Answer:
C. 2
Step-by-step explanation:
Cohen's d is a parameter used to express the standardised difference between two means. It is defined as the difference between the means divided by the pooled standard deviation.
In this case, the difference between both means (M2-M1) is 8. As for the pooled standard deviation, simply take the square root of the given pooled variance:
Therefore, the value of Cohen's d (d) is:
We want to find the answer of the following polynomial:
We can see that the last term is -72
We want to find all the possible numbers that can divide it. Those are:
{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}
We want to factor this polynomial in order to find all the possible x-values. In order to factor it we will have to find some binomials that can divide it using the set of divisors of -72.
We know that if
(x - z) is a divisor of this polynomial then z might be a divisor of the last term -72.
We will verify which is a divisor using synthetic division. If it is a divisor then we can factor using it:
Let's begin with
(x-z) = (x - 1)
We want to divide
Using synthetic division we have that if the remainder is 0 it will be a factor
We can find the remainder by replacing x = z in the polynomial, when it is divided by (x - z). It is to say, that if we want to know if (x -1) is a factor of the polynomial we just need to replace x by 1, and see the result:
If the result is 0 it is a factor
If it is different to 0 it is not a factor
Replacing x = 1
If we replace x = 1, we will have that:
Then the remainder is not 0, then (x - 1) is not a factor.
Similarly we are going to apply this until we find factors:
(x - z) = (x + 1)
We replace x by -1:
Then, (x + 1) is a factor.
Using synthetic division we have that:
Then:
Now, we want to factor the 4th grade polynomial.
Let's remember our possibilities:
{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}
Since we verified ±1, let's try with ±2 as we did before.
(x - z) = (x - 2)
We want to divide:
We replace x by z = 2:
Then (x - 2) is a factor. Let's do the synthetic division:
Then,
Then, our original polynomial is:
Now, let's prove if (x +2) is a factor, using the new 3th grade polynomial.
(x - z) = (x + 2)
We replace x by z = -2:
Since the remainder is not 0, (x +2) is not a factor.
All the possible cases are:
{±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±36, ±72}
let's prove with +4
(x - z) = (x + 4)
We want to divide:
Let's replace x by z = -4 in order to find the remainder:
Then (x + 4) is a factor. Let's do the synthetic division:
Then,
Since
x² + 9 cannot be factor, we have completed our factoring:
Now, we have the following expression:
Then, we have five posibilities:
(x - 1) = 0
or (x - 2) = 0
or (x + 4) = 0
or (x² + 9) = 0
Then, we have five solutions;
x - 1 = 0 → x₁ = 1
x - 2 = 0 → x₂ = 2
x + 4 = 0 → x₃ = -4
x² + 9 = 0 → x² = -9 → x = ±√-9 = ±√9√-1 = ±3i
→ x₄ = 3i
→ x₅ = -3i
<h2><em>Answer- the solutions of the polynomial are: x₁ = 1, x₂ = 2, x₃ = -4, x₄ = 3i and x₅ = -3i</em></h2>