A charge Q is to be divided into 2 parts, "q" and "(Q-q)" , such that for a given separation, the force between them is a Maximum. What is the value of q? .... why is it q=Q/2?
A random variable Q has pdf ?q(q) = {
− 2 ≤ Q ≤ 2; 0 else} .
(1) Find ?[Q], and Var(Q).
(2) Let Z = Q2. Find ?z(Z).
(3) Find P{Q > 1}
(4) Find conditional PDF of Q given B= {Q > 1} i.e., find
?Q|?(Q).
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Q has Beta[a,b] (a=6, b=5) (X|Q=q) i.i.d Geom[q] therefore E[X|Q] = [1-Q / Q] Var[X] = [1-Q / Q^2] 6.1 E[X] = ? 6.2 Var[X] = ?
How many of the disjunctions p∨¬q, ¬p∨q, q ∨r, q ∨¬ r, and ¬q ∨¬ r can be made simultaneously true by an assignment of truth values to p, q, and r? Please explain how to find that? thinking process!
A certain charge Q is divided into two parts q and Q-q, which are then separated by a certain distance. What must q be in terms of Q to maximise the electrostatic repulsion between the two charges?
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In class we considered TC(q) = q^2 + 4. AFC(q) is always decreasing, and AVC(q) is always increasing. Thus, two forces affect average cost: AC(q) = AVC(q) + AFC(q). Is it true that AVC(q) = AFC(q) at the minimum of AC(q) (the two "balance each other")? either prove the result for an arbitrary TC(q) function, or find a counterexample.
AAA +q +q +q r r r r r C +q r r +q q r positive to more negative 12. Rank the electric potential energy that eachpne of the arrangements have, from more UE(A)> U (B)> Up(C)> Up(D) c. Up(A)>U(B) = U,(C)> U,(D) e. Up(A)>UE(D)> U,(C)= U,(B) b. U(B) U(C)>Up(4) = U, (D) d. U(A) U(D)> U, (C) = U, (B) a.
Consider two charges of opposite signs and magnitudes Q, q, with Q> q that are at a distance a. What geometric shape does the zero potential surface have? At what point on the line connecting the charges, does the zero potential surface and the line intersect?
(a) use the logical equivalences p → q ≡∼p ∨ q and p ↔ q ≡ (∼p ∨ q) ∧ (∼q ∨ p) to rewrite the given statement forms without using the symbol → or ↔, and (b) use the logi- cal equivalence p ∨ q ≡∼(∼p∧ ∼q) to rewrite each statement form using only ∧ and ∼. * p∨∼q→r∨q