All categories

3 years ago

I need help on problem number 4

Mathematics

2 answers:

Andru [333]3 years ago

7 0

The answer is B.

Also guys, don't hate on him, he's in elementary :(

Pepsi [2]3 years ago

6 0

B is the best answer
.....

You might be interested in

(5•-3) - (4•-3) + -3 - (3•-3)

Dmitry_Shevchenko [17]

Use PEMDAS.

Parenthesis first.

(-15)-(-12)+3-(-9)

Next multiply,

(-15)+12+3+9 (two negatives get you a positive)

Finally, add and subtract:
-15+24=9

You get 9 as your answer.

Hope this helps!

4 0

4 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

PLEASE! WILL GIVE METAL AND FAN.

Pani-rosa [81]

20000*0.45 = 9000 in the bond
20000*0.15 = 3000 in the CD
20000*0.20 = 4000 in stocks
20000*0.029 = 580 in savings

A=9000(1 + 4.35%)^3 = 10,226.33
A=3000(1 + 2.90%)^3 = 3,268.64
A=4000 (1 + 8%) x (1 - 4%) x (1 + 6%) = 4,396.03
A=580(1 + 4.35%)^3 = 4,545.04
Total value = 22,436.04
Gain = 22,436.04 - 20,000 = 2,436.04

6 0

3 years ago

Read 2 more answers

A small-business Web site contains 100 pages an 60%, 30%, and 10% of the pages contain low, moderate, an high graphic content, r

Karolina [17]

Answer:

fx(x)=[(4! /(x!)*(4-x)!]*[(0.3) ^x]*[(0.7) ^ (4-x)]

Step-by-step explanation:

F(x) is been calculated by considering X (moderate) as one outcome and low and high combined as one outcome now probability function will be same as binomial distribution

F(x) = [(4! /(x!)*(4-x)!]*[(0.3) ^x]*[(0.7) ^ (4-x)]

8 0

3 years ago

Find the period of the function. y=3 sin x/8

Alexus [3.1K]

Answer:

The period of given function is $Period = 16\pi$