Answer with explanation:
Class Interval Variate Frequency
16-18 16 1
19-21 0 0
22-24 0 0
25-27 26,27 2
X=9
3x-12=x+6
2x-12=6
2x=18
X=9
Answer is below.
<em>(Note to asker: Lack of explanations can get my answer deleted)</em>
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<u><em>Step 1</em></u>
We'll to isolate the variable, but to do that, get rid of the coefficient that multiplies it.
To do that, divide both sides by 22.

<em><u>Step 2.</u></em>
Now we have divided both sides by 22, but we'll need to reduce the fraction
.
We can reduce the fraction by dividing the factors that are in the numerator and denominator.
That being said, 22 appears both in the numerator and denominator.

<u><em>Step 3</em></u>
Now, for
, we'll need to reduce it to the lowest terms. Like in the 2nd step, we'll divide factors that are in the numerator and denominator.
That being said, 2 appears both in the numerator and denominator.
Hence, M = 5/11
Since the exponent (-3) is negative, flip the expression to: 1/(5n^4)^3.
Notice how the negative exponent becomes positive as you flip it.
Now evaluate the powers in the denominator: 1/((5^3)(n^4)^3
I separated the constant (5) from the variable (n) to show you how the powers are evaluated.
1/(5x5x5)(nxnxnxn)(nxnxnxn)(nxnxnxn)
—-> the power four means that there are 4 multiples of n in the parentheses. The power 3 corresponds to how many groups.
1/(125)(n^12)
= 1/125n^12