Answer / Step-by-step explanation:
To properly understand the principle behind the solving of this question, it would be necessary to define and go through some basic terms and their definition.
Therefore to start with, a completely randomized single factor is a type of experiment in which One factor of two or more levels has been manipulated. e.g, the experiment may be investigating the effect of different levels of cost, expenditure or different advertisements
Also, it should be noted that in the area of groping of the experiment, At this point, Each group of different level receives one of the a levels of the independent variable with participants being treated identically in every other respect. The two-group experiment considered previously is a special case of this type of design.
Consequentially, the answer to the narrative of the question would then be:
N= 3 factors levels x 5 replicates = 15
Degrees of freedom for the factor: a – 1 = 3 – 1 = 2
Degrees of freedom Total = 15 – 1= 14
Degrees of freedom error = Total – factor = 14 – 2 = 12
Bounds of P-value for F = 2.91 with 2 and 12 degrees of freedom are
= 0.01 < P < 0.05
Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
N+2 divide 3 so it should be correct hope it helped
