Answer:
-14.8p+25.9
Step-by-step explanation:
got it right edge 2020.
Answer:
Probability that a car need to be repaired once = 20% = 0.20
Probability that a car need to be repaired twice = 8% = 0.08
Probability that a car need to be repaired three or more = 2% = 0.02
a) If you own two cars what is the probability that neither will need repair?
Probability that a car need to be repaired once , twice and thrice or more= 0.20+0.08+0.02=0.3
Probability that car need no repair = 1-0.3=0.7
Neither car will need repair=![0.7 \times 0.7=0.49](https://tex.z-dn.net/?f=0.7%20%5Ctimes%200.7%3D0.49)
b) both will need repair?
Probability both will need repair = ![0.3 \times 0.3=0.09](https://tex.z-dn.net/?f=0.3%20%5Ctimes%200.3%3D0.09)
c)at least one car will need repair
Neither car will need repair=![0.7 \times 0.7=0.49](https://tex.z-dn.net/?f=0.7%20%5Ctimes%200.7%3D0.49)
Probability that at least one car will need repair= 1-0.49 = 0.51
48/6=8
8*3=24
24 yards of trim
There is no picture of the parallelogram needed to answer this question.
Answer:
Step-by-step explanation:
common ratio is -3
10-3=7
7-3=4
4-3=1
etc