A single die is rolled twice. the set of 36 equally likely outcomes is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1),
Maksim231197 [3]
<span>We need to find the rolls whose sum is greater than 10. By looking at the outcomes, we see that (5,6), (6,5), and (6,6) all have a sum greater than 10. Therefore, there are 3 chances to get a sum greater than 10. Since there are 36 chances overall, the probability of rolling greater than 10 are 3/36 = 1/12.</span>
54
a rule I follow:
same signs positive,
different signs negative.
-9 * -3 = 27
27 + 27 = 54
4000.0093979 = 63.2456275
-15999998
= 8.7
Then you get 50
Multiply the first equation by 2 to then use elimination by adding the 2 equations.
So, 14x-2y=14
+. x+2y=6
Equals 15x=20 so x=20/15=4/3
Plug in to first equation to get y
7(4/3)-y=7
28/3-y=7
28/3-y=21/3
So y=7/3
Plug both x and y into second equation to check
4/3+2(7/3)=
4/3+14/3=18/3=6 it works
So x=4/3, y=7/3
Hope this helps! Have a blessed day!
<span>solve <span><span>3≤−3x+6<15</span><span>3≤−3x+6<15</span></span></span>
Answer: (−3,1](−3,1]
<span>Approximate Form: <span><span>(<span>−3,1</span>]</span></span></span>
