There are 6+5 marbles.
Of those, 5 are red.
The chances of drawing a red marble are 5/11
Since you replace the marble you drew, the chances each draw remain 5/11
You cannot draw two red marbles without drawing a red marble on your first draw, so the odds of you making a second draw are 5/11. The odds of that also being red are 5/11… but we only care about that 5/11ths the time
5/11*5/11=25/121
Convert fifths into tenths: every one fifth equals two tenths
4*2=8...8/10 + 1/10=9/10
9/10-10/10=1/10
He has 1/10 left to paint
The answer is 7/9
5/9 + 2/9 = 7/9
brainliest?
The answer for this question is 4^3=64
This problem can be converted into a linear algebra problem. The condition is that if the derminant below is not zero, then the system has one solution.
| k 1 1 |
| 1 k 1 | = k^3 - 3k + 2 = 0
| 1 1 k |
Solving for the roots, k = -2, and k = 1.
When k = 1, the three equations are the same so there are infinite solutions.
When k = -2, there are no solutions.
When k /= -2 and k /= 1, there is one solution.
