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lesantik [10]
3 years ago
8

Each of the 25 student in mr mcdounld class sold 16 tickets if each ticket costs15$ how much money did the students raise

Mathematics
1 answer:
adoni [48]3 years ago
6 0
25 x 16 = 400
400 x 15 = 6000
they raised $6000

hope this helps you
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2/6, 3/9, 4/12, 5/15, 6/18 ect.
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The circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the ma
andreev551 [17]

The maximum error in the calculated surface area is 24.19cm² and the relative error is 0.0132.

Given that the circumference of a sphere is 76cm and error is 0.5cm.

The formula of the surface area of a sphere is A=4πr².

Differentiate both sides with respect to r and get

dA÷dr=2×4πr

dA÷dr=8πr

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The circumference of a sphere is C=2πr.

From above the find the value of r is

r=C÷(2π)

By using the error in circumference relation to error in radius by:

Differentiate both sides with respect to r as

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The maximum error in surface area is simplified as:

Substitute the value of dr in dA as

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Substitute the value of r=C÷(2π) in above and get

dA=4dC×(C÷2π)

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Here, C=76cm and dC=0.5cm.

Substitute this in above as

dA=(2×76×0.5)÷π

dA=76÷π

dA=24.19cm².

Find relative error as the relative error is between the value of the Area and the maximum error, therefore:

\begin{aligned}\frac{dA}{A}&=\frac{8\pi rdr}{4\pi r^2}\\ \frac{dA}{A}&=\frac{2dr}{r}\end

As above its found that r=C÷(2π) and r=dC÷(2π).

Substitute this in the above

\begin{aligned}\frac{dA}{A}&=\frac{\frac{2dC}{2\pi}}{\frac{C}{2\pi}}\\ &=\frac{2dC}{C}\\ &=\frac{2\times 0.5}{76}\\ &=0.0132\end

Hence, the maximum error in the calculated surface area with the circumference of a sphere was measured to be 76 cm with a possible error of 0.5 cm is 24.19cm² and the relative error is 0.0132.

Learn about relative error from here brainly.com/question/13106593

#SPJ4

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