Question25 warmliquid corp., a manufacturer of water heaters, produces 30 water heaters per week. past records indicate that 10% of total water heaters produced in a week are likely to be defective. suppose a quality check was conducted on a sample of six water heaters. what is the probability that four out of six water heaters will be defective?

To solve this problem, we use the binomial probability equation.

P = [n! / (n – r)! r!] p^r q^(n – r)

where,

n is the total number of samples = 6

r is the number of defective = 4

p is the probability it will be defective = 0.10

q is the probability it wont be = 0.90

P = [6! / (6 – 4)! 4!] (0.10)^4 * (0.90)^(6 – 4)

P = 1.215 x 10^-3 = 0.1215%

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

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step-by-step explanation: