Hey :)
x= -12
Use the distributive property to multiply −4 by 2x+13
−8x−52=3x+80
Subtract 3x from both sides
−8x−52−3x=80
Combine −8x and −3x to get −11x
−11x−52=80
Add 52 to both sides
−11x=80+52
Add 80 and 52 to get 132
−11x=132
Divide both sides by −11
x= 132/-11
Divide 132 by −11 to get −12
x=−12
Hope this helps :)
C.7000
1km=1000
1000*7=7000
A reduction of 15% per year means 100-15=85% reducing factor.
1st year: 0.85×18000=$15300.
2nd year: 0.85×15300=$13005.
3rd year: 0.85×13005=$11054.25.
So after 3 years the value of the car is $11054.25 or $11054 to the nearest dollar.
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
