9(3j-6) = 27j - 54
<span>Note: a(b-c) = ab - ac</span>
If I understand what you are asking, you can use the median as an average since it is not affected by the mean ( median is a measure of center). Since a mean is affected by outliers, it might lower the score if there are very low outliers. But note that if there are outliers way <em>greater </em>it can increase the average. You can use the Inter Quartile Range if shown on a box plot ( measure of variability).
Step-by-step explanation:
Since y is directly proportional to x,
we have y = kx, where k is a real constant.
When x = 15, y = 12.
=> (12) = k(15), k = 0.8.
Therefore we have y = 0.8x.
When x = 2, y = 0.8(2) = 1.6.
The answer is y = 1.6 when x = 2.
18.39 is 5.58% of 329.64.
We have to find that 18.39 is what percent of 120?
First, make the assumption that 329.64 is 100% as it is our output value.
We next represent the value we seek with x, therefore
100% = 329.64
And, x% = 18.39
Now, we get pair of simple equations
100% = 329.64 ........(1)
x% = 18.39 ........(2)
Now by simply dividing equation 1 by equation 2 and note of the fact that the LHS of both equations have the same unit (%);
100% / x% = 329.64 / 18.39
Taking the reciprocal of both sides, we get
x% / 100% = 18.39 / 329.64
or x = 5.58%
Hence, 18.39 is 5.58% of 329.64.
To learn more about percentages, visit: brainly.com/question/14319057
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