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

europa2433

europa2433

step-by-step explanation:

hi there

the slope is 1/3.

this is because from the y-intercept, you go up one, and you go to the right three, to get to the next exact point on the graph. this is known at the “rise over run.”

p.s. (if you need it).your y-intercept is -2, because that is the point on the graph where the line crosses the y-axis.

hope that : )