Answer:
$580
Step-by-step explanation:
Step 1:
$200 + $380
Answer:
$580
Hope This Helps :)
Let p be 0.831 denote the percentage of defective welds and q be 0.169 denote the percentage of non-defective welds.
Using the binomial distribution, we want all three to be defective.
![P(X = 3) = \binom{3}{3}(p)^{3}{q}^{3 - 3}](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cbinom%7B3%7D%7B3%7D%28p%29%5E%7B3%7D%7Bq%7D%5E%7B3%20-%203%7D)
![P(X = 3) = \binom{3}{3}(0.831)^{3}(1)](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cbinom%7B3%7D%7B3%7D%280.831%29%5E%7B3%7D%281%29)
Answer: i am pretty sure the answer is either A or C
Step-by-step explanation:
It says that there was an error with my answer, but I don't know what, so I screenshotted what I typed and attached them as 4 images below:
Answer: It should be an 8th grade level
(sorry if i'm wrong)
:(