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2 years ago

Which data sets have outliers? Check all that apply

Mathematics

2 answers:

dedylja [7]2 years ago

6 0

Answer:

C.16, 32, 38, 39, 41, 42, 58

and

E.18, 30, 34, 38, 43, 45, 68

sertanlavr [38]2 years ago

3 0

its C. & E.

they both have outliers lol

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Find the volume of a cylinder with a diameter of 10 inches and a height of 20 inches. Use 3.14 for π.

Snowcat [4.5K]

Lets find the volume of a cylinder with a diameter D=10 inches and lenght L= 20 inches. First we need to notice that the diameter is twice the size of the radius: D=2r. Then we write the equation for the volume of the cylinder. it is the base times the height: V=pi*r^2 * L. Now we input the nubmers in the equation: V=3.14*(D/2)^2 * L where r=D/2=5 inches and after calculating we get: V=1570 inches^3

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2 years ago

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-10x - 135 = 117 + 2x A. 25 B. -18 C. -21 D. 22 Helps pls

Lina20 [59]

Answer:

Step-by-step explanation:

Add 10x to both sides:

-10x - 135 = 117 + 2x

-10x - 135 + 10x = 117 + 2x + 10x

-135 = 117 + 12x

Subtract 117 from both sides:

-135 - 117 = 117 + 12x - 117

-252 = 12x

Divide both sides by 12:

$\frac{-252}{12} = \frac{12x}{12}$

-21 = x

So the answer is C.

Hope this helps!

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2 years ago

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3x^2-4x+1, match the following

il63 [147K]

Answer:

x = variable

3 = coefficient

1 = constant

3x^2-4x+1 = algebraic expression

2 = degree

Step-by-step explanation:

variables are typically the letters in equations (usually x and y)

coefficients are the numbers attached to the variables

constants don't have any variables attached

algebraic expressions are the full expression

degrees are the small numbers in the top right corners on either constants, variables or coefficients

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2 years ago

(m) रेखीय फलन (Linear function)

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2 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial:

$\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3$

$\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4$

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2 years ago