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2 years ago

Which number is an integer? -34 -0.4 0.4 3.4 PLS HELP I'M TIMED EDGE 2020

Mathematics

2 answers:

Stella [2.4K]2 years ago

7 0

Answer:

Step-by-step explanation: The four numbers: - 34, -0.4, 0.4, 3.4

so if you look at it it The answer is -34

musickatia [10]2 years ago

3 0

Answer:

-34

Step-by-step explanation:

this is because an integer is a negative or positive whole number, and in this case, -0.4, 0.4 and 3.4 are decimal numbers

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Chang knows one

larisa86 [58]

9514 1404 393

Answer:

(d) 8 cm and 9 cm

Step-by-step explanation:

The sum of the other two side lengths must exceed 13 cm for a triangle to be possible.

8 cm and 9 cm

5 0

2 years ago

Read 2 more answers

Ok so I wanna if my answer is right so

sergey [27]

Yes and you can say more x€ (-inf;15)

5 0

2 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

The moment generating function for health care costs experienced by a policyholder is given as follows:

Kaylis [27]

Answer:

0.525 = 52.5%

Step-by-step explanation:

Moment generating function ( Mx(t) ) = ( 4 / 4-t)^3

Reimbursement by Insurer = 70%

Determine the expected reimbursement by insurer for policyholder

d/dx (Mx(t) ) = d/dt ( 4 / 4-t)^3 = d/dt (1 - t/4 )^-3

= 3/4 ( 1 - t/4 )^-4 = 3/4

as t → 0

Given that the insurer reimburses 70% = 0.7

expected reimbursement = 0.7 * 3/4 = 0.7 * 0.75 = 0.525

6 0

2 years ago

Solve the equation 15p2 + 7p − 4 = 0 by completing the square.

natulia [17]

Answer:

p=0 or p=-7/15

Step-by-step explanation:

15p2+7p=0

p(15p+7)=0

p=0 or 15p+7=0

4 0

1 year ago