What fraction of an hour is 18 minutes? Give your answer in its simplest form.

What number between 20 and 25 is a prime number?

Elena-2011 [213]

Prime numbers are numbers that have only one factor pair; 1 and themselves.

Let's start from 21.

21 = 1 x 21, 3 x 7.
This has two factor pairs, this is not prime.

Let's try 22.

22 = 1 x 22, 2 x 11.
This also has two factor pairs, this is not prime.

Let's try 23.

1 x 23.
This is a prime number, as it only has one factor pair; 1 and itself.

23 is your prime number.

I hope this helps!

6 0

3 years ago

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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 :

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

(c) Get the marginal density of Y by integrating with respect to x instead:

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

(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

What does (Y) equal? 2(5 + y) = 18

Bas_tet [7]

Answer:

y = 4

Step-by-step explanation:

distribute

10 + 2y = 18

minus 10 from both sides

2y = 8

divide by 2

y = 4

6 0

3 years ago

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A circle is to be cut from a 12 ft square board. What will be the area, in square feet, of the inscribed circle? Leave your answ

Kazeer [188]

Answer:

3π sq.feet

Step-by-step explanation:

Area of the square = L²

L is the length of the square

Given

Area of the square = 12 ft square

12 = L²

L = √12

L = 2√3 feet

Since the circle is inscribed in the square, the length of the square will be the diameter of the circle;

Area of the circle = πd²/4

Area of the circle = π(2√3)²/4

Area of the circle = 12π/4

Area of the circle = 3π sq.feet

5 0

3 years ago

Which expression is equivalent to 0 + 2 29 C+3 C+2 C+3 C+2

Kryger [21]

Answer:

C+3

Step-by-step explanation:

6 0

3 years ago