The correct answer for this question is this one: "<span>|-60 - 11| = |-71| = 71 units"
The statement says that "the </span><span>absolute value that show the distance between -60 and 11"
Absolute value is the distance of that specific number from zero in a straight number line. The selected answer is the correct one.</span>
Answer:
f(x) = -2.5x + 4
Step-by-step explanation:
If you go down 5 units and 2 units to the right then there is another point so the slope is -5/2 which is the same thing as -5 divided by 2. So it’s -2.5 as the slope and for the y- intercept, there is only one point on there and it is on 4. So the function is
f(x) = -2.5x + 4
I hope this helps
(I deserve to be marked the brainliest)
Find the marginal density for
by integrating the joint PDF over all possible values of
:
![f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^y2\,\mathrm dx](https://tex.z-dn.net/?f=f_Y%28y%29%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20f_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5Ey2%5C%2C%5Cmathrm%20dx)
![\implies f_Y(y)=\begin{cases}2y&\text{for }0](https://tex.z-dn.net/?f=%5Cimplies%20f_Y%28y%29%3D%5Cbegin%7Bcases%7D2y%26%5Ctext%7Bfor%20%7D0%3Cy%3C1%5C%5C0%26%5Ctext%7Botherwise%7D%5Cend%7Bcases%7D)
Then the density of
conditioned on
is
![f_{X|Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\frac1y](https://tex.z-dn.net/?f=f_%7BX%7CY%7D%28x%5Cmid%20y%29%3D%5Cdfrac%7Bf_%7BX%2CY%7D%28x%2Cy%29%7D%7Bf_Y%28y%29%7D%3D%5Cfrac1y)
for
and undefined elsewhere.
Thus
![P(0.25](https://tex.z-dn.net/?f=P%280.25%3CX%3C0.50%5Cmid%20Y%3D0.9%29%3D%5Cdisplaystyle%5Cint_%7B0.25%7D%5E%7B0.50%7Df_%7BX%7CY%7D%28x%5Cmid%20y%3D0.9%29%5C%2C%5Cmathrm%20dx%3D%5Cint_%7B0.25%7D%5E%7B0.50%7D%5Cfrac%7B%5Cmathrm%20dx%7D%7B0.9%7D%5Capprox%5Cboxed%7B0.28%7D)
= = =
I'm not seeing the TeX rendering on any of my browsers. Maybe it's a site-wide issue? In any case, I'm attaching a picture of the text above in case it's incomprehensible.
Answer:
![68m^2](https://tex.z-dn.net/?f=68m%5E2)
Step-by-step explanation:
To find the surface area of the prism, we find the area of each similar faces and multiply by 2 and then add all together
Area of square faces
Area of parallelogram faces
![=2bh](https://tex.z-dn.net/?f=%3D2bh)
![=2(2\times 4)=16m^2](https://tex.z-dn.net/?f=%3D2%282%5Ctimes%204%29%3D16m%5E2)
Area of the rectangular faces;
![=2(4\times 2.5)=20m^2](https://tex.z-dn.net/?f=%3D2%284%5Ctimes%202.5%29%3D20m%5E2)
The total surface area of the prism is 20+16+32=68 square meters