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3 years ago

9

What is the radius of a sphere with a volume of 4186 in3 to the nearest tenth of an inch?

1 answer:

Aleks04 [339]3 years ago

3 0

Answer:

10 inches

Step-by-step explanation:

r=(3V

4π)⅓=(3·4186

4·π)⅓≈9.99778in

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Find the standard deviation of

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Answer:

11.5

The sample standard deviation measures the spread of a data distribution of the sample. It is usually used to estimate the population standard deviation

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3 years ago

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

6 0

3 years ago

Need answer 25 and answer 27

UNO [17]

$\\ \bull\tt\dashrightarrow \dfrac{-3}{8}x-20+2x>6$

$\\ \bull\tt\dashrightarrow \dfrac{-3}{8}x+2x=6+20=26$

$\\ \bull\tt\dashrightarrow \dfrac{-3+16}{8}x=26$

$\\ \bull\tt\dashrightarrow \dfrac{13}{8}x=26$

$\\ \bull\tt\dashrightarrow x=26\times \dfrac{8}{13}$

$\\ \bull\tt\dashrightarrow x=8(2)=16$

#27

$\\ \bull\tt\dashrightarrow 0.5x-4-2x\leqslant 2$

$\\ \bull\tt\dashrightarrow 0.5x-2x\leqslant=2+4=6$

$\\ \bull\tt\dashrightarrow -1.5x\leqslant 6$

$\\ \bull\tt\dashrightarrow x\leqslant\dfrac{6}{-1.5}$

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3 years ago

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The answer would be a

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