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3 years ago

Is 5/6 close to 1, 0, or 1/2. Explain

1 answer:

5 0

5/6 is closest to 1 because it it 1/6 away from 1 and is 5/6 away from 0 and is 2/6 away from 1/2 (referring 1/2 as 3/6) Hope this helps!

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A line has slope

loris [4]

Y = mx+b
-1 = 13/4 (-1) +b
-4/4 + 13/4 = b
9/4 = b

3 0

2 years ago

Cubic yards of concrete will be required for 14 parking lot light pole standards

riadik2000 [5.3K]

Using conversion of units and volume of a cylinder, it is found that 23.17 cubic yards of concrete are needed, option D.

The volume of a cylinder of radius r and height h is given by:

$V = \pi r^2h$

In this problem:

We want the volume in cubic yards, thus, the units of radius and height have to be in yards.
32 inches in diameter, thus 16 inches of radius. Since each inch has 0.0278 yards, this is a radius of 0.44 yards, thus $r = 0.44$ .
Height of 8 feet, and since each feet has 0.33 yards, this is a height of 2.67 yards, thus $h = 2.67$
14 poles, thus the volume is multiplied by 14.

Then:

$V = 14\pi(0.44)^2(2.67) = 23.17$

23.17 cubic yards of concrete are needed, option D.

A similar problem is given at brainly.com/question/22789697

3 0

2 years ago

What is the slope of a line parallel to 5x-3y=-10

Llana [10]

For parallel lines, slopes are equal.
5x - 3y = -10
3y = 5x + 10
y = 5/3 x + 10/3

Required slope = 5/3

7 0

2 years ago

STUDENT HOURS KATY 5.5

mario62 [17]

The best answer to go with is b

4 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