a. In order for
to be a PDF,
needs to be non-negative (it is) and the integral over its support must evaluate to 1.
is symmetric about
, i.e. even, so
![\displaystyle\int_{-\infty}^\infty\frac\lambda2e^{-\lambda|x|}\,\mathrm dx=\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-e^{-\lambda x}\bigg|_0^\infty=1](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cfrac%5Clambda2e%5E%7B-%5Clambda%7Cx%7C%7D%5C%2C%5Cmathrm%20dx%3D%5Clambda%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx%3D-e%5E%7B-%5Clambda%20x%7D%5Cbigg%7C_0%5E%5Cinfty%3D1)
and so
is indeed a PDF.
b. Let
be a random variable with
as its PDF. Then
![\mu=E[X]=\displaystyle\int_{-\infty}^\infty xf(x)\,\mathrm dx=\frac\lambda2\int_{-\infty}^\infty xe^{-\lambda|x|}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cmu%3DE%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac%5Clambda2%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xe%5E%7B-%5Clambda%7Cx%7C%7D%5C%2C%5Cmathrm%20dx)
The integrand is odd, so the integral vanishes and the mean is
.
The variance of
is
![\sigma^2=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=%5Csigma%5E2%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
The second moment is
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f(x)\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2f%28x%29%5C%2C%5Cmathrm%20dx)
This integrand is even, so
![E[X^2]=\displaystyle2\int_0^\infty x^2f(x)\,\mathrm dx=\lambda\int_0^\infty x^2e^{-\lambda x}\,\mathrm dx](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle2%5Cint_0%5E%5Cinfty%20x%5E2f%28x%29%5C%2C%5Cmathrm%20dx%3D%5Clambda%5Cint_0%5E%5Cinfty%20x%5E2e%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx)
Integrate by parts, taking
![u=x^2\implies\mathrm du=2x\,\mathrm dx](https://tex.z-dn.net/?f=u%3Dx%5E2%5Cimplies%5Cmathrm%20du%3D2x%5C%2C%5Cmathrm%20dx)
![\mathrm dv=e^{-\lambda x}\,\mathrm dx\implies v=-\dfrac{e^{-\lambda x}}\lambda](https://tex.z-dn.net/?f=%5Cmathrm%20dv%3De%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx%5Cimplies%20v%3D-%5Cdfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%5Clambda)
so that
![\displaystyle E[X^2]=\lambda\left(-\frac{x^2e^{-\lambda x}}\lambda\bigg|_0^\infty+2\int_0^\infty\frac{xe^{-\lambda x}}\lambda\,\mathrm dx\right)=2\int_0^\infty xe^{-\lambda x}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20E%5BX%5E2%5D%3D%5Clambda%5Cleft%28-%5Cfrac%7Bx%5E2e%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5Cbigg%7C_0%5E%5Cinfty%2B2%5Cint_0%5E%5Cinfty%5Cfrac%7Bxe%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5C%2C%5Cmathrm%20dx%5Cright%29%3D2%5Cint_0%5E%5Cinfty%20xe%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx)
Integrate by parts again, this time with
![u=x\implies\mathrm du=\mathrm dx](https://tex.z-dn.net/?f=u%3Dx%5Cimplies%5Cmathrm%20du%3D%5Cmathrm%20dx)
![\mathrm dv=e^{-\lambda x}\,\mathrm dx\implies v=-\dfrac{e^{-\lambda x}}\lambda](https://tex.z-dn.net/?f=%5Cmathrm%20dv%3De%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx%5Cimplies%20v%3D-%5Cdfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%5Clambda)
![\displaystyle E[X^2]=2\left(-\frac{xe^{-\lambda x}}\lambda\bigg|_0^\infty+\int_0^\infty\frac{e^{-\lambda x}}\lambda\,\mathrm dx\right)=\frac2\lambda\int_0^\infty e^{-\lambda x}\,\mathrm dx=-\frac2{\lambda^2}e^{-\lambda x}\bigg|_0^\infty=\frac2{\lambda^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20E%5BX%5E2%5D%3D2%5Cleft%28-%5Cfrac%7Bxe%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5Cbigg%7C_0%5E%5Cinfty%2B%5Cint_0%5E%5Cinfty%5Cfrac%7Be%5E%7B-%5Clambda%20x%7D%7D%5Clambda%5C%2C%5Cmathrm%20dx%5Cright%29%3D%5Cfrac2%5Clambda%5Cint_0%5E%5Cinfty%20e%5E%7B-%5Clambda%20x%7D%5C%2C%5Cmathrm%20dx%3D-%5Cfrac2%7B%5Clambda%5E2%7De%5E%7B-%5Clambda%20x%7D%5Cbigg%7C_0%5E%5Cinfty%3D%5Cfrac2%7B%5Clambda%5E2%7D)
and so the variance is
.
Given
two angles (150° and 12°) and the shortest side (12 cm) of a triangle
Find
the second-shortest side
Solution
Strategy: Make use of the Law of Sines to find missing side lengths when angles are known. To find the third angle, make use of the sum of angles of a triangle.
The sum of angles of a triangle is 180°, so we have
... 150° + 12° + C = 180°
... C = 180° - 162° = 18°
The law of sines tells us
... c/sin(C) = b/sin(B)
... c = sin(C)·b/sin(B) = sin(18°)·(10 cm)/sin(12°)
... c ≈ 14.9 cm
_____
We are calling the sides a, b, c. We are calling the angles opposite those sides A, B, and C.
I believe the answer is Height
Answer:
Now Myra has 60 lines of code, no grid picture attached.
Step-by-step explanation:
If Myra removed 20% of the lines and she started with 75 lines of code, then she is removing 1/5 of those lines of code.
75 divided by 5 is 15, and 75 - 15 = 60
Now Myra would have 60 lines of code. It's unclear what grid you are talking about though - could you attach a picture of the grid you need to shade in?
Answer:
<em>720</em>
<em>12</em>
<em>120</em>
Step-by-step explanation:
<u>Factorials</u>
We must recall that the factorial of a number n (positive or zero) is the product of all the integers from n down to 1
![n!= n(n-1)(n-2)...1](https://tex.z-dn.net/?f=n%21%3D%20n%28n-1%29%28n-2%29...1)
Let's evaluate the given expressions
![6!=6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1=720](https://tex.z-dn.net/?f=6%21%3D6%5Ccdot%205%5Ccdot%204%5Ccdot%203%5Ccdot%202%5Ccdot%201%3D720)
Similarly
![3!\cdot 2!=(3\cdot 2\cdot 1)\cdot (2\cdot 1)=6\cdot 2=12](https://tex.z-dn.net/?f=3%21%5Ccdot%202%21%3D%283%5Ccdot%202%5Ccdot%201%29%5Ccdot%20%282%5Ccdot%201%29%3D6%5Ccdot%202%3D12)
Finally
![\displaystyle \frac{6!}{3!}=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6%21%7D%7B3%21%7D%3D%5Cfrac%7B6%5Ccdot%205%5Ccdot%204%5Ccdot%203%5Ccdot%202%5Ccdot%201%7D%7B3%5Ccdot%202%5Ccdot%201%7D)
![\displaystyle \frac{6!}{3!}= \frac{720}{6}=120](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6%21%7D%7B3%21%7D%3D%20%5Cfrac%7B720%7D%7B6%7D%3D120)