Fast plz rkoxeooq be dvbdisksndv
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It looks like you want to compute the double integral
over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.
Convert to polar coordinates, in which <em>D</em> is given by the set
<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}
and
<em>x</em> = <em>r</em> cos(<em>θ</em>)
<em>y</em> = <em>r</em> sin(<em>θ</em>)
d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>
Then the integral is
So if x people were surveyed, 
people owned a bike and 
did not.
The difference between them is 72:
x-\frac{1}{8} [/tex] =72
so this means that
x=72
let's simplify:
x=72
and divide by 3:
x=24
and mupliply by 4:
x=96
so 96 people were surveyed
The difference between them is 72:
so this means that
let's simplify:
and divide by 3:
and mupliply by 4:
x=96
so 96 people were surveyed