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2 years ago

What is the difference of the means of the distributions?

Mathematics

1 answer:

alexdok [17]2 years ago

4 0

The different of the means of the distribution will be 15. Then the correct option is A.

<h3>What is a normal distribution?</h3>

The Gaussian Distribution is another name for it. The most significant continuous probability distribution is this one. Because the curve resembles a bell, it is also known as a bell curve.

the graph is shown below.

Then the different of the means of the distribution will be

⇒ 45 - 30

⇒ 15

Then the correct option is A.

More about the normal distribution link is given below.

brainly.com/question/12421652

#SPJ1

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Explain your answer !! Have a nice day Will give braisnlt

DochEvi [55]

Answer:

(-2,3)

Step-by-step explanation:

Step 1. Isolate x for 3x + 5y = 9

3x + 5y - 5y = 9 - 5y

3x = 9 - 5y

3x/3 = 9/3 - 5y/3

x= (9-5y)/3

Step 2. Simplify

-3*((9-5y)/3) + 3y = 15

-3*((9-5y)/3) = 9 - 5y = -(-5 + 9) + 3y

-9+5y+3y

-9+8y = 15

Step 3 Isolate y for -9+8y=15

-9+8y+9=15+9

8y=24

8y/8 = 24/8

y=3

Step 4 Substitute y = 3

x = (9-5*3)/3

-(6/3) = (-2)

x = -2

7 0

2 years ago

Please help it’s very urgent and I need the right answer

8090 [49]

Answer:

3x + 15 = 3*10 + 15 = 30 + 15 = 45

3x + 15 = 45

Now,

2x + 25 = 2*10 + 25 = 20 + 25 = 45

2x + 25 = 45

Step-by-step explanation:

3x + 15 + 2x + 25 = 90

or, 5x + 40 = 90

or, 5x = 90 - 40

so, 5x = 50

so, x = 50/5 = 10

3 0

2 years ago

Simplify the fraction 6/9

irina [24]

6/9=2/3 because we divide both sides by 3
hope this helps

5 0

2 years ago

Please help w this! Its a calculus question! look at the picture for the problem,

neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

$x^2+y^2=36\implies x=\sqrt{36-y^2}$

Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18$

Then the area of the entire blue region is 4 times this, a total of $\boxed{36\pi-72}$ .

Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18$

so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is