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3 years ago

15

Find the mean and range. 20 14 29 39 33 25 19 35

2 answers:

Maksim231197 [3]3 years ago

7 0

Mean: 26.75
Range: 25

egoroff_w [7]3 years ago

4 0

Count:8Sum:214Mean:26.75Median:27Mode:20, 14, 29, 39, 33, 25, 19, 35

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What is the circumference of a circle with a diameter of 11 cm?

qaws [65]

The answer is 11 pi

I hope this helps you !
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2 years ago

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Which number line best shows the position of square root of 7?

Serggg [28]

Step-by-step explanation:

First off, you want to find the whole number in square root form

1 = √1

2 = √4

3 = √9

4 = √16

5 = √25

You want to find which to numbers √7 is in between

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The second one is your answer

I hope this helps!!!

8 0

3 years ago

Every week, Mr. Kirkson uses 3 1/6 gallon of water to water every 1/3 square foot of his garden. How many gallons of water does

nata0808 [166]

3 1/6 / 1/3 =

19/6 / 1/3 =

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5 0

3 years ago

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Evaluate this power with a base that is a negative fraction. (−1/4 )^2

Tomtit [17]

$\left(-\dfrac 1 4\right)^2 = (-1)^2 \cdot \dfrac{1}{4^2} = \dfrac{1}{16}$

Answer: 1/16

6 0

4 years ago

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Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago