Often, variability is most usefully expressed in the same units as the original data. That's what <em>standard deviation</em> (square root of variance) does. In situations where that is the case, the variance is useful only for getting to the value of standard deviation.
Answer:
First we substitute
4(-3) - 6(-1)
-7 -6
-13
Step-by-step explanation:
Answer: The required derivative is 
Step-by-step explanation:
Since we have given that
![y=\ln[x(2x+3)^2]](https://tex.z-dn.net/?f=y%3D%5Cln%5Bx%282x%2B3%29%5E2%5D)
Differentiating log function w.r.t. x, we get that
![\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B%5Bx%282x%2B3%29%5E2%5D%7D%5Ctimes%20%5Bx%27%282x%2B3%29%5E2%2B%282x%2B3%29%5E2%27x%5D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B1%7D%7B%5Bx%282x%2B3%29%5E2%5D%7D%5Ctimes%20%5B%282x%2B3%29%5E2%2B2x%282x%2B3%29%5D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B4x%5E2%2B9%2B12x%2B4x%5E2%2B6x%7D%7Bx%282x%2B3%29%5E2%7D%5C%5C%5C%5C%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7B8x%5E2%2B18x%2B9%7D%7Bx%282x%2B3%29%5E2%7D)
Hence, the required derivative is
Answer:
-300
Step-by-step explanation:
because is in the negative and is a number bigger than 23 but in negative.
Answer:
Oh if I could hug you
Step-by-step explanation:
tysm you are the best!
