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
Answer:
133°
Step-by-step explanation:
As vertically opposite angles are equal,
< B E D = < F E G
Therefore,
< B E D = 47°
We know that co - interior angles which occur between 2 parallel lines, are added up to 180°.
Given that, < A B E = x
Therefore,
< B E D + < A B E = 180 °
47 + x = 180
x = 180 - 47
x = 133°
Let me know if you have any other questions. :)
5/2
70 is the constant number that doesn’t change, but anytime you increase t you are raising it above 5/2. So every time you grow 5/2.
To solve the problem we could separate the figure into three parts. First figure is a triangle, second figure is a rectangle, third figure is a triangle. See image attached.
Solve each area of the figuresFirst figure, a triangle that have 7 units long of the base, and 2 units long of the height.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the first figure is 7 units²
Second figure is a rectangle, the length of the rectangle is 7 units, the width of the rectangle is 4 units.
a = l × w
a = 7 × 4
a = 28
The area of the second figure is 28 units²
Third figure is a triangle, the base is 7 units long and the height is 2 units long.
a = 1/2 × b × h
a = 1/2 × 7 × 2
a = 14/2
a = 7
The area of the third figure is 7 units²
The area of the three figuresarea = first figure area + second figure area + third figure area
area = 7 + 28 + 7
area = 42
The total area of the figures is 42 units²