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kaheart [24]
3 years ago
15

Take two orthogonal vectors u, v ε R". u v, and let X = (Xi, , X,) be a vector of n iid standard Gaussians. Xi ~ N(0, 1), yl E「r

). Let llr-〈II, X〉 and vr-(v, X〉, Are ux and vr independent? Hin: First try to see if they are correlated; you may use the fact that jointly normal random variables are independent iff. they are uncorrelated.

Mathematics
1 answer:
Semenov [28]3 years ago
7 0

Answer:

Attached is the complete solution.

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Dimensional <br> 528 feet to miles
Volgvan

Answer: 0.1 mile

Step-by-step explanation:

5280 feet equals one mile

5280 ÷ 10 equals 528

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3 years ago
Damian plans to do a lot of babysitting this summer. He charges $6.50 per hour and babysits for 3 hours. He has agreed to babysi
RoseWind [281]

Answer:

$195

Step-by-step explanation:

multiplying 6.50*3*10 = 195

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What is the difference of the fractions? Use the number line to help find the answer.
rjkz [21]

hope that this answers your problem !

To solve this question, we first need to layout the equation

-2 1/2 -(-1 3/4)

Step 1

-2 1/2-(-1 3/4) ... Equation

Step 2

5/2-(-1 3/4) ... Converted to improper fraction

Step 3

5/2-(-7/4) ... converted to improper fraction

Step 4

5/2--7/4 ... Got rid of parenthesis

Step 5

-3/4 ... Subtract

Answer:

-3/4 ... Answer

So the answer for this problem is C, -3/4

4 0
2 years ago
Fred sells flags for f dollars each. he decides to give a discount and sell the flags in his store for a 26% markdown. write an
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G=.74f, where f is the initial price and g is the sale price
7 0
3 years ago
If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?
Naddik [55]

Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor  (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

5 0
3 years ago
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