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I'm not sure about this...but I'll take a stab at it

The probability of red or black = 18/38 = 9/19

And the probability of green = 2/38 = 1/19

So the probability of a ball landing in red 4 times = (9/19)^4

And the probability of a ball landing in black twice is (9/19)^2

And the probability of landing in green once is (1/19)

And note that we have 7! arrangements of  these results......but some are indistiguishable from others

So, the totlal distinguishable arrangements is  7! / [4! 2! ] = 105

So the total probability is given by.... 105*(9/19)^4*(9/19)^2*(1/19) = 0.0624265233650037 = about 6.2%

Nice , Thank you seven up 

$${\mathtt{9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{10}} = {\mathtt{19}}$$

This “planet” has a gravity field higher than Jupiter. His whole body will “feel” nearly three times heaver. Unless this man was built like Arnold Schwarzenegger, he probably wouldn’t be able to lift his head without breaking his neck.

_7UP_

Yep!                                

ok i know you guys are in

The expanding spring continues to impart energy to the ball as long as there is contact. At separation, it no longer accelerates so that is the maximum velocity. After this point, the ball begins to decelerate, until it reaches the peak, where it stops, and the ball begins to accelerate in the opposite direction.

If this were a closed system, where the energy was not lost to friction, this ball would “bounce” indefinitely. It is the same reason why planets orbit a sun for billions of years. This kind of system has miniscule friction and gravitational perturbations.

_7UP_

Hello!!!!!!!!!!

Thank you! :)

Stop with these questions. You know the answer.

Yeah...if c is the hypoteneuse....that would be correct....!!

I believe side C is the hypotenuse

Would this be correct?

sin(A)=16.9/21.6 
A=sin(^-1)(16.9/21.6) 
A=51.48 degrees. 

$${\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{8}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}} = {\mathtt{100}}$$

This equals..... [1 - probability of no kings ]

So, the ways to choose no kings is to just select 5 cards from the remaining 48

This is ginen by  C(48,5) = 1712304

And the total way of selecting 5 cards from 52 is C(52,5) = 2598960

So....the probability of at least one king is

1 - 1712304/2598960 = about 34.1%

A prime number is a whole-number that is greater than 1 and only two whole-number factors are 1 and itself.

2, 3, 5, 7, 11, 13, 17

I can't play it

-2=-2

2 jokers,  4 aces, 4 2's, 4 3's, 4 4's, 4 5's, 4 6's, 4 7's, 4 8's, 4 9's, 4 10's, 4 jacks, 4 queens, and 4 kings

I'm not good with probabiltiy...but in 54 cards....there are 4 kings,  so maybe 4/25 because their are 5 cards in the hand....like I said,  not good, giving my best try

What's up with all these probability problems?

I don't get this....

What should I do?  I'm done with my schoolwork for the entire day

Solve what???....we're just taking the cosine inverse to find  an angle whose cosine is = to the expression in the parentheses......

Thank you!

This is interesting.

I went to a science symposium where one of the lecturers spoke on cloning. The scientist explained why it is easier to clone a sheep, cow or a cat compared to other animals, like a dog or primates.

Looks like Puma really loves her kitten.

_7UP_

Mmmm...I see what Melody is saying.....I think that we would need to account for the other people, as well!!

For each arrangement of the three sitting together, there are 4! ways of arranging the other 4 people.

So, the total arrangements are 30*4! = 720

So

The probability of them not sitting together is

1 - 720/5040 = about 85.7%

I  may think on this one some more !!!

What would  x^6-y^6 be when completely factored?

$$\textcolor[rgb]{1,0,0}{
a^3 + b^3 = (a + b)(a^2-ab + b^2)
}\\
\textcolor[rgb]{1,0,0}{
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
} \\
(a^6-b^6)=(a^3 + b^3)*(a^3 - b^3 )=(a + b)(a^2-ab + b^2)(a - b)(a^2 + ab + b^2)$$

$$a=x \text{ and } b=y$$$$(x^6-y^6)=(x + y)(x^2-xy + y^2)(x - y)(x^2 + xy + y^2)$$