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IRINA_888 [86]
3 years ago
6

What is the interquartile range (IQR) of the data set represented by this box plot?

Mathematics
2 answers:
Zinaida [17]3 years ago
7 0

Answer:

B

Step-by-step explanation:

The interquartile range is the difference between the upper quartile and the lower quartile.

The upper quartile is the right side value of the box

The lower quartile is the left side value of the box

upper quartile = 61 and lower quartile = 39

IQR = 61 - 39 = 22 → B

FromTheMoon [43]3 years ago
7 0

Answer:

B.23

Step-by-step explanation:

apexs

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3 years ago
can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc
mariarad [96]
The first equation is linear:

x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x

Divide through by x^2 to get

\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for y.

\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x
\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C
\implies y=-x\cos x+Cx

- - -

The second equation is also linear:

x^2y'+x(x+2)y=e^x

Multiply both sides by e^x to get

x^2e^xy'+x(x+2)e^xy=e^{2x}

and recall that (x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x, so we can write

(x^2e^xy)'=e^{2x}
\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C
\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}

- - -

Yet another linear ODE:

\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1

Divide through by \cos^2x, giving

\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}
\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x
\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x
\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C
\implies y=\cos x\tan x+C\cos x
y=\sin x+C\cos x

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

a(x)y'(x)+b(x)y(x)=c(x)

then rewrite it as

y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)

The integrating factor is a function \mu(x) such that

\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'

which requires that

\mu(x)P(x)=\mu'(x)

This is a separable ODE, so solving for \mu we have

\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx
\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx
\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)

and so on.
6 0
4 years ago
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