Answer:
Δ MNO ≅ ΔXYZ by AAS postulate.
Step-by-step explanation:
Given:
, 
, and YO = NZ.
Consider YO = NZ
Add OZ on both sides
YO + OZ = NZ + OZ
YZ = NO
Consider triangles Δ MNO and ΔXYZ.
Statement Reason
1. Given
2. Given
3. YZ = NO ∵ YO = NZ
Therefore, the two triangles Δ MNO and ΔXYZ are congruent to each other from AAS postulate as two corresponding angles and a corresponding side are equal to each other.
It is written/spoken like so:
"Twelve million three hundred and twenty four thousand nine hundred and four"
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
Multiply the $0.95 by .06 or if you have a percentage button then do $0.95 by 60% and you’ll get your answer of 0.57
