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Orlov [11]
3 years ago
5

Giả thiết X là biến ngẫu nhiên X~N(µ, σ^2) thì P( |X-µ| >3,05*σ ) sẽ là

Mathematics
1 answer:
svp [43]3 years ago
7 0

Answer:

,luvxieixeqbxeixbwibiblccucucx

Step-by-step explanation:

  1. h hccylbs f
  2. qvdjvjeflnd
  3. nwlxlwbxaabx
  4. emvfvten
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12+11+24 is 43 and subtract that by 3 which is 40

Answer is C
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What is 2E*-4V*:34>2+sv5
Mrac [35]

Answer:yes

Step-by-step explanation:

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3 years ago
Prove the following
DerKrebs [107]

\bf [cot(\theta )+csc(\theta )]^2=\cfrac{1+cos(\theta )}{1-cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand side}}{[cot(\theta )+csc(\theta )]^2}\implies cot^2(\theta )+2cot(\theta )csc(\theta )+csc^2(\theta ) \\\\\\ \cfrac{cos^2(\theta )}{sin^2(\theta )}+2\cdot \cfrac{cos(\theta )}{sin(\theta )}\cdot \cfrac{1}{sin(\theta )}+\cfrac{1}{sin^2(\theta )}\implies \cfrac{cos^2(\theta )}{sin^2(\theta )}+\cfrac{2cos(\theta )}{sin^2(\theta )}+\cfrac{1}{sin^2(\theta )}

\bf \cfrac{\stackrel{\textit{perfect square trinomial}}{cos^2(\theta )+2cos(\theta )+1}}{sin^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the right-hand-side}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}}\implies \stackrel{\textit{multiplying by the denominator's conjugate}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}\cdot \cfrac{1+cos(\theta )}{1+cos(\theta )}}

\bf \cfrac{[1+cos(\theta )]^2}{\underset{\textit{difference of squares}}{[1-cos(\theta )][1+cos(\theta )]}}\implies \cfrac{[cos(\theta )+1]^2}{1^2-cos^2(\theta )} \\\\\\ \cfrac{[cos(\theta )+1]^2}{1-cos^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}}

recall that sin²(θ) + cos²(θ) = 1, thus sin²(θ) = 1 - cos²(θ).

7 0
3 years ago
What is the minimum value of the objective function C = 4x + 9y?
Elodia [21]

Answer:-24

explain how you know:

Since xy = 4, we have y=4/x. Substituting in (1) we get f(x) = 4x+(36/x) …….(2)

Differentiating we get f ' (x) = 4 - (36/x^2)…..(2)

For maxima or minima, f ‘ (x) = 0

=> 4 - (36/x^2) =0 => (4x^2 - 36)/x^2 = 0 which gives x= +/- 3

Differentiating (2), f ‘’ (x) = 72/x^3

When x= +3 , f ‘’ (x) is clearly +ve. Therefore f(x) at (2) gives minima and the minimum value is 24.

When x= -3, f ‘’ (x) is -ve. Therefore, f(x) has maxima at x= -3. The maximum value is -24.

Note: You may be surprised to observe that the minima is more than the maxima. This is due to the fact that the function is discontinuous at x= 0, the graph of f(x) comprising of two branches, one in the first quadrant giving the minima (= +24) and another in the third quadrant that gives maxima (= -24)

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3 years ago
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6(x-3)=3x+8 solve for x give an answer in inproper fraction linear equation with x on both sides
8_murik_8 [283]
6x-18=3x+8
3x=26
x= 26/3
8 0
2 years ago
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