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3 years ago

Determine whether n^2+20n+100 is a perfect square trinomial. Yes No

Mathematics

1 answer:

Black_prince [1.1K]3 years ago

7 0

N²+20n+100 = (n+10)(n+10)
Yes is it a perfect square trinomial

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Find the mean and standard deviation for the set of data.

dem82 [27]

The mean and standard deviation for the set of data will be 16 and 5.87. Then the correct option is A.

<h3>What is a standard deviation?</h3>

It is the measure of the dispersion of statistical data. Dispersion is the extent to which the value is in a variation.

The set of data is given below.

16, 22, 8, 5, 20, 18, 14, 17, 24

The mean of the data set will be

μ = (16 + 22 + 8 + 5 + 20 + 18 + 14 + 17 + 24) / 9

μ = 144 / 9

μ = 16

Then the standard deviation of the data set will be

$\rm \sigma = \sqrt{\dfrac{\Sigma (x_i - \mu)^2}{n}} \\\sigma = \sqrt{\dfrac{(16 - 16)^2 +(22 - 16)^2 + (8-16)^2 + ......+(24-16)^2 }{9}} \\\sigma = \sqrt{\dfrac{310}{9}}\\$

Simplify the expression further, then we have

σ = √34.44

σ = 5.87

The mean and standard deviation for the set of data will be 16 and 5.87.

Then the correct option is A.

More about the standard deviation link is given below.

brainly.com/question/12402189

#SPJ1

3 0

1 year ago

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$