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Mathematics

1 answer:

eimsori [14]1 year ago

4 0

Answer

Elena must have substracted 1/2x from both sides of the equation.

Lin must have multiplied both sides of the equation by 2

Explanation

The equation given is

$\frac{1}{2}x+3=\frac{7}{2}x+5$

For Elena to have arrived at

$3=\frac{7}{2}x-\frac{1}{2}x+5$

Then Elena must have substracted 1/2x from both sides of the equation.

That is;

$\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Substracting }\frac{1}{2}x\text{ from both sides of the equation will give Elena first step} \\ \frac{1}{2}x+3-\frac{1}{2}x=\frac{7}{2}x+5-\frac{1}{2}x \\ 3=\frac{7}{2}x-\frac{1}{2}x+5 \end{gathered}$

For Lin to have arrived at

$x+6=7x+10$

It shows Lin must have multiplied both sides of the equation by 2

That is;

$\begin{gathered} \frac{1}{2}x+3=\frac{7}{2}x+5 \\ \text{Multiply both sides of the equation by the lowest common mutiple } \\ \text{of the denominator which is 2.} \\ \frac{1}{2}x(2)+3(2)=\frac{7}{2}x(2)+5(2) \\ x+6=7x+10 \end{gathered}$

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A six-sided die is rolled two times. What is the theoretical fractional probability that the first number is even and the second

Nostrana [21]

Answer:

The theoretical probability that the first number is even and the second number is odd is $\frac{1}{4}$ .

Step-by-step explanation:

Since each dice has six sides, and there are two die, the total amount of different outcomes are $6^2$ or 36. The probability of getting an even number on the first roll is $\frac{3}{6}$ or $\frac{1}{2}$ , and the probability of getting an odd number on the second roll is also $\frac{3}{6}$ or $\frac{1}{2}$ , so the probability of getting an even number on the first roll and an odd number on the second roll is $\frac{1}{2} * \frac{1}{2} = \frac{1}{4}$ .