Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

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I overlooked "clockwise". My answer is for counterclockwise 270 degrees rotation.

Nice.

Not from Bengaluru but from Kumta of Karwar district.

mathicINDIA wrote:

If 1 & -1 are the zeroes of the polynomial p(x)=Lx^4+Mx^3+Nx^2+Rx+P=0, prove that L+M+P=M+R=0. [JEE mains 2014, AIEEE 2006, IIT 1998]

I think it should be

If 1 & -1 are the zeroes of the polynomial

The problem is just a hoax, simply to test presence of mind.

I understand Samuel's point of view.The Chinese among themselves are indistinguishable. Similarly others.It is just like saying 5Cs,5Rs and 5Ms are to be arranged in 15 slots with the given conditions. let us see in first 5 slots.

(1) 0 russian and 5 Mexicans can be seated in only 1 way. Now seats 6 to 10 can accommodate 5 Chinese in 1 way and last 5 seats also 1 way. It is 1*1*1=1 way.

(2) 1 russian and 4 Mexicans can be seated in 5 different ways. so 5*5*5=125 ways.

(3)2 russian and 3 Mexicans can be seated in

What is cooking? No response from any body including Samuel!

The sum of coefficients , obviously is

4 times hotter is meaningless.

for (3) simply ascertain the radii of the circles first and then put the equation.

My answer for EF is

Due to symmetry.

bob bundy wrote:

The correct formula is

If we assume that P(A) = P(B) then this reduces to thickhead's formula and the result 3/5 follows.

But the question does not say that a class is chosen first with equal probability.

If a West High precalculus student chosen at random

The classes are unequal so P(A) is not equal to P(B).

The correct calculation is

Bob

Just for fun let us change number of boys in each class.

Abel' class boys=20 girls=12 total 32

Bonitz's class boys=25 girls=10 total 35

P(G/Bonitz)=10/25=2/5=0.4

P(Abel/G)=P(G/Abel)/{P(G/abel)+P(G/Bonitz)}=0.6/(0.6+0.4)=0.6

This is established rule.

(1) My answer is 4/3.