I will give brainiest to whoever answers correctly !!

Which expression shows 18 + 15 written as a product of two factors?

____ [38]

1st one is ur answer

6 0

3 years ago

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Suppose that 12% of test scores for the unit 1 test were above 85 and 8% were below 70. Assuming a normal

Doss [256]

Answer:

Mean = 78.2

Standard deviation = 5.8

Step-by-step explanation:

Mathematically z-score;

= (x-mean)/SD

From the question;

12% of test scores were above 85

Thus;

P( x > 85) = 12%

P(x > 85) = 0.12

Now let’s get the z-score that has a probability of 0.12

This can be obtained from the standard normal distribution table and it is = 1.175

Thus;

1.175 = (85 - mean)/SD

let’s call the mean a and the SD b

1.175 = (85-a)/b

1.175b = 85 - a

a = 85 - 1.175b ••••••••(i)

Secondly 8% of scores were below 70

Let’s find the z-score corresponding to this proportion;

We use the standard normal distribution table as usual;

P( x < 70) = 0.08

z-score = -1.405

Thus;

-1.405 =( 70-a)/b

-1.405b = 70-a

a = 70 + 1.405b ••••••(ii)

Equate the two a

70 + 1.405b = 85 - 1.175b

85 -70 = 1.405b + 1.175b

15 = 2.58b

b = 15/2.58

b = 5.81

a = 70 + 1.405b

a = 70 + 1.405(5.81)

a = 78.16

So mean = 78.2 and Standard deviation is 5.8

4 0

3 years ago

Could someone answer these 3 questions?h

Aleks04 [339]

9) x= -6
10) x= 3/2
11) x= -4,0,5
12) no solution

8 0

3 years ago

What is a fraction equivalent to -6/1

pashok25 [27]

Answer:

-18/3

Step-by-step explanation:

Multiply the denominator and numerator by the same whole number.

8 0

3 years ago

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Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago