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3 years ago

A water pipe has a circular cross section of radius 0.75 cm. Water flows through the pipe at a rate of 16 cm/s.

Mathematics

1 answer:

Alexeev081 [22]3 years ago

8 0

Step-by-step explanation:

volume of water flows through the pipe per second= π r² h = 22/7 × ( 3/4)² × 16

= 22/7 × 9/16 × 16

= 22/7 × 9

= 3.14 × 9 = 28.26 cm ³= 28.26/ 1000= 0.02826 litre

0.02826litre = 1 second

1 litre = 1 / 0.02826 = 100000/2826

= 35 .38 seconds

answer = 35.38 seconds

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For this case we can use the moment generating function for the normal model given by:

$\phi(t) = E[e^{tX}]$

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

$\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx$

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And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

$E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...$

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And then we can find the $E[X^2]$

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And we can find the variance like this :

$Var(X^2) = E[X^4]-[E(X^2)]^2$

And first we find:

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And then the variance is given by:

$Var(X^2)= 3-(1)^2 =2$

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