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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Answer:
False
Step-by-step explanation:
To find the hypothenuse use Pythagorean theorem: 13²+5²=h²,
h=√13²+5²=13.93cm = hypothenuse.
(3^8 ⋅ 2^-5 ⋅ 9^0)^-2 ⋅ (2^ -2 / 3^3) ^4 ⋅ 3^28 =
(6561 * 0.03125 * 1)^2 * (0.00925)^4 * 22876792454961 =
42037.81348 * 0.00000000732094 * 22876792454961 =
7040477235.56798349
round answer as needed