Probability that 2 of the 10 chargers will be defective =0.35
Number of ways of selecting 10 chargers from 20 chargers is 20C10
20C10 = 184756
Number of ways of selecting 10 chargers from 20 = 184756
Number of ways of selecting 2 defective chargers from 5 defective chargers = 5C2
5C2 = 10
Since 2 defective chargers have been chosen, there remains 8 to choose
Number of ways of selecting 8 good chargers from 15 remaining chargers = 15C8
Number of ways of selecting 8 good chargers from 15 remaining chargers = 6435
Probability that 2 of the 10 will be defective =
(10x6435)/184756
Probability that 2 of the 10 will be defective = 64350/184756
Probability that 2 of the 10 chargers will be defective =0.35
Learn more on probability here: brainly.com/question/24756209
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hope this helps
Ms. Cassidy instructed Miguel to change one sign of the graph of y < 2x – 4 so that point (2, 3) can be included in the solution set.
To check which of the given options might Miguel write we check the inequality that holds true for the point (2,3).Substituting x=2 ,y=3 we have:
1) y < 2x – 1
3<2(2)-1
3<3 Not True.
2)y ≤ 2x – 4
3≤ 2(2) -4
3≤ 0 .Not true.
3) y > 2x – 4
3> 2(2)-4
3> 0 True.
4) y < 2x + 4
3<2(2)+4
3<8 True
5) .y < 3.5x – 4
3< 3.5(2)-4
3<3 Not true
6) y < 4x – 4
3<4(2)-4
3<4 True.
Options 3 ,4 ,6 holds true for the point (2,3)
The exact form is x= -3/14
Answer:
;;6
Step-by-step explanation:
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