For starters, the probability of finding any of these colors is a 50/50 type of thing, or in this case, an equal quarter. So each would be 1/4, or 25%. Unless, of course, they're lying about having one or more colors of marbles in their possession. But I'm answering under the assumption that they're not.
bearing in mind that, on the III Quadrant, sine as well as cosine are both negative, and that hypotenuse is never negative, so, if the sine is -4/5, the negative number must be the numerator, so sin(x) = (-4)/5.
![\bf sin(x)=\cfrac{\stackrel{opposite}{-4}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2-(-4)^2}=a\implies \pm\sqrt{9}=a\implies \pm 3=a \\\\\\ \stackrel{III~Quadrant}{-3=a}~\hfill cos(x)=\cfrac{\stackrel{adjacent}{-3}}{\stackrel{hypotenuse}{5}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-4%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B5%5E2-%28-4%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B9%7D%3Da%5Cimplies%20%5Cpm%203%3Da%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7BIII~Quadrant%7D%7B-3%3Da%7D~%5Chfill%20cos%28x%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B-3%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B5%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf tan\left(\cfrac{\theta}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos(\theta)}{1+cos(\theta)}} \\\\ \cfrac{sin(\theta)}{1+cos(\theta)}\qquad \leftarrow \textit{let's use this one} \\\\ \cfrac{1-cos(\theta)}{sin(\theta)} \end{cases} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20tan%5Cleft%28%5Ccfrac%7B%5Ctheta%7D%7B2%7D%5Cright%29%3D%20%5Cbegin%7Bcases%7D%20%5Cpm%20%5Csqrt%7B%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Ctheta%29%7D%7B1%2Bcos%28%5Ctheta%29%7D%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Blet%27s%20use%20this%20one%7D%20%5C%5C%5C%5C%20%5Ccfrac%7B1-cos%28%5Ctheta%29%7D%7Bsin%28%5Ctheta%29%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

Answer:
y=2/3x - 6
Step-by-step explanation:
use the formula y=mx+b. the m of this equation is equal to the slope and you know that the slope is 2/3. the b of the equation is equal to the y-intercept so, to find b you simply look at the points given. the points are written in (x,y) form so if -14 is x, -6 must be y. remember to KEEP CHANGE CHANGE your signs so that plus sign becomes negative 6. Hope this helps!
Answer:
Step-by-step explanation:
y = 3x - 5
6x + 3y = 15
Substitute the first equation into the second
6x + 3(3x - 5) = 15
6x + 9x - 15 = 15
15x = 30
x = 2
Then use the answer in the first equation and solve
y = 3(2) - 5
y = 6 - 5
y = 1
There are 100 centimeters in one meter
therefore .1 meters = 10 centimeters
.01 meters = 1 centimeter
to convert form centimeters to meters you will move the decimal to the left 2 times starting from the far most right digit.