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3 years ago

List the range pls help

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

5 0

Answer:

range {1,3,-3,4,1} yes that is it

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I don’t know this one

torisob [31]

Answer:

3077.2

Step-by-step explanation:

A circle's mass is found by

Pie times R^2

7^2 is 49, times 3.14 is 153.86

then you Multiply 153.86 by 20 which gives you 3077.2

5 0

3 years ago

Read 2 more answers

Is 23 prime or composite ?

dalvyx [7]

It is prime, it has no factors besides 1,23

3 0

3 years ago

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12.4. Measure of Dispersion 2076 Q.No. 15 Find the standard deviation of: 4, 6, 8, 10, 12.

Alenkinab [10]

Answer:

Standard deviation of given data = 3.16227

Step-by-step explanation:

Step(i):-

Given sample size 'n' = 5

Given data 4, 6,8,10,12

$Mean = \frac{4+6+8+10+12}{5} = 8$

Mean of the sample x⁻ = 8

Standard deviation of the sample

$S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}$

Step(ii):-

Given data

x : 4 6 8 10 12

x-x⁻ : 4 - 8 6-8 8-8 10-8 12-8

(x-x⁻) : -4 -2 0 2 4

(x-x⁻)² : 16 4 0 4 16

$S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}$

$S.D = \sqrt{\frac{16+4+0+4+16}{4}}$

S.D = √10 = 3.16227

Final answer:-

The standard deviation = 3.16227

8 0

2 years ago

Answer if you do know

Alex777 [14]

Answer:

Find the hypotenuse:

3^3 + 10^2 = 109

Square root of 109 is around 10.4

8 0

3 years ago

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Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .