Can anyone help me with this math question

What's the inverse of 5y+4 = (x+3)2 + 1/2

kondor19780726 [428]

F^-1 (x) = 5x/2 - 5/4

8 0

2 years ago

What is the equation of the line that has a slope of -4 and goes through the point (4,-6)

Elis [28]

It would be Y=-4x+10. Please mark me brainliest. It’s fine if you don’t want to tho.

6 0

3 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 :

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

(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

A circle is growing so that the radius is increasing at the rate of 2cm/min. How fast is the area of the circle changing at the

poizon [28]

The answer to this question is:

A circle is growing so that the radius is increasing at the rate of 2cm/min. How fast is the area of the circle changing at the instant the radius is 10cm? Include units in your answer.?
✔️I assume here the linear scale is changing at the rato of 5cm/min
✔️dR/dt=5(cm/min) (R - is the radius.... yrs, of the circle (not the side)
✔️The rate of area change would be d(pi*R^2)/dt=2pi*R*dR/dt.
✔️At the instant when R=20cm,this rate would be,
✔️2pi*20*5(cm^2/min)=200pi (cm^2/min) or, almost, 628 (cm^2/min)

Hoped This Helped, Cello10
Your Welcome :)

5 0

3 years ago

Read 2 more answers

Which expression is equivalent to (18)2⋅(19)2

sladkih [1.3K]

The expression that is equivalent to the given expression where the expression is given as (18)2⋅(19)2 is (18 * 19)^2

<h3>How to determine which expression is equivalent to the given expression? </h3>

The expression is given as

(18)2⋅(19)2

Rewrite the above expression properly

So, we have

(18)^2 * (19)^2

The factors in the above expression have the same exponent.

So, the expression can be rewritten as

(18 * 19)^2

Hence, the expression that is equivalent to the given expression where the expression is given as (18)2⋅(19)2 is (18 * 19)^2

Read more about equivalent expression at

brainly.com/question/2972832

#SPJ1

5 0

1 year ago