What is the median of the data set 3 35 23 37 45 5 49 27 48?

1 answer:

7 0

First u put the numbers in order.....very important to do this
3 , 5 , 23, 27, 35, 37, 45, 48, 49

the median is the middle number. So the median is 35.

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Complete Question

The complete question is shown on the first uploaded image

Answer:

The solution is $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

Step-by-step explanation:

From the question we are told that

$x = e^{-2a} cos (5v)$

and $y = e^{-2a} sin(5v)$

Generally the absolute value of the determinant of the Jacobian for this change of coordinates is mathematically evaluated as

$| \frac{\delta (x,y)}{\delta (u, v)} | = | \ det \left[\begin{array}{ccc}{\frac{\delta x}{\delta u} }&{\frac{\delta x}{\delta v} }\\\frac{\delta y}{\delta u}&\frac{\delta y}{\delta v}\end{array}\right] |$

$= |\ det\ \left[\begin{array}{ccc}{-2e^{-2u} cos(5v)}&{-5e^{-2u} sin(5v)}\\{-2e^{-2u} sin(5v)}&{-2e^{-2u} cos(5v)}\end{array}\right] |$

$Let \ a = -2e^{-2u} cos(5v), \\ b=-2e^{-2u} sin(5v),\\c =-2e^{-2u} sin(5v),\\d=-2e^{-2u} cos(5v)$

$\frac{\delta (x,y)}{\delta (u, v)} | = |det \left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right] |$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | a * b - c* d |$

substituting for a, b, c,d

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 (e^{-2u})^2 cos^2 (5v) - 10 e^{-4u} sin^2(5v)|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = | -10 e^{-4u} (cos^2 (5v) + sin^2 (5v))|$

=> $\frac{\delta (x,y)}{\delta (u, v)} | = 10e^{-4u}$

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