Answer:
I thinks its T (3)
Step-by-step explanation:
Answer:
2 and 2/3
Step-by-step explanation:
2/3 multiplied by 4
Answer:
18
Step-by-step explanation:
Denominators are 3, 6 and 9
So 18 is the required number
a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have
![\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20f%28y%29%20%5C%2C%20dy%20%3D%20%5Cfrac14%20%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cfrac%20y4%7D%20%5C%2C%20dy%20%3D%20-%5Cleft%28%5Clim_%7By%5Cto%5Cinfty%7De%5E%7B-%5Cfrac%20y4%7D%20-%20e%5E0%5Cright%29%20%3D%20%5Cboxed%7B1%7D)
so both conditions are met and f(y) is indeed a PDF.
b) The probability P(Y > 4) is given by the integral,
![\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E4%20f%28y%29%20%5C%2C%20dy%20%3D%20%5Cfrac14%20%5Cint_0%5E4%20e%5E%7B-%5Cfrac%20y4%7D%20%5C%2C%20dy%20%3D%20-%5Cleft%28e%5E%7B-1%7D%20-%20e%5E0%5Cright%29%20%3D%20%5Cfrac%7Be%20-%201%7D%7Be%7D%20%5Capprox%20%5Cboxed%7B0.632%7D)
c) The mean is given by the integral,
![\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%20f%28y%29%20%5C%2C%20dy%20%3D%20%5Cfrac14%20%5Cint_0%5E%5Cinfty%20y%20e%5E%7B-%5Cfrac%20y4%7D%20%5C%2C%20dy)
Integrate by parts, with
![u = y \implies du = dy](https://tex.z-dn.net/?f=u%20%3D%20y%20%5Cimplies%20du%20%3D%20dy)
![dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}](https://tex.z-dn.net/?f=dv%20%3D%20e%5E%7B-%5Cfrac%20y4%7D%20%5C%2C%20dy%20%5Cimplies%20v%20%3D%20-4%20e%5E%7B-%5Cfrac%20y4%7D)
Then
![\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%20f%28y%29%20%5C%2C%20dy%20%3D%20%5Cfrac14%20%5Cleft%28%5Cleft%28%5Clim_%7By%5Cto%5Cinfty%7D%5Cleft%28-4y%20e%5E%7B-%5Cfrac%20y4%7D%5Cright%29%20-%20%5Cleft%28-4%5Ccdot0%5Ccdot%20e%5E0%5Cright%29%5Cright%29%20%2B%204%20%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cfrac%20y4%7D%20%5C%2C%20dy%5Cright%29)
![\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccdots%20%3D%20%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Cfrac%20y4%7D%20%5C%2C%20dy)
![\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccdots%20%3D%20-4%20%5Cleft%28%5Clim_%7By%5Cto%5Cinfty%7D%20e%5E%7B-%5Cfrac%20y4%7D%20-%20e%5E0%5Cright%29%20%3D%20%5Cboxed%7B4%7D)
Answer:
The values of
k are
(-4,2)
Step-by-step explanation:
sorry if wrong