Notebooks cost $1.20 each this weekend they will be on sale for $.80 what percentage is the sale

Size of u.s. states the total surface area (in square miles) for each of six selected eastern states is listed here. 28,995 pa 3

KonstantinChe [14]

<span>Standard deviation of first data set = 5879.1 Standard deviation of second data set = 14768.78 The second data set is more variable. The basic definition of standard deviation is the square root of the mean of the squares of the difference from the mean. It's a bit of a mouthful, but easy enough to do. For the first data set, first calculate the mean. (28995 + 37534 + 31361 + 27087 + 20966 + 37741) / 6 = 30614 Now calculate the square of the differences from the mean (28995 - 30614)^2 = 2621161 (37534 - 30614)^2 = 47886400 (31361 - 30614)^2 = 558009 (27087 - 30614)^2 = 12439729 (20966 - 30614)^2 = 93083904 (37741 - 30614)^2 = 50794129 And now the average of the squares (2621161 + 47886400 + 558009 + 12439729 + 93083904 +50794129) / 6 = 34563888.67 And finally, take the square root to get the standard deviation. sqrt(34563888.67) = 5879.1 Now for the second data set of western states. First, the mean (72964 + 70763 + 101510 + 62161 + 66625 + 54339) / 6 = 71393.67 Now the squares of the differences (72964 - 71393.67)^2 = 2465946.778 (70763 - 71393.67)^2 = 397740.4444 (101510 - 71393.67)^2 = 906993533.4 (62161 - 71393.67)^2 = 85242133.78 (66625 - 71393.67)^2 = 22740181.78 (54339 - 71393.67)^2 = 290861655.1 And the average of the squares is 218116865.2 Finally, the square root of the average is 14768.78 So the standard deviation of the 2nd data set is 14768.78 And since the standard deviation of the 2nd data set is larger than the standard deviation of the 1st data set, that means that the 2nd data set is more variable.</span>

7 0

3 years ago

What is the following product? <br>(4x square root 5x^2 + 2^2 square root 6)^2

tangare [24]

The product is $104 x^{4}+16 \sqrt{30} x^{4}$

Explanation:

The given expression is $\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}$

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2$

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2$

Expanding the expression, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)$

Now, we shall apply FOIL, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}$