Well i think it is true,because when you Multiplying you will not have a one decimal place
Answer:
a) 45 possible outcomes
b) 55 possible outcomes
Step-by-step explanation:
Given:
- Total cavities = 12
- Selection = 3 parts
- Non-conforming cavities = 2
Find:
a) How many samples contain exactly 1 nonconforming part?
b) How many samples contain at least 1 nonconforming part?
Solution:
- The question asks for the use of combinations to express the outcomes for each scenario.
- For first part, we want the inspector to pick exactly one non-conforming part among 3 selected. So let us say that he has already chosen that one non conforming cavity. Now he has to make 2 more selections out of total conforming cavities = 12 - 2 = 10 conforming cavities. Hence, the total possible outcome is to chose 2 randomly from 10 conforming cavities.
( Exactly 1 ) 10C2 = 45 possible outcomes
- The second part entails that at-least 1 non-conforming cavity is selected. To choose exactly 1 non conforming we calculated above. In the similar way calculate for selecting exactly 2 non-conforming cavities. The total possible outcome would be to choose from 10 conforming and we choose 1 from it:
( Exactly 2 ) 10C1 = 10 possible outcomes
- Hence, for at-least 1 non conforming cavity being selected we same the above two cases calculated:
(At-least 1 ) = ( Exactly 1 ) + ( Exactly 2 )
(At-least 1 ) = 45 + 10 = 55 possible outcomes
Step-by-step explanation:
2 × 10⁸ - 7 × 10⁷
= 200,000,000 - 70,000,000
=130,000,000
= 1.3 × 10⁸
5m − 3h = 45
Subtract 5m from both sides
-3h = 45 - 5m
Divide both sides by -3
h = (45 - 5m)/ -3
h = -15 + 5m/3
Answer:
Hope this helps...
Step-by-step explanation:
P stands for polynomial time. NP stands for non-deterministic polynomial time. Definitions: Polynomial time means that the complexity of the algorithm is O(n^k), where n is the size of your data (e. g. number of elements in a list to be sorted), and k is a constant.
Now, a German man named Norbert Blum has claimed to have solved the above riddle, which is properly known as the P vs NP problem.
