It's not specified whether 1 is the 1st or 2nd roll: HOWER:
The 1st Roll is "1": P(odd sum/the 1st Roll is 1)
What is the sample space of all numbers starting with "1":
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),} = 6
the couple of add sum=(1,2), (1,4), (1,6), =3
P(odd sum/ 1st is 1) = 3/6 =1/2
or in applying the formula:
P(odd sum/the 1st Roll is 1) =P(odd sum ∩ 1) / P(getting "1") it will give the same probability = 1/2
NOW if the 2nd Roll is "1", it 's still 1/2
3b+6
If you make both equations equal each other, you can then solve the system of equations.
If they all lined up and held hands, the line will be 6 billion meter apart.
Answer:
(x + 2/3)^2 + (y + 3/4)^2 = 5^2 (or 25).
Step-by-step explanation:
Start with the general equation of a circle with center at (h, k) and radius r:
(x - h)^2 + (y - k)^2 = r^2
Substitute -2/3 for h and -3/4 for k, and also 5 for r:
(x - [-2/3])^2 + (y - [-3/4])^2 = 5^2, or
(x + 2/3)^2 + (y + 3/4)^2 = 5^2 (or 25).
