14/99
Select 1 marble; the chance that it is white is 4/12. Select a 2nd marble; the chance that it is white is 3/11. Select a 3rd; the chance it is white is 2/10. Select a 4th; the chance it is red is 8/9. Select a 5th; the chance it is red is 7/8. The chance of getting this specific set of 5 marbles in this order is (4/12)×(3/11)×(2/10)×(8/9)×(7/8)=(2×7)/(11×10×9).
This specific set could occur in the permutation of 5 things taken 5 at a time where 3 are identical (white), and the other 2 are also identical (red). The formula for this is 5!/(3!2!)=10.
Combining the chance of getting white, white, white, red, red with the number of ways 3 white and 2 red could have been distributed in the draw of 5 marbles gives the answer:
{(2×7)/(11×10×9)}×10=14/99
A similar process will show that the chance of getting 5 red marbles is 7/99; 4 white and 1 red is 1/99; 2 white and 3 red is 42/99; and 1 white and 4 red is 35/99.
Answer:
295000
Step-by-step explanation:
It would 295,000
Answer: D
Step by Step: Even though the y value repeats, to be considered not a function is when the X value repeats
Answer:
Step-by-step explanation:
![\sf 3(t+3)+5(3+2t)-76 = 0\\\\Expanding \ Parenthesis\\\\3t + 9 + 15 + 10t -76 = 0\\\\13t + 24 -76 = 0\\\\13t - 52 = 0\\\\Add \ 52 \ to \ both \ sides\\\\13t = 52\\\\Dividing\ both\ sides\ by\ 13\\\\t = 52/13\\\\t = 4\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%203%28t%2B3%29%2B5%283%2B2t%29-76%20%3D%200%5C%5C%5C%5CExpanding%20%5C%20Parenthesis%5C%5C%5C%5C3t%20%2B%209%20%2B%2015%20%2B%2010t%20-76%20%3D%200%5C%5C%5C%5C13t%20%2B%2024%20-76%20%3D%200%5C%5C%5C%5C13t%20-%2052%20%3D%200%5C%5C%5C%5CAdd%20%5C%2052%20%5C%20to%20%5C%20both%20%5C%20sides%5C%5C%5C%5C13t%20%3D%2052%5C%5C%5C%5CDividing%5C%20both%5C%20sides%5C%20by%5C%2013%5C%5C%5C%5Ct%20%3D%2052%2F13%5C%5C%5C%5Ct%20%3D%204%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
