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kirill115 [55]
3 years ago
12

A large serving of soup from a take-out restaurant is 4/5 liter. The restaurant has 3 liters of soup. The diagram below models t

he number of servings of soup.
How many serving of soup does the restaurant have?

A. 2 2/5

B. 3 3/4

C. 3 3/5

D. 2 1/5

Mathematics
1 answer:
Airida [17]3 years ago
5 0

Answer:

The quantity of total serving soup does restaurant has = T = 2 \dfrac{2}{5} liters .

Step-by-step explanation:

Given as :

The quantity of large serving soup = \dfrac{4}{5}  liters

The total quantity of soup does restaurant has = 3 liters

Let the quantity of total serving soup does restaurant has = T liters

So, According to question

The quantity of total serving soup does restaurant has = The quantity of large serving soup × total quantity of soup does restaurant has

Or, T =  \dfrac{4}{5} × 3

Or, T =  \frac{4\times 3}{5}

Or, T =  \dfrac{12}{5}

∴   T = 2 \dfrac{2}{5} liters

So, quantity of total serving soup does restaurant has = T = 2 \dfrac{2}{5} liters

Hence,The quantity of total serving soup does restaurant has = T = 2 \dfrac{2}{5} liters . Answer

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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

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)

1=dC÷(2πdr)

dr=dC÷(2π)

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

dA=4rdC

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

dA=4dC×(C÷2π)

dA=(2CdC)÷π

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

3 0
2 years ago
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