Answer:
See explanation.
Step-by-step explanation:
We are looking at a geometric distribution.
The probability of selecting a brown peanut is .12 = p
The probability of not selecting a brown peanut is .88 = q
The probability mass function is p(y) = (.88)^(y-1) * (.12)
a) p(7) = (.88)^6 * .12 = .0557
b) p(7 <= y <= 8) = p(7) + p(8)
= .0557 + (.88)^7 * .12 = .1048
c) p(y <= 7) = p(0) + p(1) + ... + p(7)
= .12 + (.88)^1 * .12 + (.88)^2 * .12 + ... + (.88)^6 * .12 = .4713
d) The expect value is 1/p. So, 1/(.12) = 8.33 M&M's
Answer:
Your answer is absolutely correct
Step-by-step explanation:
The work would be as follows:
![\int _0^{\sqrt{\pi }}4x^3\cos \left(x^2\right)dx,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=> 4\cdot \int _0^{\sqrt{\pi }}x^3\cos \left(x^2\right)dx\\\\\mathrm{Apply\:u-substitution:}\:u=x^2\\=> 4\cdot \int _0^{\pi }\frac{u\cos \left(u\right)}{2}du\\\\\mathrm{Apply\:Integration\:By\:Parts:}\:u=u,\:v'=\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\int \sin \left(u\right)du\right]^{\pi }_0\\\\](https://tex.z-dn.net/?f=%5Cint%20_0%5E%7B%5Csqrt%7B%5Cpi%20%7D%7D4x%5E3%5Ccos%20%5Cleft%28x%5E2%5Cright%29dx%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%3E%204%5Ccdot%20%5Cint%20_0%5E%7B%5Csqrt%7B%5Cpi%20%7D%7Dx%5E3%5Ccos%20%5Cleft%28x%5E2%5Cright%29dx%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Au-substitution%3A%7D%5C%3Au%3Dx%5E2%5C%5C%3D%3E%204%5Ccdot%20%5Cint%20_0%5E%7B%5Cpi%20%7D%5Cfrac%7Bu%5Ccos%20%5Cleft%28u%5Cright%29%7D%7B2%7Ddu%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3AIntegration%5C%3ABy%5C%3AParts%3A%7D%5C%3Au%3Du%2C%5C%3Av%27%3D%5Ccos%20%5Cleft%28u%5Cright%29%5C%5C%3D%3E%204%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cint%20%5Csin%20%5Cleft%28u%5Cright%29du%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C)
![\int \sin \left(u\right)du=-\cos \left(u\right)\\=> 4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0\\\\\mathrm{Simplify\:}4\cdot \frac{1}{2}\left[u\sin \left(u\right)-\left(-\cos \left(u\right)\right)\right]^{\pi }_0:\quad 2\left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0\\\\\mathrm{Compute\:the\:boundaries}:\quad \left[u\sin \left(u\right)+\cos \left(u\right)\right]^{\pi }_0=-2\\=> 2(-2) = - 4](https://tex.z-dn.net/?f=%5Cint%20%5Csin%20%5Cleft%28u%5Cright%29du%3D-%5Ccos%20%5Cleft%28u%5Cright%29%5C%5C%3D%3E%204%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cleft%28-%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C%5Cmathrm%7BSimplify%5C%3A%7D4%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29-%5Cleft%28-%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%3A%5Cquad%202%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29%2B%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%5C%5C%5C%5C%5Cmathrm%7BCompute%5C%3Athe%5C%3Aboundaries%7D%3A%5Cquad%20%5Cleft%5Bu%5Csin%20%5Cleft%28u%5Cright%29%2B%5Ccos%20%5Cleft%28u%5Cright%29%5Cright%5D%5E%7B%5Cpi%20%7D_0%3D-2%5C%5C%3D%3E%202%28-2%29%20%3D%20-%204)
Hence proved that your solution is accurate.
Five to the power of 10 would be 9765625, so the scientific notation would be
9.765625 * (10^6), I think.
(I haven't done scientific notation in a long time, but I'm relatively sure that is what it is.)
Hope this kind of helped!
THE SUM IS 89+27+11=127
IF WE WILL USE THE DISTRIBUTIVE PROPERTY,
(89+11)+27=100+27=127
I AM NOT SURE IF YOU NEED TO FIND THE PRODUCT,
89*27*11=26,433
Answer:
Use shell method
And chose the cross section perpendicular to the x-axis
