A. C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.C(13, 10) = 13! = 13·12·11 = 13 · 2 · 11 = 286.
B. P(13,10)= 13! =13! =13·12·11·10·9·8·7·6·5·4.
(13−10)! 3!
C. f there is exactly one woman chosen, this is possible in C(10, 9)C(3, 1) =
10! 3!
9!1! 1!2!
10! 3!
8!2! 2!1!
10! 3!
7!3! 3!0!
= 10 · 3 = 30 ways; two women chosen — in C(10,8)C(3,2) =
= 45·3 = 135 ways; three women chosen — in C(10, 7)C(3, 3) =
= 10·9·8 ·1 = 120 ways. Altogether there are 30+135+120 = 285
1·2·3
<span>possible choices.</span><span>
</span>
Answer:
Problem 10:
Input X: -2, -1, 0, 1, 2
Output Y: -2, 1, 2, 1, -2
Nonlinear, exponential because there's an exponent. Plug the X values in place of X and solve solution to get Y values. Then graph
Problem 11: Linear because there's no exponent
