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 =...
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...
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A Story Proof is a method of proving an identity by showing that both sides of the equation are solutions to the same problem. This...
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...
Recall the slope formula: m = (y2 - y1) / (x2 - x1) (a) m = (-2 - 1) / (3 - -2) = -3/5 (b) m = (-1 - -2) / (5 - 0) = 1/5
Yes. This represents a function with domain {-1,0,2,3} because in this set, knowing x uniquely gives f(x). x = -1 ⇒ f(x) = 4 x = 0 ⇒ f(x) = 0 x = 2 ⇒ f(x) = 5 x = 3 ⇒ f(x) = 5
We first need the mean μ and standard deviation σ. We also know that since we are testing at significance α=1-0.95 =0.05. Thus we have: n = 12 xbar = Σxi / n = (4045 + ... + 4324) / 12 = 4341.25 s = √ [Σ(xi - xbar)2 / (n-1)] = √{ [(4045 -...