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Rama09 [41]
3 years ago
6

A ratio that is equivalent to 60:25

Mathematics
2 answers:
gulaghasi [49]3 years ago
8 0
A ratio that is equivalent to 60:25.
You can simplify it by finding the Greatest Common Factor and dividing the numbers by the GCF.
60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 30, 60
25: 1, 5, 25
The greatest common factor is 5.
Divide both sides by 5.
60/5=12
25/5=5
A ratio that is equivalent is 12:5

Hope I helped!

Margarita [4]3 years ago
3 0
Divide both sides by 5 to get:

12:5

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tatiyna

0.6 cup of gliter is used for 1 cup of glue and 1.6 cup of gliter glue is made with 1 cup of glue.

Part A

Given that,

15 cup of gliter is for 25 cup of glue ,

Therefore, "X" cup of gliter is for 1 cup of glue

X = (1*15)/25

X= 0.6 cup of gliter

Part B,

Given that,

From part A 0.6 cup of gliter is for 1 cup of glue

Thus, 1 cup of glue and 0.6 cup of gliter when mixed will form "Y" cup of gliter glue

Now, Y= 0.6 cup of gliter + 1 cup of glue

Y= 1.6 cup of gliter glue

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1 year ago
What is the range of y = sin θ ?
Tamiku [17]
The range of a function is the set of all the possible function outputs. We know that sin theta varies between -1 and 1 if we draw a graph. According to the graph, the largest number that is an output of g is -1, and the smallest number is 1. Every number between them is also an output of g for some input. Therefore, the range of g is [-1,1]. 
5 0
3 years ago
Simplify the algebraic expression: 4(3x + y) – 2(x – 5y)
VladimirAG [237]

Answer:

C

Step-by-step explanation:

First use distribution

4(3x+y)= 12x+4y

and

-2(x-5y)= -2x+10y

combine the 2 answers

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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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I thought it was -4 I calculated it and that’s what I got?¿
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