Name the sets of numbers to which -1/2 belongs

561 rounded to the nearest hundred

insens350 [35]

561 rounded to the nearest hundred is 600

4 0

3 years ago

Multiply and give the answer in scientific notation:

kakasveta [241]

Step-by-step explanation:

given

$(1.5 \times {10}^{4} )(8 \times {10}^{8} )$

in order to make multiplication easier we need to change the 1.5 into a whole number form.

thus

$(15 \times {10}^{ - 1} \times {10}^{4} )(8 \times {10}^{8} )$

$= (15 \times {10}^{4 - 1} )(8 \times {10}^{8} )$

First law of indices applied there

=(15×10^3)(8×10^8)

=(15×8)(10^3×10^8)

=120×10^3+8 ( first law of indices, which states that , numbers of the same base multiplying each other, take one of the base and add the exponent. and clearly both 15 and 8 are in base 10

=120×10^11

=1.20×10^2 ×10^11

=1.20×10^11+2

=1.20×10^13

so the answer is alt B

7 0

3 years ago

Bean plants grow rapidly. A bean plant is 16 inches tall. Tomorrow it will be 24 inches tall, the next day it will be 32 inches

vodka [1.7K]

Every day, it increases by 8 inches. We know that it started at 16 inches.

We can construct an equation with this information.

Equation: 16+8x=h
h=Height
x=Days

Plug in 12 for x:
16+96=h
h=112 inches

Answer: The plant will be 112 inches tall in 12 days.

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).

7 0

3 years ago

Evaluate the following expression. 1/5^-2

Romashka [77]

This is how the whole equation goes:
1/x^-a = x^a
So, 1/5^-2 is the same as 5^2, which is 25.

8 0

3 years ago

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