100 POINTS Do these problems, show your work, and check your solution.
1. -x(1+x)=3(x+2)
2. 7 + y = 5(2y – 1) + 3y
3. 4(y + 2) – 3 = –y
4. –9 + 2(n + 3) = 4n – 7y
5. –10 – 2g = –2(3g – 1) + 4
6. 6x– 3(x – 1) = 2(x + 8) – 6
7. 6(r + 5) – 3 = 2r + 7
8. 0.5(2 – x) = 9.5 – (x + 1)
9. 0.3(y + 2) = 1.7 + (8 – y)
10. 0.4(g – 2) + 1.2 = –(g – 1) + 2.9
11. 3.1 – (x + 4.7) = 5.2(x – 2) – 3.6
12. 2y + 10(y– 0.5) = 2.5y – (0.5y + 1)
13. 7.5n – 4(n – 3) = 2n + 9
14. 0.1(y + 2) = 4(0.2y – 0.3)
15. 7+2/5(5y−10)=45(y+5)+1/5y
16. −9+1/4(8n+16)=1/4(20n+4)
17. 1/4p+3/4(p−8)=11+1/4(12p+4)
18. 5+1/3(9y−12)=5/6(y+18)+1/6y
19. 3/4(12x−8)+3=10−1/2(6x+2)
20. 1/4(2n−8)+1/3n=1/2(n+6)

well

Step-by-step explanation:

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a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.