When two dice are rolled, what is the probability the two numbers will have a sum of 5?
1 answer:
When two dice are rolled, then possible outcomes are
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Total possible outcomes = 36
In that, we count the ordered pairs that give us sum 5
(1,4) (2,3) (3,2) and (4,1) = 4 outcomes
Probability = (number of times the event occurs)/ (total number of outcomes)
P( sum of 5) = =
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<h2>Step-by-step explanation:</h2>
Given equations;
y₁ = 3x - 8 -------------------(i)
y₂ = 0.5x + 7 --------------------(ii)
To fill the table, substitute the values of x into equations (i) and (ii)
=> At x = 0
y₁ = 3(0) - 8 = -8
y₂ = 0.5(0) + 7 = 7
=> At x = 1
y₁ = 3(1) - 8 = -5
y₂ = 0.5(1) + 7 = 7.5
=> At x = 2
y₁ = 3(2) - 8 = -2
y₂ = 0.5(2) + 7 = 8
=> At x = 3
y₁ = 3(3) - 8 = 1
y₂ = 0.5(3) + 7 = 8.5
=> At x = 4
y₁ = 3(4) - 8 = 4
y₂ = 0.5(4) + 7 = 9
=> At x = 5
y₁ = 3(5) - 8 = 7
y₂ = 0.5(5) + 7 = 9.5
=> At x = 6
y₁ = 3(6) - 8 = 10
y₂ = 0.5(6) + 7 = 10
=> At x = 7
y₁ = 3(7) - 8 = 13
y₂ = 0.5(7) + 7 = 10.5
=> At x = 8
y₁ = 3(8) - 8 = 16
y₂ = 0.5(8) + 7 = 11
=> At x = 9
y₁ = 3(9) - 8 = 19
y₂ = 0.5(9) + 7 = 11.5
=> At x = 10
y₁ = 3(10) - 8 = 22
y₂ = 0.5(10) + 7 = 12
The complete table is attached to this response.
(ii) To find the solution of the system of equations using the table, we find the value of x for which y₁ and y₂ are the same.
As shown in the table, that value of <em>x = 6</em>. At this value of x, the values of y₁ and y₂ are both 10.