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Grace [21]
3 years ago
7

What are the answers to these thank you and please tell me how to get it

Mathematics
1 answer:
mote1985 [20]3 years ago
4 0
5-3(1/2x+2)=-7
-3(1/2x+2)=-7-5
(1/2x+2)=4
1/2x=2
x=4
just to give u an idea
You might be interested in
Is this correct? If not<br> please tell me the answer!
attashe74 [19]
Yes, you are absolutely correct.
3 0
2 years ago
An aircraft factory manufactures airplane engines. The unit cost (the cost in dollars to make each airplane engine) depends on t
Juliette [100K]

Answer:

Minimum unit cost = 5,858

Step-by-step explanation:

Given the function : C(x)=x^2−520x+73458

To find the minimum unit cost :

Take the derivative of C(x) with respect to x

dC/dx = 2x - 520

Set = 0

2x - 520

2x = 520

x = 260

To minimize unit cost, 260 engines must be produced

Hence, minimum unit cost will be :

C(x)=x^2−520x+73458

Put x = 260

C(260) = 260^2−520(260) + 73458

= 5,858

7 0
3 years ago
II. Let f(x) = 9 – x , g(x) = x2<br> + 4, and h(x) = x – 2. Compute the following:<br> 7. g(f(12))
Alex17521 [72]
G(f(12))
f(12) = 9 - 12 = -3
g(-3) = (-3)^2 + 4 = 9 + 4 = 13
4 0
2 years ago
Please help me thanks
BabaBlast [244]

Answer:

16.7

Step-by-step explanation:

We can use the Pythagorean theorem, a^2 + b^2 = c^2.

a = 11

c = 20

11^2 + b^2 = 20^2

121 + b^2 = 400

b^2 = 279

b = sqrt(279) = 16.7

5 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).

7 0
3 years ago
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