The absolute deviation for 3 is

Mathematics

1 answer:

luda_lava [24]3 years ago

3 0

Step-by-step explanation:

To find the mean absolute deviation of the data, start by finding the mean of the data set.

Find the sum of the data values, and divide the sum by the number of data values.

Find the absolute value of the difference between each data value and the mean: |data value – mean|.

Find the sum of the absolute values of the differences.

Divide the sum of the absolute values of the differences by the number of data values.

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Using the Factor Theorem, which of the polynomial functions has the zeros 2, radical 3 , and negative radical 3 ? f (x) = x3 – 2

Basile [38]

Answer:

$f(x) = x^3 - 2x^2 -3x + 6$

Step-by-step explanation:

According to the Factor Theorem, if (x - k) is a factor of a polynomial P(x), then P(k) must equal zero.

We are given that a polynomial function has the zeros 2, √3, and -√3. So, we can let k = 2, √3, -√3.

So, according to the Factor Theorem, P(2), P(√3) and P(-√3) must equal 0.

Testing each choice, we can see that only A is true:

$\displaystyle f(x) = x^3 - 2x^2 - 3x + 6$

Testing all three values yields that:

$\displaystyle \begin{aligned} f(2) &= (2)^3 - 2(2)^2 -3(2) + 6 \\ &= (8) - (8) -(6) + (6) \\ &= 0\stackrel{\checkmark}{=}0 \\ \displaystyle f(\sqrt{3}) &= (\sqrt{3})^3 - 2(\sqrt{3})^2 - 3(\sqrt{3}) + 6 \\ &=(3\sqrt{3}) -(6)-(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \\ f(-\sqrt{3}) &= (-\sqrt{3})^3 - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 \\ &=(-3\sqrt{3}) -(6)+(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \end{aligned}$