At the beginning you have 10 balls, 5 of which are red. So, the chance of picking a red ball with the first pick is
![\dfrac{5}{10}=\dfrac{1}{2}](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B10%7D%3D%5Cdfrac%7B1%7D%7B2%7D)
When it comes to the second pick, you have 9 balls (because you didn't replace the first red ball), 3 of which are blue. So, you pick a blue ball with the second pick with probability
![\dfrac{3}{9}=\dfrac{1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7B3%7D%7B9%7D%3D%5Cdfrac%7B1%7D%7B3%7D)
The two events will occur one after the other with the product of the probabilities:
![\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%5Cdfrac%7B1%7D%7B3%7D%3D%5Cdfrac%7B1%7D%7B6%7D)
Answer:
w = -16
Step-by-step explanation:
w + (-2) = -18
Add 2 to each side
w-2+2 = -18+2
w = -16
Answer:
x=19
Step-by-step explanation:
Those two angles are vertical so they are congruent, equal the same thing. 8*19+2 = 154.
The answer is B, 5/6, because when you multiply 4 by 5 it is 20, and when you multiply 5 by 6, you get 30, so 20/30. And, 20/30=2/3. I hope I answered your question!
Answer:
The price of 1 adult ticket is 12 dollars, and the price of a ticket for one student is 7 dollars
Step-by-step explanation:
Make a system of equations for the two days that the play was shown.
Let x = the price of an adult ticket
Let y = the price of a student ticket
For the first day:
<h3>9x+8y=164</h3>
For the second day:
<h3>2x+7y=73</h3>
Now, we can solve using the elimination method. Multiply the first equation by 2 and the second equation by 9. Then swap the order of the equations.
<h3>18x+63y= 657</h3><h3>-</h3><h3>18x+16y= 328</h3><h3>0x+ 47y= 329</h3><h3>divide both sides by 47</h3><h3>y = 7</h3><h3>Plug in 7 for y for the 2nd equation</h3><h3>2x+7(7)=73</h3><h3>2x+49=73</h3><h3>subtract 49 from both sides</h3><h3>2x= 24</h3><h3>divide both sides by 2</h3><h3>x = 12 </h3><h3>Check:</h3><h3>2(12)+7(7)=73</h3><h3>24+49= 73!</h3>