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liberstina [14]
3 years ago
10

Nylan robotics has a budget of 99.00 to buy sprockets axles and gears. If they spend 2/3 of the budget on sprockets how much mon

ey from the budget remains to buy axles and gears
Mathematics
2 answers:
grin007 [14]3 years ago
8 0

99*1/3=33

so 2/3=66

so 33 dollars remain

insens350 [35]3 years ago
8 0

Answer:

$33.00

Step-by-step explanation:

Nylan robotics has a budget to buy sprockets axles and gears = $99.00

They spend \frac{2}{3} of the budget on sprockets = \frac{2}{3}\times 99

                                              = \frac{198}{3}

                                              = $66

They spend $66 on sprockets.

Now the money remains = 99 - 66 = $33.00

They remains $33.00 to buy axles and gears.

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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

dA=8πr×dr

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

dr÷dr=dC÷(2πdr)

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The maximum error in surface area is simplified as:

Substitute the value of dr in dA as

dA=8πr×(dC÷(2π))

Cancel π from both numerator and denominator and simplify it

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

dA=4dC×(C÷2π)

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Substitute this in above as

dA=(2×76×0.5)÷π

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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

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