If $x$ is the average of $13$, $-16$, and $6$ and if $y$ is the cube root of $8$, find $x^2 + y^3$. Please Help

If we have a set of N elements:

{x₁, x₂, ..., xₙ}

The mean value (also called the average value) Is calculated as:

Average = (x₁ + x₂ + ... + xₙ)/n

So if x is the average of 13, -16 and 6 ( a total of 3 values)

x will be equal to:

x = (13 + (-16) + 6)/3 = (19 - 16)/3 = 3/3 = 1

x = 1

And we know that:

y = ∛8

Remember that:

2*2 = 4

and

4*2 = 8

then

2*2*2 = 2^3 = 8

then ∛8 = 2.

So we have:

y = ∛8 = 2

Now we can replace these values in the equation:

x^2 + y^3

replacing:

x = 1

y = 2

we get:

1^2 + 2^3 = 1 + 8 = 9

stops on an even number.

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step-by-step explanation:

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