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3 years ago

What is the interquartile range of this data 0, 5, 10, 15, 20,

Mathematics

1 answer:

pentagon [3]3 years ago

6 0

Answer:

Step-by-step explanation:

The range from Q1 to Q3 is the interquartile range (IQR).

The interquartile range is the difference between the third and first quartiles.

The third quartile is 17.5.

The first quartile is 2.5.

The interquartile range = <u>17.5 - 2.5</u> = 15.

You might be interested in

The volume is 120.9 in. What is the missing dimension (X)

Elina [12.6K]

Answer:

X = 2.5 in

Step-by-step explanation:

volume = L x W x H

120.9 = 9.3 x 5.2 x X

X = 2.5 in

5 0

3 years ago

Read 2 more answers

Z^4-5(1+2i)z^2+24-10i=0

mixer [17]

Using the quadratic formula, we solve for $z^2$ .

$z^4 - 5(1+2i) z^2 + 24 - 10i = 0 \implies z^2 = \dfrac{5+10i \pm \sqrt{-171+140i}}2$

Taking square roots on both sides, we end up with

$z = \pm \sqrt{\dfrac{5+10i \pm \sqrt{-171+140i}}2}$

Compute the square roots of -171 + 140i.

$|-171+140i| = \sqrt{(-171)^2 + 140^2} = 221$

$\arg(-171+140i) = \pi - \tan^{-1}\left(\dfrac{140}{171}\right)$

By de Moivre's theorem,

$\sqrt{-171 + 140i} = \sqrt{221} \exp\left(i \left(\dfrac\pi2 - \dfrac12 \tan^{-1}\left(\dfrac{140}{171}\right)\right)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= \sqrt{221} i \left(\dfrac{14}{\sqrt{221}} + \dfrac5{\sqrt{221}}i\right) \\\\ ~~~~~~~~~~~~~~~~~~~~= 5+14i$

and the other root is its negative, -5 - 14i. We use the fact that (140, 171, 221) is a Pythagorean triple to quickly find

$t = \tan^{-1}\left(\dfrac{140}{171}\right) \implies \cos(t) = \dfrac{171}{221}$

as well as the fact that

$0$

$\sin\left(\dfrac t2\right) = \sqrt{\dfrac{1-\cos(t)}2} = \dfrac5{\sqrt{221}}$

(whose signs are positive because of the domain of $\frac t2$ ).

This leaves us with

$z = \pm \sqrt{\dfrac{5+10i \pm (5 + 14i)}2} \implies z = \pm \sqrt{5 + 12i} \text{ or } z = \pm \sqrt{-2i}$

Compute the square roots of 5 + 12i.

$|5 + 12i| = \sqrt{5^2 + 12^2} = 13$

$\arg(5+12i) = \tan^{-1}\left(\dfrac{12}5\right)$

By de Moivre,

$\sqrt{5 + 12i} = \sqrt{13} \exp\left(i \dfrac12 \tan^{-1}\left(\dfrac{12}5\right)\right) \\\\ ~~~~~~~~~~~~~= \sqrt{13} \left(\dfrac3{\sqrt{13}} + \dfrac2{\sqrt{13}}i\right) \\\\ ~~~~~~~~~~~~~= 3+2i$

and its negative, -3 - 2i. We use similar reasoning as before:

$t = \tan^{-1}\left(\dfrac{12}5\right) \implies \cos(t) = \dfrac5{13}$

$1 < \tan(t) < \infty \implies \dfrac\pi4 < t < \dfrac\pi2 \implies \dfrac\pi8 < \dfrac t2 < \dfrac\pi4$

$\cos\left(\dfrac t2\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\dfrac t2\right) = \dfrac2{\sqrt{13}}$

Lastly, compute the roots of -2i.

$|-2i| = 2$

$\arg(-2i) = -\dfrac\pi2$

$\implies \sqrt{-2i} = \sqrt2 \, \exp\left(-i\dfrac\pi4\right) = \sqrt2 \left(\dfrac1{\sqrt2} - \dfrac1{\sqrt2}i\right) = 1 - i$

as well as -1 + i.

So our simplified solutions to the quartic are

$\boxed{z = 3+2i} \text{ or } \boxed{z = -3-2i} \text{ or } \boxed{z = 1-i} \text{ or } \boxed{z = -1+i}$

3 0

1 year ago

Vineisha earned 22 out of

GarryVolchara [31]

Answer:

I believe it is 110%.

Step-by-step explanation

Explanation: 20x100=2200

2200 divided by 100=22

Hope this helps

3 0

3 years ago

Differentiate: f(x) = 3x ln(6x) − 3x

galben [10]

Answer is in the attachment below. Please open it up in a new window to see it in full.

5 0

4 years ago

Help me please<br>help.me

levacccp [35]

Answer:

The answers of the questions are given below :

a) = 4096
b) = 1.25
3) = m²
4) = r⁴s³
5) = a⁸/b¹²

Step-by-step explanation:

$\large{\tt{\underline{\underline{\red{QUESTION}}}}}$

$\begin{gathered}\footnotesize\boxed{\begin{array}{c|c|c}\bf\underline{Given}&\bf\underline{Solution}&\bf{\underline{Simple\: Form}}\\\\\rule{60pt}{0.5pt} &\rule{70pt}{0.5pt}& \rule{70pt}{0.5pt}\\\\ 1.\: {4}^{6} & & \\\\ 2.\: \bigg(\dfrac{2^6}{5^3} \bigg)& &\\\\ 3. \: \Big({m}^{\frac{2}{3}}\Big)\bull \Big({m}^{\frac{4}{3}}\Big) & &\\\\4. \: \big({r}^{12} {s}^{9}\big)^{ - \frac{1}{3}} &&\\\\ 5.\bigg(\dfrac{a^4}{a^6}\bigg)^{2}& &\end{array}}\end{gathered}$

$\begin{gathered}\end{gathered}$

$\large{\tt{\underline{\underline{\red{SOLUTION}}}}}$

Question. 1

>> 4⁶

$\begin{gathered}\qquad{= 4 \times 4 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 16 \times 4 \times 4 \times 4 \times 4} \\ \qquad{= 64 \times 4 \times 4 \times 4} \\ \qquad{= 256\times 4 \times 4} \\ \qquad{= 1024 \times 4} \\ \qquad{= 4096} \end{gathered}$

Hence, the answer is 4096.

$\begin{gathered}\end{gathered}$

Question. 2

>> (2⁶/5³)^-⅓

$\begin{gathered} \qquad\implies{\bigg(\frac{2^6}{5^3}\bigg)^{ - \frac{1}{3}}}\\ \\ \qquad\implies{\bigg(\frac{64}{125}\bigg)^{ - \frac{1}{3}}}\\ \\\qquad\implies{\bigg( \frac{1}{\frac{64}{125}}\bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( 1 \times \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\ \qquad\implies{\bigg( \frac{125}{64} \bigg)^{ \frac{1}{3}}} \\ \\\qquad\implies{\bigg( \sqrt[3]{ \frac{125}{64}}\bigg)} \\ \\ \qquad\implies{\bigg( \dfrac{5}{4} \bigg)} \\ \\ \qquad\implies{\Big( 1.25\Big)}\end{gathered}$

Hence, the answer is 1.25.

$\begin{gathered}\end{gathered}$

Question. 3

>> (m^2/3)•(m^4/3)

$\begin{gathered} \qquad{= \Big({m}^{\frac{2}{3}}\Big) \bull \Big({m}^{ \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2}{3} + \frac{4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{2 + 4}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{\frac{6}{3}}\Big)} \\ \\ \qquad{= \Big({m}^{2}\Big)}\end{gathered}$

Hence, the answer is m².

$\begin{gathered}\end{gathered}$

Question. 4

>> (r¹² s⁹)^⅓

$\begin{gathered} \qquad\implies{\Big( {r}^{12} \: {s}^{9}\Big)^{\frac{1}{3}}}\\\\ \qquad\implies{\Big({r}^{\frac{12}{3} } \: {s}^{\frac{9}{3}}\Big)} \\ \\ \qquad\implies{\Big({r}^{\cancel{\frac{12}{3}}} \: {s}^{\cancel{\frac{9}{3}}}\Big)} \\ \\ \qquad\implies{\Big({r}^{4} \: {s}^{3}\Big)} \end{gathered}$

Hence, the answer is r⁴s³.

$\begin{gathered}\end{gathered}$

Question. 5

>> (a⁴/b⁶)^2

$\begin{gathered} \qquad{ = \Big(\frac{a^4}{b^6}\Big)^{2}} \\ \\ \qquad{ = \Big(\frac{a^{4 \times 2}}{b^{6 \times 2}}\Big)} \\ \\ \qquad{ = \Big(\frac{a^{8}}{b^{12}}\Big)} \end{gathered}$

Hence, the answer is a⁸/b¹².

$\underline{\rule{220pt}{3pt}}$

4 0

2 years ago