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8 days ago

Two number cubes are rolled to determine how a token moves on a game board. The sides of each number cube are numbered 1, 2, 3,

4, 5 and 6.
There are two options describing how the token will be moved.

Option A: If the product is even, move forward 4 spaces. Otherwise, move backward 3.

Option B: If the product is even, move forward 5 spaces. Otherwise, move backward 4.
Use mathematical expectation to determine which option offers the greater likelihood of moving closer to the finish line on the game board.

Which statement best explains the better option?

The mathematical expectation of Option A is 2.75. The mathematical expectation of Option B is 2.25. Option A offers a greater likelihood of advancing to the finish line.

The mathematical expectation of Option A is 2.75. The mathematical expectation of Option B is 2.25. Option B offers a greater likelihood of advancing to the finish line.

The mathematical expectation of Option A is 2.25. The mathematical expectation of Option B is 2.75. Option B offers a greater likelihood of advancing to the finish line.

The mathematical expectation of Option A is 2.25. The mathematical expectation of Option B is 2.75. Option A offers a greater likelihood of advancing to the finish line.

Mathematics

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Question

Which expression is equivalent to $\frac{4mn}{m^{-2}n^6}$ . Assuming $m \neq 0; n\neq 0$

Answer:

$\frac{4m^3}{n^5}$

Step-by-step explanation:

Given

$\frac{4mn}{m^{-2}n^6}$

Required:

Simplify

To achieve simplification, we begin by dividing each function individually

$\frac{4mn}{m^{-2}n^6} = \frac{4m}{m^{-2}} * \frac{n}{n^6}$

According to the properties of indices

$\frac{a^x}{a^y} = a^{x-y}$

Thus, the aforementioned expression can be restated similarly

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{1-(-2)} * n^{1-6}$

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{1+2)} * n^{1-6}$

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{3} * n^{-5}$

From the indices laws

$a^{-x} = \frac{1}{a^x}$

So,

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = 4m^{3} * \frac{1}{n^5}$

$\frac{4m}{m^{-2}} * \frac{n}{n^6} = \frac{4m^3}{n^5}$

Therefore, $\frac{4mn}{m^{-2}n^6}$ can be regarded as $\frac{4m^3}{n^5}$

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