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3 years ago

Write the number 31 in tens and ones. 2 tens

2 answers:

5 0

31 =
3 tens or 30
and 1 ones or 1 and 30 + 1 = 31
"AB84"

sammy [17]3 years ago

3 0

The tens digit of 31 is 3. There are 3 tens in this number and that is equal to 30. Also, the ones digit of this number is equal to 1. The answer to this item are therefore 30 and 1. Hope This Helps! :)

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Year 11 math methods, help pls

Ludmilka [50]

Answer:

$\frac{1}{180} (x - 90) {}^{2} + 30 = y$

Step-by-step explanation:

Please see the attachment:

4 0

3 years ago

What is the volume of the pyramid in cubic millimeters? a rectangular pyramid with a base of 21 millimeters by 18 millimeters an

grandymaker [24]

The volume of the rectangular pyramid with the given dimensions is: 1,890 cubic millimeters.

<h3>What is the Volume of a Rectangular Pyramid?</h3>

Volume of rectangular pyramid = 1/3(length × width × height)

Given the following:

Length = 21 mm
Width = 18 mm
Height = 15 mm

Volume of the rectangular pyramid = 1/3(21 × 18 × 15)

Volume of rectangular pyramid = 1,890 cubic millimeters.

Learn more about volume of rectangular pyramid on:

brainly.com/question/924442

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8 0

3 years ago

If 3% of ties sold in the United States are bow ties, what is the probability of randomly selecting three ties that were purchas

Luda [366]

Given that 3 ties were selected and the probability of selecting a bow tie is 3%, the probability of selecting 3 bow ties will be:
P(selecting 3 ties)=P(selecting a bow tie)*P(selecting a bow tie)*P(selecting a bow tie)
=(0.03)*(0.03)*(0.03)
=0.000027

7 0

4 years ago

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

3 years ago

How many solutions -3x+9-2x=-12-5X

vesna_86 [32]

Answer:

Step-by-step explanation:

5 0

3 years ago