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Oduvanchick [21]
3 years ago
5

8th grade math

Mathematics
2 answers:
UkoKoshka [18]3 years ago
6 0

Answer:

10 Milk chocolates

8 dark chocolates

6 white chocolates

24 TOTAL

\frac{10}{24} Probability that Hanissa choose a milk chocolate

Now, since she has already eaten one chocolate, then what's left is 23 chocolates.

So, since the second one should be a white chocolate and there are 6 of them, then

\frac{6}{23} - probability of choosing a white chocolate.

Then, multiply,

( \frac{10}{24} )( \frac{6}{23} )= \frac{60}{552} or \frac{5}{46}

Answer: 5/46 or 10.87%

Step-by-step explanation:

Vlad1618 [11]3 years ago
3 0

Answer: there is 1/2 chance of your first being milk chocolate, 1/3 chance your second will be drake and 1/6 chance it will be white chocolate

Step-by-step explanation:

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Tyrone mixed 7/12 quart of red psont with 1/12 quart of yellow paint how much paint does tyrone have in the mixture
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Hello there.

<span>Tyrone mixed 7/12 quart of red paint with 1/12 quart of yellow paint how much paint does Tyrone have in the mixture?
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Answer: In this question you need to add together the fractions provided.
The sum would look like this:
</span>\frac{7}{12} + \frac{1}{12} =
<span>Since the denominators are the same , it is simple to add.
All you need to do is 7 + 1 = 8.
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3 0
3 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).

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Step-by-step explanation:

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Step-by-step explanation:

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