>treatment circumstances twice, first in one sequence, then in the other (AB, BA) by subject counterbalancing.</em>
If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.
Answer:
The dependent variable is academic performance
The independent variable is the presence/absence of tutorial support
The control group are students who did not get the tutorial support.
The experimental group were students that got the tutorial support
Explanation:
In every experiment, there is a dependent and independent variable as well as an experimental and a control group.
The experimental group receive the treatment while the control group do not receive the treatment. The independent variable is manipulated and its impact on the dependent variable is evaluated.
The control group are students who did not receive the tutorial support while the experimental group are students that received the tutorial support.
The dependent variable in this case is academic performance. Its outcome depends on the presence or absence of tutorial support (independent variable).
John carry the heaviest load.
<h3>How to find out who is carrying the heavy load?</h3>
Write down given data from questions:
Board=510cm X 510mm.
Cylinder head with dimensions=43cm X 250mm.
Cylinder lies across the board 210cm from john.
Find out: Who is carry the heaviest load?
Calculation:
We assume that mass of cylinder head = x kg
Then weight=x x 9*81
W=9.81x Newton.
Weight per unit length= Weight/Total leanth
Weight per unit length= 9.81x/43
(w/l)=0.23x N/cm
From equation contition: 
(210+21.5)




Therefore 
To learn more about mass per unit length, refer to:
brainly.com/question/24180692
#SPJ9
Its simple use formuila ,
PV=nRT
n,R is constant as the both have same moles.
so,
(p1v1)/T1 = (p2v2)/T2
so, 128.53338kpa
To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective

Replacing with our values


Therefore the focal length of th eobjective lenses is 27.75cm