You take the number of possible 13 card hands with no 9 in them and subtract that from the total number of possible 13 card hands 9's included.
so C(52,13) - C(48,13) The number of all possible 13 card hands is: 52!/13!(52-13)! or 52!/13!*39! which is 635,013,559,600
The number of all possible 13 card hands with no 9s is:
48!/13!(48-13)! or 48!/13!*35! = 192,928,249,296
The difference is 635,013,559,600 - 192,928,249,296 = 442,085,310,304
So 442,085,310,304 out of 635,013,559,600 hands will have at least 1 nine. The ratio of 442,085,310,304 to 635,013,559,600 is 0.696182472... So roughly 70% of the time
27,630,331,894/39,688,347,475 is the best I could do for a whole number ratio.
Note: I'm afraid I can't put the work for the answers, for lack of time =/ ============================================================
1. 0 or -3
2. 0 or 4
3. -2 or 3
4. 0 or 1/2
5. -2 or 3
6. 0 or 5
7. 0 or 7
8. 0 or -2 ============================================================
9. Substitute 0 for h in "h = -16x^2 + 16x" then solve for x 0 = -16x^2 + 16x
<span>Subtract -16x^2+16x from both sides. </span><span>0−<span>(<span><span>−<span>16x2</span></span>+16x</span>)</span></span>=<span><span><span>−<span>16<span>x2</span></span></span>+<span>16x</span></span>−<span>(<span><span>−<span>16x2</span></span>+16x</span>) </span></span><span><span>16<span>x2</span></span>−<span>16x</span></span>=<span>0 </span> Factor the left side of the equation <span><span>16x</span><span>(<span>x−1</span>)</span></span>=<span>0 </span> Set factors to equal 0 <span><span>16x</span>=<span><span><span>0<span> or </span></span>x</span>−1</span></span>=<span>0 </span>x = 0 or x = 1