Answer:
The probability that a container will be shipped even though it contains 2 defectives if the sample size is 88, will be 
Step-by-step explanation:
The first step is to count the number of total possible random sets of taking a sample size of 88 engines over 1212 engines of the population, so ![\left[\begin{array}{ccc}1212\\88\end{array}\right] =1212C88=4.7205x10^{135}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1212%5C%5C88%5Cend%7Barray%7D%5Cright%5D%20%3D1212C88%3D4.7205x10%5E%7B135%7D)
The second step is to count the number of total possible random sets of taking a sample size of 88 engines over 1210 engines (discounting the 2 defective engines) as the possible ways to succeed, so ![\left[\begin{array}{ccc}1210\\88\end{array}\right] =1212C88=4.0596x10^{135}](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1210%5C%5C88%5Cend%7Barray%7D%5Cright%5D%20%3D1212C88%3D4.0596x10%5E%7B135%7D)
Finally we need to compute
, therefore the probability that a container will be shipped is
Answer:
I think 13
Step-by-step explanation:
P(7) = 57
Porque :
P(x) = x^2 + x + 1
P(7) = 7^2 + 7 + 1
P(7) = 49 + 7 + 1
P(7) = 57 ✔️
Let x = first number, and y = second number.
x + y = 40 (1)
x - y = 6.5 (2)
Rearrange (2)
y = x - 6.5 (3)
Sub (3) in (1)
x + (x - 6.5) = 40
2x - 6.5 = 40
2x = 33.5
x = 16.75 (4)
Sub (4) in (1)
16.75 + y =40
y = 23.25
x * y = 16.75 * 13.25 = 221.9375
The product is 221.9375
Answer:
anything above 12
Step-by-step explanation:
would be greater than 12. 13, 14, 16...etc