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2 years ago

Mean variance standard deviation discussions questions

Mathematics

1 answer:

Simora [160]2 years ago

5 0

The answer will be Mean = 135 Variance = 62 and Standard Deviation = 8

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

Standard deviation is defined as the amount of variation or the deviation of the numbers from each other.

Given that :

Proportion (p) = 54%

Sample size (n) = 250

Using Normal approximation :

Mean = np

Mean = 250 * 0.54 = 135

Variance = npq ; q = 1 - p = 1 - 0.54 = 0.46

Variance = (250 * 0.54 * 0.46) = 62.1 = 62

Standard deviation = √variance = √62.1

Standard deviation = 7.8803553 = 8

To know more about standard deviation follow

brainly.com/question/475676

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Kobotan [32]

Answer:

x > 100

Step-by-step explanation:

Consider x as "Chicago Cubs".

The alligator sign always points/goes to the number (or food) with the highest value (or bigger meal).

Since Chicago Cubs wins is bigger than 100, you can write the inequality as follows:

x > 100

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Lexi made 14 sandwiches for a group picnic. The number of ham sandwiches she made was 4 less than the twice the number of turkey

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x = # of ham sandwiches

14-x = # of turkey sandwiches ( as Lexi made 14 sandwiches)

as the question said that the number of ha sandwiches are 4 less than twice number of turkey sandwiches,

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3 years ago

An equilateral triangle has

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3 years ago

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Please solve this question

777dan777 [17]

The first figure, on left, is the composition of two areas, the area of a 7cm circle and the area of a 7 cm square so:

A=7^2+7^2π/4

A=49+49π/4

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The second figure is the area of a 15cm by 8cm rectangle minus the area of a 3cm circle so:

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4 years ago

Evaluate the integral ∫2032x2+4dx. Your answer should be in the form kπ, where k is an integer. What is the value of k? (Hint: d

faltersainse [42]

Here is the correct computation of the question;

Evaluate the integral :

$\int\limits^2_0 \ \dfrac{32}{x^2 +4} \ dx$

Your answer should be in the form kπ, where k is an integer. What is the value of k?

(Hint: $\dfrac{d \ arc \ tan (x)}{dx} =\dfrac{1}{x^2 + 1}$ )

k = 4

(b) Now, lets evaluate the same integral using power series.

$f(x) = \dfrac{32}{x^2 +4}$

Then, integrate it from 0 to 2, and call it S. S should be an infinite series

What are the first few terms of S?

Answer:

(a) The value of k = 4

(b)

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

Step-by-step explanation:

(a)

$\int\limits^2_0 \dfrac{32}{x^2 + 4} \ dx$

$= 32 \int\limits^2_0 \dfrac{1}{x+4}\ dx$

$=32 (\dfrac{1}{2} \ arctan (\dfrac{x}{2}))^2__0$

$= 32 ( \dfrac{1}{2} arctan (\dfrac{2}{2})- \dfrac{1}{2} arctan (\dfrac{0}{2}))$

$= 32 ( \dfrac{1}{2}arctan (1) - \dfrac{1}{2} arctan (0))$

$= 32 ( \dfrac{1}{2}(\dfrac{\pi}{4})- \dfrac{1}{2}(0))$

$= 32 (\dfrac{\pi}{8}-0)$

$= 32 ( (\dfrac{\pi}{8}))$

$= 4 \pi$

The value of k = 4

(b) $\dfrac{32}{x^2+4}= 8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -... \ \ \ \ \ (Taylor\ \ Series)$

$\int\limits^2_0 \dfrac{32}{x^2+4}= \int\limits^2_0 (8 - \dfrac{3x^2}{2^1}+ \dfrac{3x^4}{2^3}- \dfrac{3x^6}{2x^5}+ \dfrac{3x^8}{2^7} -...) dx$

$S = 8 \int\limits^2_0dx - \dfrac{3}{2^1} \int\limits^2_0 x^2 dx + \dfrac{3}{2^3}\int\limits^2_0 x^4 dx - \dfrac{3}{2^5}\int\limits^2_0 x^6 dx+ \dfrac{3}{2^7}\int\limits^2_0 x^8 dx-...$

$S = 8(x)^2_0 - \dfrac{3}{2^1*3}(x^3)^2_0 +\dfrac{3}{2^3*5}(x^5)^2_0- \dfrac{3}{2^5*7}(x^7)^2_0+ \dfrac{3}{2^7*9}(x^9)^2_0-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3-0^3)+\dfrac{3}{2^3*5}(2^5-0^5)- \dfrac{3}{2^5*7}(2^7-0^7)+\dfrac{3}{2^7*9}(2^9-0^9)-...$

$S= 8(2-0)-\dfrac{1}{2^1}(2^3)+\dfrac{3}{2^3*5}(2^5)- \dfrac{3}{2^5*7}(2^7)+\dfrac{3}{2^7*9}(2^9)-...$

$S = 16-2^2+\dfrac{3}{5}(2^2) -\dfrac{3}{7}(2^2) + \dfrac{3}{9}(2^2) -...$

$S = 16-4 + \dfrac{12}{5}- \dfrac{12}{7}+ \dfrac{12}{9}-...$

$a_0 = 16\\ \\ a_1 = -4 \\ \\ a_2 = \dfrac{12}{5} \\ \\a_3 = - \dfrac{12}{7} \\ \\ a_4 = \dfrac{12}{9}$

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3 years ago