6.12 x 10^3 = 6120 .........<span> in standard notation
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hope it helps</span>
Answer:
Yes
Step-by-step explanation:
Yes, because each x has one specific y.
On the opposite side of Point A
Answer:

Step-by-step explanation:
The Universal Set, n(U)=2092


Let the number who take all three subjects, 
Note that in the Venn Diagram, we have subtracted
from each of the intersection of two sets.
The next step is to determine the number of students who study only each of the courses.
![n(S\:only)=1232-[103-x+x+23-x]=1106+x\\n(F\: only)=879-[103-x+x+14-x]=762+x\\n(R\:only)=114-[23-x+x+14-x]=77+x](https://tex.z-dn.net/?f=n%28S%5C%3Aonly%29%3D1232-%5B103-x%2Bx%2B23-x%5D%3D1106%2Bx%5C%5Cn%28F%5C%3A%20only%29%3D879-%5B103-x%2Bx%2B14-x%5D%3D762%2Bx%5C%5Cn%28R%5C%3Aonly%29%3D114-%5B23-x%2Bx%2B14-x%5D%3D77%2Bx)
These values are substituted in the second Venn diagram
Adding up all the values
2092=[1106+x]+[103-x]+x+[23-x]+[762+x]+[14-x]+[77+x]
2092=2085+x
x=2092-2085
x=7
The number of students who have taken courses in all three subjects, 
Answer:
4221 birhs
Step-by-step explanation:
. Find the mean number of births per day, then use that result to find the probability that in a day, there are 15 births.