You have a first order nonlinear system, which can be written as X' = f(X), where X is a vector.
In such systems, the phase portrait around a hyperbolic equilibrium point (an equilibrium point X* such that the real part of every eigenvalue of Df(X*) is nonzero) is the same as the phase portrait...
Stating that a_n is bounded means that you can find a value of M which is greater than all the terms in the sequence. You must prove there is no such M. Reduction to the absurd works well in this situation.
You suppose there exists M >= a_n for all n. But there is a problem with this...
Suppose an is bounded. Then there exists M > 0 such that an <= M for all n. However this is absurd since you only have to pick n greater than M^2 to have an > M.
To illustrate: pick M = 120. 120^2 = 14400. a_14401 > 120.
Makes sense?
The negative number appeared since you are considering the distribution to be discrete when it's not. P(T<=3) = 1 - P(T<3) = 1-P(T<=3) since the distribution is continuous.
In 2. I think you are not calculating what you think you are... First of all P(T=1) = P(X<=1)-P(X<1) = 0.
You really...
It's actually the other way around, i.e., x <= 3.
You can always test your answer. For example if x = 1, then 3x <= 9. But x is not greater than 3. Thus your answer is wrong.
You only change the inequality when the coefficient is negative ( -x > 0 <=> x < 0).
The equations are correct. Notice the first and second equations are equivalent so you'll only really use the second and third equation and will get an undetermined solution.
Write x and y as functions of z for, example. You will hopefully arrive at a solution where (3/2)*a + (1/2)*b - 1*c = 0.
Well, the expression really is x - x0 = v0*t + 0.5*g*t^2.
Now the signs will depend on the choice of the x-axis direction. If it points downwards you will assume x = d and x0 = 0 and g = 9.8 (because the gravity pull is in the positive x-axis direction); if you choose an x-axis pointing...