Answer:
7890
Step-by-step explanation:
(5000*4%)+(x*5%)=594.50
200+(x*5%)=594.50
x*5%=394.50
x=7890
Answer:

Step-by-step explanation:
For the random variable
we define the possible values for this variable on this case
. We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable
like this:

For the distribution of
we need to take in count that we are finding a conditional distribution.
given
, for this case we see that
, so then exist
ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined:

And if we want to find the joint probability we just need to do this:

And if we multiply the probabilities founded we got:

Answer:
500, or 400 if its 4
Step-by-step explanation:
First, you'll need to find the marginal distributions of
. By the law of total probability,

which translates to

Similarly,

Compute the expectations for both random variables:
![E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5E12x%281-x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac13)
![E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\int_0^12y^2\,\mathrm dy=\frac23](https://tex.z-dn.net/?f=E%5BY%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%5C%2Cf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cint_0%5E12y%5E2%5C%2C%5Cmathrm%20dy%3D%5Cfrac23)
Compute the variances and thus standard deviations:
![V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5B%28X-E%5BX%5D%29%5E2%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
where
![E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20x%5E2%5C%2Cf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_0%5E12x%5E2%281-x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac16)
![\implies V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}\implies\sqrt{V[X]}=\dfrac1{3\sqrt2}](https://tex.z-dn.net/?f=%5Cimplies%20V%5BX%5D%3D%5Cdfrac16-%5Cleft%28%5Cdfrac13%5Cright%29%5E2%3D%5Cdfrac1%7B18%7D%5Cimplies%5Csqrt%7BV%5BX%5D%7D%3D%5Cdfrac1%7B3%5Csqrt2%7D)
![E[Y^2]=\displaystyle\int_{\infty}^\infty y^2f_Y(y)\,\mathrm dy=\int_0^12y^3\,\mathrm dy=\frac12](https://tex.z-dn.net/?f=E%5BY%5E2%5D%3D%5Cdisplaystyle%5Cint_%7B%5Cinfty%7D%5E%5Cinfty%20y%5E2f_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cint_0%5E12y%5E3%5C%2C%5Cmathrm%20dy%3D%5Cfrac12)
![\implies V[Y]=\dfrac12-\left(\dfrac23\right)^2=\dfrac1{18}\implies\sqrt{V[Y]}=\dfrac1{3\sqrt2}](https://tex.z-dn.net/?f=%5Cimplies%20V%5BY%5D%3D%5Cdfrac12-%5Cleft%28%5Cdfrac23%5Cright%29%5E2%3D%5Cdfrac1%7B18%7D%5Cimplies%5Csqrt%7BV%5BY%5D%7D%3D%5Cdfrac1%7B3%5Csqrt2%7D)
Compute the covariance:
![\operatorname{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]](https://tex.z-dn.net/?f=%5Coperatorname%7BCov%7D%5BX%2CY%5D%3DE%5B%28X-E%5BX%5D%29%28Y-E%5BY%5D%29%5D%3DE%5BXY%5D-E%5BX%5DE%5BY%5D)
We have
![E[XY]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^1\int_0^y2xy\,\mathrm dx\,\mathrm dy=\frac14](https://tex.z-dn.net/?f=E%5BXY%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xy%5C%2Cf_%7BX%2CY%7D%28x%2Cy%29%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D%5Cint_0%5E1%5Cint_0%5Ey2xy%5C%2C%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D%5Cfrac14)
and so
![\operatorname{Cov}[X,Y]=\dfrac14-\dfrac13\dfrac23=\dfrac1{36}](https://tex.z-dn.net/?f=%5Coperatorname%7BCov%7D%5BX%2CY%5D%3D%5Cdfrac14-%5Cdfrac13%5Cdfrac23%3D%5Cdfrac1%7B36%7D)
Finally, the correlation:
![\operatorname{Corr}[X,Y]=\dfrac{\operatorname{Cov}[X,Y]}{\sqrt{V[X]}\sqrt{V[Y]}}=\dfrac{\frac1{36}}{\left(\frac1{3\sqrt2}\right)^2}=\dfrac12](https://tex.z-dn.net/?f=%5Coperatorname%7BCorr%7D%5BX%2CY%5D%3D%5Cdfrac%7B%5Coperatorname%7BCov%7D%5BX%2CY%5D%7D%7B%5Csqrt%7BV%5BX%5D%7D%5Csqrt%7BV%5BY%5D%7D%7D%3D%5Cdfrac%7B%5Cfrac1%7B36%7D%7D%7B%5Cleft%28%5Cfrac1%7B3%5Csqrt2%7D%5Cright%29%5E2%7D%3D%5Cdfrac12)
<span>If the part needs to be 5.72 inches long, with a tolerance of 0,02 inches (I cannot read the symbol before 0.02, so will assume it's meant to be +/-), the part length must be in the range 5.70 to 5.74. If the part is now 5.85 inches, the most it can be shortened by is 0.15 inches.</span>