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: D. (8, 4)
Step-by-step explanation:
Answer:
x =7
Step-by-step explanation:
Solve for x:
5 x - 10 + 65 = 90
Add like terms. 65 - 10 = 55:
5 x + 55 = 90
Subtract 55 from both sides:
5 x + (55 - 55) = 90 - 55
55 - 55 = 0:
5 x = 90 - 55
90 - 55 = 35:
5 x = 35
Divide both sides of 5 x = 35 by 5:
(5 x)/5 = 35/5
5/5 = 1:
x = 35/5
The gcd of 35 and 5 is 5, so 35/5 = (5×7)/(5×1) = 5/5×7 = 7:
Answer: x = 7
Answer:
8
Step-by-step explanation:
2 times 4 is 8
Answer:
$600
Step-by-step explanation:
Omar spent $600.
Essentially, we divide his total number of points:
1800
by the amount earned
300
=
6
we take that and multiply by 100 (our amount needed to be spent to earn points) and boom, problem solved.
