I believe A and B are the ones. As for explaining? hmm. I would honestly say that any number to the right of a number is a tenth of the place above it. But then again, it's optional to explain ;P
notice, the "common ratio" is -3, thus their sum is just
1 + (-3) + 9 + (-27) + 81.
Answer:
14 ft
Step-by-step explanation:
a² + b² = c²
a=5
b= unknown
c= hypotnuse (13)
5² =25
13²=169
subtract 25 from 169 to get 144. Then take the square root of 144.
You should get 12. Dont forget to add the extra length
14 is the answer
Answer:
The answer is D
Step-by-step explanation:
Each blue dot over the number represents that number one time. There is one bot over 145 so there is only one 145. There are 5 dots over 165, so you write 165 five times and so on.
Answer:
1716 ;
700 ;
1715 ;
658 ;
1254 ;
792
Step-by-step explanation:
Given that :
Number of members (n) = 13
a. How many ways can a group of seven be chosen to work on a project?
13C7:
Recall :
nCr = n! ÷ (n-r)! r!
13C7 = 13! ÷ (13 - 7)!7!
= 13! ÷ 6! 7!
(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)
1235520 / 720
= 1716
b. Suppose seven team members are women and six are men.
Men = 6 ; women = 7
(i) How many groups of seven can be chosen that contain four women and three men?
(7C4) * (6C3)
Using calculator :
7C4 = 35
6C3 = 20
(35 * 20) = 700
(ii) How many groups of seven can be chosen that contain at least one man?
13C7 - 7C7
7C7 = only women
13C7 = 1716
7C7 = 1
1716 - 1 = 1715
(iii) How many groups of seven can be chosen that contain at most three women?
(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)
Using calculator :
(15 * 35) + (6 * 21) + (1 * 7)
525 + 126 + 7
= 658
c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?
(First in second out) + (second in first out) + (both out)
13 - 2 = 11
11C6 + 11C6 + 11C7
Using calculator :
462 + 462 + 330
= 1254
d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?
Number of ways with both in the group = 11C5
Number of ways with both out of the group = 11C7
11C5 + 11C7
462 + 330
= 792
