Recent content by balakrishnan_v

It can be proved by induction. Consider any vertex i. Let it be connected to k other vertices in the bulbs section. If you switch vertex i, we switch on k bulbs. Consider the subgraph H formed by the other n-k switches and n-k bulbs. Since this subgraph also satisfies the condition that...

Since the distribution is not given,a fair assumption is to take exactly 6 bookings that day Let us say he books 1,2,3 as his time slot(of 1,2,3,4,5,6,7,8) We first compute the probability that all the 3 time-slots are filled with all the dentists Pr(of the 6 appointments,3 appointments fall...

The answer is P(\text{getting n})=\frac{r_1^n+r_2^n+r_3^n+r_4^n+r_5^n}{7} +\frac{2}{7}where r_1,r_2,r_3,r_4,r_5 are the roots of the equation 6x^5+5x^4+4x^3+3x^2+2x+1=0 The roots are: r_1 = 0.29419455636014125+0.66836709744330092i r_2=0.29419455636014125000+-0.66836709744330092000i...

If f converges to a point say f_{\infty} then the answer would be \frac{f_{\infty}}{(b-y)^2}

Just do some integration . You will get \frac{\pi R}{4} where R is the radius of the sphere

The net acceleration is the coriolis acceleration and the centrepetal acceleration. So So the radial part is -rw^2 angular part is 2wv0 So the net acceleration is sqrt(r^2.w^4+4w^2.v0^2)=mu.g which gives r=v0.t=sqrt(mu^2g^2-4w^2v0^2)/w^2 So t=sqrt(mu^2.g^2-4w^2v0^2)/(v0.w^2)It is clear that...

Any bonafide moment generating function always maps to a unique distribution. The random variable for the moment generating functi`on that you have given takes values at 1,2,3...infinity with probability 1/2,1/4,1/8...ie p(k)=1/2^k for k in natural numbers and zero otherwise.

I did not know that.I just looked up the net and somewhere it was written cards have 4 colors. In any case for a 2 color suite,I guess the answer would be 3/4

you have 2 cards left. Since you already know the colors of all other cards,you know the colors that the last 2 cards can take each of the card can take one of the 4 possible colors and you know the color pair C1 C1 C1 C2 . . .16 pairs for C1,C1 C2C2 C3C3 C4C4 you will judge it...

@Leach I think you misread the question.The problem is not to find the time of catch when the dog is always heading towards the frisbee. The dog is chasing the frisbee such that the angle formed by the line joining frisbee and the dog with the east west line is a constant as the equation...

It will follow the equation y=\frac{16+\sqrt{91}}{9} x ie y=x*{16+sqrt(9)}/9 which is a straight line Since the equation is a straight line,it is the optimal way of catching the Frisbee.So the time is minimum.

You are almost done.Note that n^2-1 is (n-1)(n+1) So product(n^2-1) for n=2 to n is 1.2.3^2.4^2...=product(n^2)/2 Product(n^2+1) for n=1 to n =2.Product(1+n^2) for n=2 to n So from the above expression pi*cosech(pi)=2*product(n^2-1)/2*product(n^2+1)=product(n^2-1/n^2+1) for n=2 to infty

Let us find the fraction of the number line(from 0 to 1) that will be filled: Now we will find E(1) which is nothing but the expected fraction of the line(x=0 to x=1) given that each filling width is a small h (h=1/N) E(1)=h+E(1-h) E(1-h)=(1-h){h+E(1-2h)} and so on hence we get...

~^{n-r+1} C_{r}

Now turn the plane such that the 2 electrons are symmetrically moving. ie e1 starts from -infty with speed v_0 at angle a/2 and e2 start at +infty at angle a/2 The vertical component does not change as there is no force in that direction Energy(total) at infinity=mv^2/2+mv^2/2=mv^2 AlsoAt...