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## Roman C.'s Resources

This is just P(X ≤ 25) where X ~ Bin(n=50,p=0.65)   Using R by typing pbinom(25,size=50,prob=0.65), we have:   P(X ≤ 25) = 0.02066839   We can also give a normal approximation   Let Y ∼ N(μ,σ2) where   μ = np =...

We can multiply both sides by 2. This will give us the answer. Note that since 2>0, the inequality sign remains the same.   x/2 ≤ 3   2(x/2) ≤ 2(3)   x ≤ 6

If we draw one card, and all cards are equally likely, then the probability of choosing a card in any subset is   |Subset| / |Deck|.   For an ace, this is 4/52 = 1/13.   For an ace of hearts, this is 1/52.

We must have at least 2 calls to meet the stopping condition. But once 4 calls are made, the stopping condition is automatically met. Therefore, the sample space is {2,3,4}.

Multiply by √x to get x2 - 14x + 49=0.   This new equation then factors as (x-7)2 = 0   Therefore the solution is x=7   You can easily verify that this solution is not extraneous.

Recall that the formula for the angle θ between two vectors in U,V ∈ Rn is cos θ = (U·V)/(||U||*||V||)   You probably made a typo when writing what V is.   For U = 〈√5,-8〉 and V = 〈5,1〉, we have   U·V = -8+5√5 ||U|| = √(5+64) = √69 |...

I think you missed the minus sign on the right hand side. It should be (14x+3) y' - y'' = -14y   Also, the H should be Euler's number "e".   Solution: Just plug it in.   y = e7x^2+3x +2 ⇒ y' = (14x+3) e7x^2+3x +2   ⇒...

When I learned Markov Chains, the rows were the ones representing the current state and the columns represented the next one, which is the traditional representation. Your matrix P is the transpose of that format, but it doesn't affect analysis much. Here it goes.   1. You need P2 to...

Part 1: (0,3) and (5,-1):   m = (y2 - y1) / (x2 - x1) = (-1 - 3) / (5 - 0) = -4/5   b = 3 because of the (0,3) point being the site of the y-intercept.   Thus y = (-4/5)x + 3.     Part 2: (3,-5) and parallel to y= -x - 5.   m...

Note that a full binary tree of height h has exactly 2h leaves.   Proof:   Base: or h=0, the full tree has just the root node, which is a leaf. This is 20 leaves.   Induction: Assume for some k≥0, that the height h full tree has 2h leaves when h=k. Then...

Let n>1 and 0<k<n.   ⇒ Clearly n divides into n!   If n is prime, then by Euclid's Lemma, n can't divide k! or (n-k)!   Thus since C(n,k) = n!/[k!(n-k)!] is an integer and so n must divide into it.   ⇐   We...

∫(4+eyz) dx = 4x + xeyz + f(y,z)   ∫(xzeyz + 2yz) dy = xeyz + y2z + g(x,z)   ∫(6z2 + xyeyz + y2) dz = 2z3 + xeyz + y2z + h(x,y)   This is a conservative field...

Recall that the work done against uniform gravity when lifting is W = ∫ F dh = ∫ mg dh where the potential energy at ground level is taken to be 0.   Note how the mass of the bucket at height h is   m(h) = (75 + 500 - 2h)  kg = (575...

d/dx x2/x   = d/dx exp(ln x2/x)   = d/dx e(2 ln x)/x   = 2e(ln x)/x d/dx [(ln x)/x]   = 2x2/x [(1/x)x - (ln x)1]/x2   = 2x2/x - 2 (1 - ln x)   This function is related to an interesting function f(x) = x1/x for which I...

The only way this can be elementary integral is if the function is   f(x) = x1/3 + 5/x6   since for all other combinations, Wolfram|Alpha gives an answer in terms of the Hypergeometric function 2F1(a,b;c;x).   ∫ (x1/3 + 5/x6)∫ (x1/3 + 5x-6) dx...

Let I be his age at inauguration, and D be the age of death.   The first clue is 2I - D = 45.   You get:   D = 2I - 45 = 2·69 - 45 = 138 - 45 = 93   So he died at age 93.   This is indeed his age of death: https://en.wikipedia...