Find the variance of the set of data to the nearest tenth: 5, 8, 2, 9, 4

Mathematics

1 answer:

Orlov [11]3 years ago

7 0

Answer:

Therefore the variance on the data set is 8.3

Step-by-step explanation:

In order to find the variance of the set of data we first need to calculate the mean of the set, which is given by:

mean = sum of each element / number of elements

mean = (5 + 8 + 2 + 9 + 4)/5 = 5.6

We can now find the variance by applying the following formula:

$s^{2} = \frac{\sum_{(i=1)}^n(x_i - mean)^2}{n -1}$

So applying the data from the problem we have:

s² = [(5 - 5.6)² + (8 - 5.6)² + (2 - 5.6)² + (9 - 5.6)² + (4 - 5.6)²]/(5 - 1)

s² = [(-0.6)² + (2.4)² + (-3.6)² + (3.4)² + (-1.6)²]/4

s² = [0.36 + 5.76 + 12.96 + 11.56 + 2.56]/4 = 8.3

Therefore the variance on the data set is 8.3

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Write the following quotient in the form a+bi.<br> -7i/2-7i

Helga [31]

Answer:

$\large\boxed{\dfrac{49}{53}-\dfrac{14}{53}i}$

Step-by-step explanation:

$\text{Use}\\\\i=\sqrt{-1}\to i^2=-1\\\\(a-b)(a+b)=a^2-b^2\to(a-bi)(a+bi)=a^2+b^2\\\\\text{distributive property}\ a(b+c)=ab+ac\\---------------------\\\\\dfrac{-7i}{2-7i}=\dfrac{-7i}{2-7i}\cdot\dfrac{2+7i}{2+7i}=\dfrac{-7i(2+7i)}{2^2+7^2}=\dfrac{(-7i)(2)+(-7i)(7i)}{4+49}\\\\=\dfrac{-14i-49i^2}{53}=\dfrac{-14i-49(-1)}{53}=\dfrac{-14i+49}{53}\\\\=\dfrac{49}{53}-\dfrac{14}{53}i$