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stealth61 [152]
2 years ago
12

Use the following dot plot to answer the questions.

Mathematics
1 answer:
cupoosta [38]2 years ago
8 0

Answer:

3.4

Decrease ; 3.375

Step-by-step explanation:

Mean = Σfx / Σf

Σfx = (1 * 2) + (2 * 2) + (3 * 3) + (4 * 9) = 51

Σf = 1 + 2 + 3 + 9 = 15

Mean = 51 / 15 = 3.4

If 1 more dot is added to the 3rd column:

Σfx = (1 * 2) + (2 * 2) + (3 * 4) + (4 * 9) = 54

Σf = 1 + 2 + 4 + 9 = 16

Mean = 54 / 16 = 3.375

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[Easy Question] Answer this with evidence of the answer that you provided with me!
monitta

Answer:

48

To start you will find the area as if it was a rectangle. To do that you have to do 8x12. 8x12 = 96, but to find the area of a triangle you take the area of the rectangle and divide it by 2. 96 divided by 2 is equal to 48.

There are other methods of solving this but I think this is the easiest way!

5 0
3 years ago
What are the surface area and volume of a regular triangular pyramid with a height of 9 cm and a slant height of 15 cm?
d1i1m1o1n [39]
I believe it is 135.
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3 0
3 years ago
F(x)=-5x. g(x)=8x^2-5x-9. Find: (f*g) (x)
Troyanec [42]
F(x) = -5x
g(x) = 8x² - 5x - 9

(f×g)(x) = ?
= ((-5x)(8x²-5x-9))
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5 0
3 years ago
The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell
emmasim [6.3K]

Answer:

a)

Highest (-3,-3)

Lowest (2,2)

b)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

a)

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12

completing the squares:

\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of  

\bf f(x,y)=x^2+y^2

subject to the constraint

\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0

Now we have

\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)

Let \bf \lambda be the Lagrange multiplier.

The maximum and minimum must occur at points where

\bf \nabla f=\lambda\nabla g

that is,

\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y

Replacing in the constraint

\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2

from this we get

<em>x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3 </em>

<em> </em>

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

b)

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0
3 years ago
The parking spaces in a city lot were created by painting rows of parallel line segments, then painting a line that cuts through
damaskus [11]

the answer is 65 degrees because the line in the middle is a 180 degree angle and 180-115=65. hope this helps :)

4 0
3 years ago
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