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postnew [5]
3 years ago
12

A scatterplot consists of (1, 4.0), (2, 3.3), (3, 3.8), (4, 2.6), and (5, 2.7). The line of best fit used to model the data is y

= –0.33x + 4.27. Which residual plot is correct? A graph shows the horizontal axis numbered 1 to 5 and the vertical axis numbered negative 2 to positive 2. Points are scattered on and below the horizontal axis. A graph shows the horizontal axis numbered 1 to 5 and the vertical axis numbered negative 2 to positive 2. Points are scattered on the horizontal axis. A graph shows the horizontal axis numbered 1 to 5 and the vertical axis numbered negative 2 to positive 2. Points are scattered on and above the horizontal axis.
Mathematics
2 answers:
zhenek [66]3 years ago
8 0

Answer:

b

Step-by-step explanation:

Phantasy [73]3 years ago
3 0

Answer:

Its the second one

Step-by-step explanation:

I got it correct

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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then  

\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

So

\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

and so

\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0
3 years ago
Help with this question anyone?
Art [367]
Adding 49 will explain it properly just do that and you will get the answers
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For the first one the answer is x = -3/2y - 7 and the second answer is x = 1/6y + -11/3
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Answer:

14.64

Step-by-step explanation:

A negative times a negative equals a positive and there are 2 decimal places.

8 0
3 years ago
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