consider the following changes in the distribution...
consider the following changes in the distribution of nine particles into three interconnected boxes.which of the following has the most negative δs?
negative δs → system gets orderly[readmore]notice for b, the 2nd panel has particles which are compact.therefore, the most negative δs is b.
given : five defective bulbs are accidently mixed with 20 good ones.to find : probability distribution from this lotformula used :five defective bulbs are accidently mixed with 20 good ones.total number of bulbs = 25x denote the number of defective bulbs drawnx can draw the value 0 , 1 , 2 , 3 , 4.since the number of bulbs drawn is 4, n = 4p(0) = p(getting a no defective bulb) = = = p(1) = p(getting 1 defective bulb and 3 good ones) = = p(1) = = p(2) = p(getting 2 defective bulbs and 2 good one) = p(2) = = = p(3) = p(getting 3 defective bulbs and 1 good one) = p(3) = = p(4) = p(getting all defective bulbs) = = = p(4) = the probability distribution table is as follows, a - ii, b - iv, c - i, d - iii there are 3 types of hybridisation for carbon in organic compounds, sp, sp2 and sp3 depending on the number of sigma and pi bonds between the carbon atoms. more the ‘s’ character more is the electronegativity of the carbon as ‘s’ orbitals are closer to the nucleus, hence electrons held more closely. dear student,

atomic radii in general decreases left to right in a period and
increases down the group
order of atomic radii is p >s > c > n
decrease in atomic radii on moving from c to n and p to s in a
period is because as we move left to right in a period as the
number of electron increases the number of protons in the nucleus
also increases due to which the valance shell electron experience
greater effective nuclear charge ; therefore more strongly pulled
by the nucleus , thus decreasing their sizes shielding effect
decreases in the order s > p > d orbital electrons s orbital
electrons have greater shielding effect on the nucleus on moving
left to right in a period shielding effect is operable but is same
across a period, therefore here we consider only the increase in
the effective nuclear charge experienced by the valance shell
electrons

regards information is below. you should have three choices.

 thanks for the update...i'll check it out. applying and factorization, we get, also changing by reversing n and dnote: - try to factorize that term which help in reducing expression. as the distance between the source and the metal is increased, the intensity of light decreases becausealso, the photocurrent is directly proportional to the intensity of light and also to the number of electrons emitted.hence, the curve d is the appropriate curve representing above equation. more importantly is the size of the foggle toggle box which is connected to the foofoo valve! 🤣🤣🤣 a, b and c