We define the probability of a particular event occurring as:

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are <em>independent</em>, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for <em>each </em>of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.
Now that we've found the number of possible outcomes, we need to find the number of <em>desired</em> outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is <em>at least one 5 rolled</em>. It turns out, there are only 3:
(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5
So, we have

probability of rolling at least one 5.
To determine which system of equations would have the same solution, we evaluate each system of equations.
System 1 4x − 5y = 2, 3x − y = 8
x = 38/11
y = 26/11
<span>System 2 4x − 5y = 2, 3x − 2y = 1
x = 1/7
y = -2/7
System 3 4x − 5y = 2, 3x − 8y = 4
x = -4/17
y = -10/17
System 4 4x − 5y = 2, 10x − 9y = 4
x = 1/7
y = -2/7
</span><span>
Therefore, the correct answer is option 3. </span><span>System 2 and system 4 are equal, because the second equation in system 4 is obtained by adding the first equation in system 2 to two times the second equation in system 2.
4x− 5y = 2 2( 3x − 2y = 1)
----------------------- 10x - 9y = 4</span>
i think it's t=0.2 Hope it helps:)
Answer:
7/15 de la tablilla de chocolate le va a quedar.
Step-by-step explanation:
2/3 se converte en 10/15
y 1/5 se converte en 3/15
Entonces le va a quedar 10/15-3/15 = 7/15 de la tablilla de chocolate.