First we have to find the mean (average)
mean = (564 + 1000 + 848 + 1495 + 1348) / 5 = 5255 / 5 = 1051

now we subtract the mean from every data point, then square it
564 - 1051 = -487......-487^2 = 237169
1000 - 1051 = -51......-51^2 = 2601
848 - 1051 = -203......-203^2 = 41209
1495 - 1051 = 444......444^2 = 197136
1348 - 1051 = 297......297^2 = 88209

now find the mean of the results.....but know when ur dealing with a sample instead of the whole population, u divide by 1 number less...so instead of dividing by 5, u divide by 4.
(237169 + 2601 + 41209 + 197136 + 88209) / 4 = 566324 / 4 =
141581.....this is called ur variance

now take the square root of the variance and u have ur standard deviation
sqrt (141581) = 376.272 rounds to 376.27 <==

5 0

3 years ago

The sum of two integers is 121. The larger integer is 7 more than twice the smaller integer. Find the two integers.

iren [92.7K]

X + (2x + 7) = 121 : Write the expression.
3x + 7 = 121 : Combine like terms.
3x + 7 - 7 = 121 - 7 : Isolate x.
3x = 114 : Combined like terms.
3x / 3 = 114 / 3 : Find the value of x.
x = 38 : This is the smaller integer.
(2x + 7) = (2(38) + 7) = 76 + 7 = 83. : Find the value of the larger number by plugging in the value of x to the larger number's quantity.
The larger number is 83, and the smaller number is 38.

5 0

3 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

$=\displaystyle36^2\int_0^{\pi/2}\int_0^{2\pi}\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}$

6 0

3 years ago