Please help with this question, giving brainliest

Round each decimal 32.62 tenths place

Tresset [83]

Answer:

32.6

Step-by-step explanation:

Look at the decimal in the hundredths place in 32.62

If the number is 0-4, round down. If the number is 5-9, round up.

In 32.62, the number in the hundredths place is 2. So, we round down to 32.6

3 0

3 years ago

You are giving 24 linear feet of fencing all 3 feet tall you wish to make the largest rectangular enclosure possible to maximize

earnstyle [38]

Answer:

The largest area enclosed is A = xy = 6 feet $\times$ 6 feet = 36 $feet^{2}$

Step-by-step explanation:

i) The perimeter of the area is 2 $\times$ (x + y) =24 ∴ x + y = 12 ∴ y = 12 - x

ii) The area of rectangle enclosed A = xy ⇒ A = x ( 12 - x) = 12x - $x^{2}$

iii) differentiating both sides of the equation in ii) we get

$\dfrac{dA}{dx} = 12 - 2x = 0$ ⇒ x = 6 feet

iv) Differentiating both sides of equation in iii) we get $\frac{d^{2}A}{dx^{2} }$ = -2

Therefore the area enclosed is maximum as the double derivative is negative

v) therefore for largest area enclosed x = 6 feet and y = 12 - 6 = 6 feet

vi) therefore the largest area enclosed is

A = xy = 6 feet $\times$ 6 feet = 36 $feet^{2}$

7 0

3 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago

A dime has a radius of 8.5 millimeters. Find the circumference of a dime to the nearest tenth

g100num [7]

Well C=Dπ so C=17 x 3.14 = 53.38 which is rounded to 53.4

3 0

3 years ago

Can someone help me with this plz I am so confused??

ruslelena [56]

F= -1 2/3 or 5/3

In order to find the answer all you have to do is multiply both sides by -4/3.

-3/4f × -4/3 = f

-4/3 × 5/4 = -20/12

-20/12 simplified is -5/3 or -1 2/3

6 0

3 years ago

Read 2 more answers