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:
t > 1380
Step-by-step explanation:
Subtract 7200 from both sides
10t + 7200 - 7200 > 21000 - 7200
Simplify
10t > 13800
Divide both sides by 10
10t/10 > 13800/10
Simplify
t > 1380
I believe the answer is -33
The product would expand to
This is a trinomial, and the only way to make it a binomial is to cancel out a coefficient using our variable .
So, we can cancel either the linear term or the constant term.
In the first case, we require
In the second case, we require
But must be a non-zero rational number, so this solution is not feasible.
C, they are the output values and output= ranget