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Serga [27]
3 years ago
6

10 girls...and 8 boys... teacher picks 3....probability of teacher picking all 3 girls?

Mathematics
1 answer:
taurus [48]3 years ago
7 0

Answer:

There is a 14.71% probability of teacher picking all 3 girls.

Step-by-step explanation:

Initially, there are 18 students, of which 10 are girls. So there is a 10/18 = 5/9 probability that the first student chosen is a girl.

Now, since the first student chosen was a girl, there are 17 students, of which 9 are girls. So there is a 9/17 of the second student chosen being a girl.

Both students picked were girls. Now there are 16 students, of which 8 are girls. So there is a 8/16 = 1/2 probability of the third student chosen in this situation being a girl.

Probability of teacher picking all 3 girls?

We multiply these probabilities:

P = \frac{5}{9}*\frac{9}{17}*\frac{1}{2} = 0.1471

There is a 14.71% probability of teacher picking all 3 girls.

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Which of the equations below contains the two points listed? (-8,-3) and (-8,1) A. y=8 B. x=-8 C. x=8 D. y=-8
victus00 [196]

Answer:

Option B.

Step-by-step explanation:

If a line passing through two points, then equation of line is

y-y_1=\dfrac{y_2-y_1}{x_2-x_1}(x-x_1)

It is given that the line passing through (-8,-3) and (-8,1). So, the equation of line is  

y-(-3)=\dfrac{1-(-3)}{-8-(-8)}(x-(-8))

y+3=\dfrac{4}{0}(x+8)

The given line is a vertical line because slope =\dfrac{4}{0}=\infty.

0=4(x+8)

0=(x+8)

x=-8

So, the equation of line is x=-8.

Therefore, the correct option is B.

6 0
3 years ago
Read 2 more answers
For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

which means \vec f is indeed conservative and we can find f.

Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

\implies g(x,z)=4\sin(xz^2)+h(z)

Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

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3 years ago
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valentinak56 [21]

The answer would be

y-5=2(x-5)

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What is the solution to the square root of 3x-6 equals 3
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Answer:

x=12

Step-by-step explanation:

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