1+3i=5
3i=5-1
3i=4
this is in standard form
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
<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>✌</em>
I think that the answer is 51/54.
First you need to find the common denominator of both fractions, 54.
1/9 is 6/54 and 5/6 is 45/54. 45+6=51. And the denominator goes on the bottom,making the answer 51/54.
Hope this helps!