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
Answer:
The answer is A
Step-by-step explanation:
Answer:
a = 60
b = 90
c = 150
Step-by-step explanation:
Numerele a, b și c sunt direct proporționale cu 2, 3 și 5.
Unde k este constantă de proporționalitate
a ∝ 2
a = 2k
b ∝ 3
b = 3k
c ∝ 5
c = 5k
Dacă media aritmetică a celor trei numere este egală cu 100, determinați numerele a, b și c
= 2k + 3k + 5k / 3 = 100
= 10k / 3 = 100
Cross Multiply
= 10k = 3 × 100
= 10k = 300
Împărțiți ambele părți la 10
k = 300/10
k = 30
Pentru numărul a
a = 2k
a = 2 × 30
a = 60
Pentru numărul b
b = 3k
b = 3 × 30
b = 90
Pentru numărul c
c = 5k
c = 5 × 30
c = 150
Prin urmare, a = 60, b = 90, c = 150
T,A,Y are 40 and 2T,X,B,Z are 80
Annie's total earnings from her initial savings, $12, and from babysitting should be equal or more than 30. Annie's total earnings from babysitting may be expressed as $6n. The inequality should be,
12 + 6n <span>≥ 30
Solving for x,
6n </span><span>≥ 30 - 12
</span><span>
6n </span>≥ 18 ; n <span>≥ 3
</span><span>
Thus, the answer is the second among the choices.
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