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%

b

step-by-step explanation:

the total is 23.97. 0.25% is off so you have to subtracting 25% from 100% and you will get 75% if you convert it to decimal you will get 0.75. you have to find the original because the formula is % times original = total.so you have the total and the % so you just have to plug it in.

$113.42

i hope that is right.

step-by-step explanation: