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The width of a triangle is 6 cm less than its length. If its perimeter is 80 cm then its lengths (in cm) and area (in cm²) are respectively A) 16,351 B) 23,391 C) 17,391 D) 20,341 E) None
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i spose the key to this is in determining how to find a standard deviation.
if we know the mean of a set of numbers, then the average of the square of the difference from the mean define variation, and sqrt(variation) = stand dev
(a-20)^2 + (b-20)^2 ------------------ = 5^2 2
First of all the AM of 5, 15 is not 20, so we are left with the other options.
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