Let x be the number of defective cell phones. It is given that in batch of 100, on average 5 are defective.
let p be the probability of defective cell phone.
p = 5/100 = 0.05
Let n be size of random sample, n=30
Here out of 30 we want to find probability that 2 will be defective. It means 30-2 =28 cell phones will be non defective.
The probability of getting non defective cell phone is 1- p=1-0.05 =0.95
The probabability of getting 2 defective is
P(X=2) = number of ways selecting 2 from 30 * probability 2 defective * probability of 28 non defective
Now number of ways of selecting 2 cell phone from 30 is
30C2 = 
= 
= 
= (30*29) /2
30C2 = 435
P(X=2) = 30C2 * (0.05)^2 * (0.95)^28
= 435 * 0.0025 * 0.2378
P(X=2) = 0.2586
Probability of getting 2 defective out of 30 is 0.2586
Answer:
Look up on the Internet you’re welcome
Step-by-step explanation:
78.2 years
2005 - 1995 = 10
.23 * 10 = 2.3
75.9 + 2.3 = 78.2
Answer:

Step-by-step explanation:
Since Θ is in fourth quadrant then cosΘ > 0, as is secΘ
Given
sinΘ = -
, then
cosΘ = 
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Hence
secΘ =
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