Membres: Alan

$${\left({\frac{{\mathtt{12}}{!}}{{\mathtt{8}}{!}{\mathtt{\,\times\,}}({\mathtt{12}}{\mathtt{\,-\,}}{\mathtt{8}}){!}}}\right)} = {\mathtt{495}}$$ This is Math(Input=Result)

Can't see how Math(Input) works!

Alan13 avr. 2014

+33661

CPhill:
I'm not a probability "guru," but I believe that we are just counting "sets" of things. (BTW...the marbles could have hit the ground one by one, but maybe not....maybe 6 of them (or any subset of such) struck the ground almost simultaneously !!!........ )

So, the total number of "sets" that we could have (the sample space) is just C(19, 6).
And we're interested in choosing 4 of the 6 blue ones and the other 2 out of the remaining 13.

So we have

C(6 , 4) * C( 13, 2) / C(19, 6)

Which I believe is the same thing that Melody found, so I won't bother with any computations !!!

Anyone else have any thoughts on this one ???

Yes, Melody is right. The probability of selecting a blue one changes each time a smartie is removed from the bag. Sometimes both the number of blues and the total number reduces; sometimes the number of blues stays the same and just the total reduces. Rom's method assumes the probability, p, is constant.

By the way, to answer Melody's query about using combination notation here (though it still appears in factorial form in the black bar):

nCr(6,2)*nCr(13,2)/nCr(19,6)

Alan12 avr. 2014

+33661

Qualababy18:
what is the definition of csc?

The cosecant is 1/sin

Alan11 avr. 2014

+33661

Qualababy18:
if the point (3/4,4/5) corresponds to an angle in the unit circle what is sec?

I think you must mean the point (3/5, 4/5) as (3/4, 4/5) is not on the unit circle. The distance of this point from the origin is 1 by definition of "unit circle". If the angle you are talking about is that between the radial-line-from-the-origin-to-this-point and the x-axis, then the secant (sec) of this angle, which is 1/cos of this angle, is given by 1/xvalue = 1/(3/5) or 5/3.

Alan11 avr. 2014

+33661

Qualababy18:
if the point (2/3,y) is on the unit circle, what is y?

The equation for the unit circle is x ² + y ² = 1, so if x = 2/3, we have (2/3) ² + y ² = 1. We can rearrange this to get y ² = 1-(2/3) ²,, by taking the square root of both sides we can find y (there will be two values)

solve(y^2=1-(2/3)^2)

Alan11 avr. 2014

#13

+33661

DavidQD:

...
By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).
...
~~D~~

Unfortunately, the original question doesn't contain the words "spherical segment", nor any other words that ensure the circle is entirely on the surface of the sphere, so it is possible to interpret it as an incomplete problem with an infinite number of solutions, depending on the assumptions made! However, it seems clear to me that the intention is, indeed, for the assumption to be that the circle lies completely on the surface of the sphere, if only because we get the very nice 5, 12, 13 right-angled triangle that way!

Alan11 avr. 2014

+33661

Melody:
...

. - or am I missing something?

I think so! See my reply immediately above yours.

A closer reading of the original shows that the circle is not specified to be wholly on the sphere, in which case Melody and CPhill are correct that the problem is indeterminate without further information. However, I doubt that the problem would have been set without the specification that the circle lies completely on the sphere.

Alan10 avr. 2014

+33661

CPhill:
kayla121497 wrote:point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

Mmmmm....I don't know about this.....here's a cross-section of a sphere where A is the center, C lies on the sphere's surface and AB =5, BC = 12 and BC lies in Plane "P"
(I think all of these meet the requirements of the problem).

Circle.JPG

Note that the radius of the sphere is ≈ 10 (At least as close to 10 as I could make it!!)

Maybe I'm reading too much into this problem, but I claim that we don't actually know that BC is at a right angle to AB

I guess my point is that we could have a right triangle, but as you see, it might not have to be. Note that ∠BAC is greater than a right angle

In hindsight, the problem only "works" if we assume that AB is perpendicular to BC.......and since we're not given any other information, then DavidQD and Alan would certainly be correct !!!

You need to complete the circle centred on B. You can see from the picture below (drawn to scale in Geogebra) that a circle based on BC as radius (as constructed by CPhill above) would not be wholly on the sphere. (BE is the extension of BC such that CE is the diameter of the circle)
sphere2.PNG

Alan10 avr. 2014

+33661

kayla121497:
point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

AB (5mm) and BC (12mm) are two sides of a right-angled triangle of which the radius of the sphere is the hypotenuse. The triangle is the well -known 5, 12, 13 triangle, so the radius of the sphere is 13mm. The volume follows from V = (4/3)*pi*r ³ as in DavidQD's reply above.

See the picture below.
Sphere.PNG

Alan10 avr. 2014

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