X actually equals the cube root of 6
![{x}^{3} - 6 = 0 \\ {x}^{3} = 6 \\ x = \sqrt[3]{6}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B3%7D%20%20%20-%206%20%3D%200%20%5C%5C%20%20%7Bx%7D%5E%7B3%7D%20%20%3D%206%20%5C%5C%20x%20%3D%20%20%5Csqrt%5B3%5D%7B6%7D%20)
I’ve provided both the answer and explanation below! Hope this helps :)
C. 7.2°F is warmer than 7.0°F but colder than 7.6°F.
What you want is P(6∩1) or P(1∩6) or P(2∩5) or P(5∩2) or P(3∩4) or P(4∩3).
The events of rolling the dice are independent (i.e. they don't affect one another) so:
E.g.
P(6∩1) = P(6) * P(1)
P(2∩5) = P(2) * P(5)
The probability of getting a given number on a roll is 1/6 for both dice.
So:
P(6∩1) = 1/6 * 1/6 = 1/36
This is the same for any arrangement of numbers you could get from rolling two dice.
So, we can see that there are 6 arrangements of numbers that will give a sum of 7 and so that is 6 * 1/36 = 6/36 = 1/6
