Two six-sided dice are tossed 288 times. how many times would you expect to get a sum of 5?
1 answer:
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Answer:
3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).
Step-by-step explanation:
Let the First positive number be <em>x</em> and second positive number be <em>y.</em>
Triple of first number = 3<em>x</em>
double of second number = 2<em>y</em>
According to question,
3<em>x </em>+ 2<em>y</em> = 34
Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34
So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34
Now, 3<em>x </em>+ 2<em>y</em> = 34
3<em>x </em>= 34 - 2<em>y</em>
Let x = 10 and by substituting its value in above expression,
Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)
In the same way put x = 4 then,
3<em>x </em>= 34 - 2<em>y</em>
Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)
Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.