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3 years ago

If there are 21 values in a data set in order from smallest to largest, what is the first quartile of the data set?

Mathematics

2 answers:

Pavlova-9 [17]3 years ago

7 0

Answer with explanation:

Total Number of Variate in the Data set = 21

Since number of Variate is odd, median is given by the formula:

$=\frac{\text{Total number of terms}+1}{2}\\\\=\frac{21+1}{2}\\\\=11^{\text{th term}}$

$Q_{1}$

= Mid value between lowest Variate and Median value.

→→Mid value of first 11 Terms is given by

$=\frac{\text{Total number of terms}+1}{2}\\\\=\frac{11+1}{2}\\\\=6^{\text{th term}}$

Most Appropriate Option:⇒

Option A: →The mean of the 6th and 7th values

jek_recluse [69]3 years ago

4 0

I think the answer is B) The mean of the 5th and 6th values

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Find the integration of (1-cos2x)/(1+cos2x)

slega [8]

Given:

The expression is:

$\dfrac{1-\cos 2x}{1+\cos 2x}$

To find:

The integration of the given expression.

Solution:

We need to find the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ .

Let us consider,

$I=\int \dfrac{1-\cos 2x}{1+\cos 2x}dx$

$I=\int \dfrac{2\sin^2x}{2\cos^2x}dx$ $[\because 1+\cos 2x=2\cos^2x,1-\cos 2x=2\sin^2x]$

$I=\int \dfrac{\sin^2x}{\cos^2x}dx$

$I=\int \tan^2xdx$ $\left[\because \tan \theta =\dfrac{\sin \theta}{\cos \theta}\right]$

It can be written as:

$I=\int (\sec^2x-1)dx$ $[\because 1+\tan^2 \theta =\sec^2 \theta]$

$I=\int \sec^2xdx-\int 1dx$

$I=\tan x-x+C$

Therefore, the integration of $\dfrac{1-\cos 2x}{1+\cos 2x}$ is $I=\tan x-x+C$ .

8 0

2 years ago

Help ASAP solve: -3/4n = -6

anyanavicka [17]

Answer:

$\huge\boxed{n=8}$

Step-by-step explanation:

$-\dfrac{3}{4}n=-6\qquad|\text{multiply both sides by (-4)}\\\\(-4\!\!\!\!\diagup)\left(-\dfrac{3}{4\!\!\!\!\diagup}n\right)=(-4)(-6)\\\\3n=24\qquad|\text{divide both sides by 3}\\\\\dfrac{3\!\!\!\!\diagup n}{3\!\!\!\!\diagup}=\dfrac{24\!\!\!\!\!\diagup}{3\!\!\!\!\diagup}\\\\n=8$