The astronomer has to choose 3 star images out of 8.This is a problem of combinations. This can be expressed as 8C3 i.e. combination of 8 objects taken 3 at a time.
This means, if 8 images are nearby the astronomer will have 56 choices <span>
of the three stars.</span>
I believe that the answer to this is b
111 tickets, because 660 divided by 6 equals 110 which is the number of tickets you need to make no profit and if you sell one more ticket you make a profit.
Answer:
The 2 numbers are
102 and 51
Step-by-step explanation:
A + B = 153
A=2B
So 2B+B=153
3B=153
B=153/3=51
A=51*2=102
45 will be the answer of 3 because 30 divide in 2 is 15. and 15x3 is 45.
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