Item 2

Here are two column vectors

Contact [7]

Answer:

$a = ( \frac{ - 6}{2} )$

Step-by-step explanation:

-2g=?

$- 2( \frac{3 }{ - 1} ) \\ \\ ( \frac{ - 6}{2} )$

8 0

2 years ago

TunsAXP what is the value of x when x = 13

Advocard [28]

Answer:

the value of x would be 13

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

How much is 2000 × 36 =? help my please

PIT_PIT [208]

Answer:

72000 should be your answer

Step-by-step explanation:

Got it off of g o o g l e so it should be right!

4 0

2 years ago

0.33333333333333333333333333333333333333333333333 simplified at a fraction

USPshnik [31]

Answer:

$\frac{1}{3}$

Step-by-step explanation:

Your original number to convert is 0.333333. Let's slide the decimal point in this number to the right 1 place(s) (the same number of digits in the number 3).

If we do this, we'll get a 3.333333 (slide the decimal in the 0.333333 right 1 places, you'll get 3.333333).

So what? Well now, we have two numbers with the same repeating decimal parts, 3.333333 and 0.333333.

Now let's just work a little algebra into all of this. Let's call your original number x. And in this case, x = 0.333333. The number with the decimal point slid over can be called 10x, because 10x = 3.333333

10x = 3.33333

-x = 0.333333 $\frac{10x = 3.33333 -x = 0.333333}{9x = 3}$

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Now, solving 9x=3 for x by dividing both sides of it by 9, we'll get that x=3/9. And this is your answer.

How is this your answer? Well remember that above, x was originally set equal to 0.333333 via x = 0.333333, and now we have that x is also equal to 3/9, so that means 0.333333 = 3/9..and there's 0.333333 written as a fraction.

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<u>Simplify 3/9 into the lowest terms:</u>

$\frac{1}{3}$

4 0

2 years ago

If s is an increasing function, and t is a decreasing function, find Cs(X),t(Y ) in terms of CX,Y .

Sedbober [7]

Answer:

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X,Y), X and Y :

F(X,Y)(x,y)=P(X≤x,Y≤y)

FX(x)=P(X≤x)
FY(y)=P(Y≤y).

Likewise for (s(X),t(Y)), s(X) and t(Y) :

F(s(X),t(Y))(u,v)=P(s(X)≤u

t(Y)≤v)
Fs(X)(u)=P(s(X)≤u)
Ft(Y)(v)=P(t(Y)≤v).

Now, First establish the relationship between F(X,Y) and F(s(X),t(Y)) :

F(X,Y)(x,y)=P(X≤x,Y≤y)=P(s(X)≤s(x),t(Y)≥t(y))

The last step is obtained by applying the functions s and t since s preserves order and t reverses it.

This can be further transformed into

F(X,Y)(x,y)=1−P(s(X)≤s(x),t(Y)≤t(y))=1−F(s(X),t(Y))(s(x),t(y))

Since our random variables are continuous, we assume that the difference between t(Y)≤t(y) and t(Y)<t(y)) is just a set of zero measure.

Now, to transform this into a statement about copulas, note that

C(X,Y)(a,b)=F(X,Y)(F−1X(a), F−1Y(b))

Thus, plugging x=F−1X(a) and y=F−1Y(b) into our previous formula,

we get

F(X,Y)(F−1X(a),F−1Y(b))=1−F(s(X),t(Y))(s(F−1X(a)),t(F−1Y(b)))

The left hand side is the copula C(X,Y), the right hand side still needs some work.

Note that

Fs(X)(s(F−1X(a)))=P(s(X)≤s(F−1X(a)))=P(X≤F−1X(a))=FX(F−1X(a))=a

and likewise

Ft(Y)(s(F−1Y(b)))=P(t(Y)≤t(F−1Y(b)))=P(Y≥F−1Y(b))=1−FY(F−1Y(b))=1−b

Combining all results we obtain for the relationship between the copulas

C(X,Y)(a,b)=1−C(s(X),t(Y))(a,1−b).

7 0

2 years ago