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Kaylis [27]
3 years ago
13

What is the sample space for the probability model?

Mathematics
2 answers:
Sonja [21]3 years ago
7 0

Answer:

sample space for a probability model is the set of all possible outcomes

muminat3 years ago
3 0
Sample space for the probability model is the set of all possible outcomes. ❤️
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Will mark bianleast
Gekata [30.6K]

Answer:r=16 ft , d =32 ft (a)

Step-by-step explanation:

c=2 pi r

32pi= 2 x pi x r

r=32 pi/2pi

r= 16

diameter = radius x 2

d= 16 x 2

d = 32

3 0
2 years ago
In the translation T of the graph below, use the figure to describe the following transformation. T T-1: (x, y) ( x, y) ( x + 3,
Serhud [2]

Answer:

(X - 3,  Y - 1).

3 0
3 years ago
Please help answer all of the following multi-step equations
Triss [41]

Answer: first one x <-2

second one x>1

third one Use the given functions to set up and simplify

(1-4x)-x.32+4x<-3=(1-4x)-x= -5x+1

fourth one

13/14

Step-by-step explanation:

I can't rlly explain I just do, I hope this helps

8 0
2 years ago
A six-sided die is rolled 50 times.
grandymaker [24]

Answer:

0.36

Step by-step explanation:

18 appearances of the number 4

------------------------------------------------ = 18/50 = 36/100 = 0.36

                 50 rolls

That experimental probability is 0.36.

3 0
3 years ago
38. Evaluate f (3x +4y)dx + (2x --3y)dy where C, a circle of radius two with center at the origin of the xy
lina2011 [118]

It looks like the integral is

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy

where <em>C</em> is the circle of radius 2 centered at the origin.

You can compute the line integral directly by parameterizing <em>C</em>. Let <em>x</em> = 2 cos(<em>t</em> ) and <em>y</em> = 2 sin(<em>t</em> ), with 0 ≤ <em>t</em> ≤ 2<em>π</em>. Then

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}

Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on <em>C</em> nor in the region bounded by <em>C</em>, so

\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy

where <em>D</em> is the interior of <em>C</em>, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: -2\times \pi\times2^2 = -8\pi.

3 0
2 years ago
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