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- anonymous

If the radius of a sphere is measured as 7 m with an error of 0.02 m, then find the
approximate error in calculating its volume.

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- anonymous

- schrodinger

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- anonymous

Why would I help you?

- anonymous

The answer is 6

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- anonymous

you both look so sweet sid by side :) GOD Bless you...

- anonymous

Bye...

- anonymous

Thanks @chad

- anonymous

http://media.tumblr.com/tumblr_mdyb2b8jJu1rp7g3w.gif

- anonymous

is that you @becca ?

- anonymous

It's not me in the Gif, no.

- anonymous

lol

- anonymous

The approximate error would actually be 12 cubic meters.
~~~~~~~
Here's the formula for finding the volume of a sphere:
\[\frac{ 4 }{ 3 }\pi r^{3}\]
That's "four-thirds, times pi, times the radius CUBED."
(Not squared, even though the little 3 looks like a 2.)
~~~~~~~~~~~~~~
If we assume that the radius of the sphere is 7,
then the volume would be 1432.44 cubic meters.
But since the error is 0.02,
the ACTUAL radius of the sphere could be 7.02, or 6.98.
If we assume that the radius is 7.02,
then the volume would be 1444.76 cubic meters.
And if we assume that the radius is 6.98,
then the volume would be 1420.20 cubic meters.
~~~~~~~~~~~~~
The difference between 1444.76 and 1432.44 is 12.32.
And the difference between 1432.44 and 1420.20 is 12.24.
So you see,
the error is about 12.00 cubic meters, if we round it.

- anonymous

Damn you just wasted all your time writing that lol

- anonymous

Better than wasting my time on giving someone a wrong answer. Right?

- goformit100

- anonymous

He didnt really want an answer lol he was just trying to troll me and Becca

- goformit100

No...

- goformit100

- anonymous

@goformit100
"No"?

- anonymous

Depending on whether I choose to multiply pi with 1.33,
or to multiply pi with 1.33333,
my spherical volume could be off by anywhere from 3 to 5 meters.
But in the end, the error is still about 12 cubic meters.

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