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e-lub [12.9K]
3 years ago
7

I need help on 32 please

Mathematics
1 answer:
saveliy_v [14]3 years ago
8 0
That is how you solve it

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Solve for x.
Ivahew [28]
ANSWER

x=\frac{2-\sqrt{10}} {3}

or

x=\frac{\sqrt{10}+2} {3}

We have

3x^2-4x-2=0

Since we cannot factor easily, we complete the square.

Adding 2 to both sides give,

3x^2-4x=2

Dividing through by 3 gives

x^2-\frac{4}{3}x= \frac{2}{3}

Adding (-\frac{2}{3})^2 to both sides gives

x^2-\frac{4}{3}x+(-\frac{2}{3})^2= \frac{2}{3}+(-\frac{2}{3})^2

The expression on the Left Hand side is a perfect square.

(x-\frac{2}{3})^2= \frac{2}{3}+\frac{4}{9}

\Rightarrow (x-\frac{2}{3})^2= \frac{10}{9}

\Rightarrow (x-\frac{2}{3})=\pm \sqrt{\frac{10}{9}}

\Rightarrow (x)=\frac{2}{3} \pm {\frac{\sqrt{10}}{3}

Splitting the plus or minus sign gives

x=\frac{2- \sqrt{10}} {3}

or

x=\frac{\sqrt{10}+2} {3}
3 0
3 years ago
Read 2 more answers
The antiderivative of 5x
Anni [7]

The antiderivate of 5x is 5x^2 / 2 .
5 0
3 years ago
Gracie plans to buy a new pair of shoes for $75.00, however, the manager sells them to her for
Cerrena [4.2K]
15% reason 75 minus 60=15
6 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
3 years ago
What is the area of sector​ GPH? the angle inside the circle measures 80º and the radius from p to h is 15
faltersainse [42]
Area = π r²

area of the required sector = (80/360) π * 15²
                                           ≈ 157.08
6 0
3 years ago
Read 2 more answers
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