Answer:
1 a.) 2t – 2
1 b.) –9z + 4
2 a.) –2v + 3
2 b.) –13k – 9
3 a.) 7w − 2
3 b.) 10s
4 a.) 5
4 b.) –3p + 11
5 a.) –29x + 9
b.) 15v – 60x – 2
a.) 25n + 2
b.) –15n
a.) –71x + 4
b.) –15b
Step-by-step explanation:
- <em>NOTE: I am kind of confused about the way this question is formatted, so I tried to answer everything I could decipher.</em>
1 a.) –2t + 5t – 2 – t → 2t – 2
1 b.) –5z + 4 – 4z → –9z + 4
2 a.) 8v + 3 – 10v → –2v + 3
2 b.) –6 – 6k – 3 – 7k → –13k – 9
3 a.) –6 – 3w + 4 + 10w → 7w − 2
3 b.) 9s + s → 10s
4 a.) 6 + 8b – 1 – 8b → 5
4 b.) 1p + 6 + 5 – 4p → –3p + 11
5 a.) 4x – 3x + 9 – 6x 5 → –29x + 9
b.) 9v + 6v – 2 – 10x 6 → 15v – 60x – 2
a.) n + 2 – (–4)n 6 → 25n + 2
b.) 6n – 3n 7 → –15n
a.) –4x – 4x + 4 – 9x 7 → –71x + 4
b.) –9b + (–6)b → –15b
Answer:
200.86
Step-by-step explanation:
Using the formulas:
A=πr2
C=2πr
Solving for A:
A=C2
4π=50.242
4·π≈200.85812
I hope this helps.
The answer is 0.4%. The probability that a unit end in a rework can be calculated using the following equation:
P(rework)= P(defects found) * P(defect)
= .8 * .005
= .004
= 0.4%
Thank you for posting your question. Feel free to ask me more.
