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What is the volume of a sphere with a radius of 7mm

1) The answer, in terms of "π", is:
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"  ( $\frac{1372 mm^3}{3}$ ) * π mm³ " .
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2) The answer, using "3.14" for "π", is:
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" 1436.0266666666666667 mm³ " ; round to: " 1436.027 mm³ " ;

or; write as: " 1436 $\frac{2}{75}$ mm³ " .
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Explanation:
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The formula for the volume, "V" of a sphere is:

V = $\frac{4}{3}$ * π* r³ ;

in which "r" = radius = "7 mm" {given}.

We can solve using "3.14" for "π" ; or we can solve in terms of "π" ;
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1) In terms of "π" :
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V = $\frac{4}{3}$ * π * r³ ;

= $\frac{4}{3}$ * π * (7 mm)³ ;

= $\frac{4}{3}$ * π * (7 * 7 * 7 * mm³) ;

= $\frac{4}{3}$ * π * (49 * 7 * mm³) ;

= $\frac{4}{3}$ * π * (343 mm³) ;

= $\frac{4* 343 mm^3}{3}$ ) * π mm³

= ( $\frac{1372 mm^3}{3}$ ) * π mm³ ;

________________________________________________________
2) Using "3.14" for "π" :
________________________________________________________
    V = $\frac{4}{3}$ * π * r³ ;

= $\frac{4}{3}$ * (3.14)* (7 mm)³ ;

    = $\frac{4}{3}$ * (3.14) * (7 * 7 * 7 * mm³) ;

= $\frac{4}{3}$ * (3.14) * (49 * 7 * mm³) ;

= $\frac{4}{3}$ * (3.14) * (49 * 7 * mm³) ;

= $\frac{4}{3}$ * (3.14) * (343 mm³) ;

= { $\frac{4}{3}$ } * (3.14 * 343 mm³) ;

= { $\frac{4}{3}$ } * (1077.02 mm³) ;

= (4 * 1077.02 mm³) / 3 ;

= (4308.08 mm³) / 3 ;
__________________________________________________________
= 1436.0266666666666667 mm³ ; round to: " 1436.027 mm³ " ;

or; write as: " 1436 $\frac{2}{75}$ mm³ " .
__________________________________________________________

6 0

4 years ago

Suppose theta(????) measures the minimum angle between a clock’s minute and hour hands in radians. What is theta′(????) at 4 o’c

makvit [3.9K]

Answer:

$\frac{\pi }{30}$ radians per minute.

Step-by-step explanation:

In order to solve the problem you can use the fact that the angle in radians of a circumference is 2π rad.

The clock can be seen as a circumference divided in 12 equal pieces (because of the hour divisions). Each portion is $\frac{1}{12}$

So, you have to calculate the angle between each consecutive hour (Let ∅ represent it). It can be calculated dividing the angle of the entire circumference by 12.

∅= $\frac{2\pi }{12} = \frac{\pi }{6} rad$

Now, you have to find how many pieces of the circumference are between 12 and 4 to calculate the angle (Because 4 o'clock is when the minute hand is in 12 and the hour hand is in 4)

There are 4 portions from 12 to 4, so the angle (Let α represent it) is:

α= $(4)\frac{\pi }{6} = \frac{2\pi }{3}$

But the answer is asked in radians per minute. So you have to divide the angle by the amount of minutes between the hands of the clock at 4 o'clock.

There are 60 divisions in a clock for representing minutes, therefore in every portion there are:

$\frac{60}{12} = 5$ minutes

So, from the 12 mark to the 4 mark there are 20 minutes

The angle per minute is:

α= $\frac{2\pi/3 }{20} = \frac{2\pi }{(20)(3)} = \frac{\pi }{30}$ rad/min

Notice that the minimum angle is the angle mesured clockwise.

4 0

3 years ago

In order to circumscribe a circle on a triangle, which line must you construct?

SIZIF [17.4K]

<span>D) perpendicular bisector <em>I believe.
</em></span>

6 0

4 years ago

Let X have the uniform distribution U(0, 2) and let the conditional distribution of Y , given that X = x, be U(0, x). Find the j

Elina [12.6K]

Answer:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

Step-by-step explanation:

We have two random variables X and Y. $X \sim Unif(0,2)$ and given that X=x, Y has uniform distribution (0,x)

From the definition of the uniform distribution we have the densities for each random variable given by:

$f_X (x) =\frac{1}{2} , 0\leq x\leq 2$

$f_{Y|X} (y|x) = \frac{1}{x}, 0\leq y \leq x$

And on this case we can find the joint density with the following formula:

$f(x,y) = f_{Y|X}(y|x) f_X (x)$

And multiplying the densities we got this:

$f(x,y) = \frac{1}{x} \frac{1}{2}= \frac{1}{2x} , 0\leq x \leq 2 , 0\leq y \leq x$

Now with the joint density we can find the expected value E(Y|x) with the following formula:

$E(Y|x) = \int y f_{Y|X}(y|x)dx$

And replacing we got:

$E(Y|x) = \int_{x=y}^2 y \frac{1}{x} dx= y ln x \Big|_{x=y}^2 =y ln 2 -y ln y = y(1-lny) \$

5 0

4 years ago

A badge costs 78p.

Inga [223]

Sam would have 41 pence change

6 0

3 years ago

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