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ozzi
4 years ago
13

Four girls and eight boys are running for president or vice president of the student council. find each probability: find the pr

bability that two girls are elected
Mathematics
1 answer:
pochemuha4 years ago
6 0
The probability is 1/11.

The probability that a girl is elected for one office is 4/12, since there are 4 girls out of 12 students.

The probability that a girl is elected for the second office is 3/11.  This is because after one girl is elected, there are 3 girls left out of 11 students left.

Together, this gives us
4/12(3/11) = 12/132 = 1/11
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SpyIntel [72]
There are no numbers on the graph, so we cannot tell for sure the value of f(3).

If the grid squares are each one unit, then the graph of f(x) appears to go through the grid point (3, -1). This means f(3) = -1.

Using this value in the expression for g(x), we have
.. g(3) = -4*f(3) +7
.. g(3) = -4*(-1) +7
.. = 4 +7
.. = 11

The value of g(3) is 11.

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3 years ago
A student received a coupon for ​%18 off the total purchase price at a clothing store. Let y be the original price of the purcha
kiruha [24]

Answer:

Final cost of item = initial price(1 – rate)

Step-by-step explanation:

y(1 – .18) = .82y

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3 years ago
Can someone please help me
ale4655 [162]

Answer:

-4

Step-by-step explanation:

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3 years ago
Read 2 more answers
If c is the line segment connecting (x1,y1) to (x2,y2), show that the line integral of xdy-ydx=x1y2-x2y1......this is in a chapt
MatroZZZ [7]
Green's theorem doesn't really apply here. GT relates the line integral over some *closed* connected contour that bounds some region (like a circular path that serves as the boundary to a disk). A line segment doesn't form a region since it's completely one-dimensional.

At any rate, we can still compute the line integral just fine. It's just that GT is irrelevant.

We parameterize the line segment by

\mathbf r(t)=\langle x_1,y_1\rangle(1-t)+\langle x_2,y_2\rangle t
\implies\mathbf r(t)=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle

with 0\le t\le1. Then we find the differential:

\mathbf r(t)\equiv\langle x,y\rangle=\langle x_1+(x_2-x_1)t,y_1+(y_2-y_1)t\rangle
\implies\mathrm d\mathbf r\equiv\langle\mathrm dx,\mathrm dy\rangle=\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt

with 0\le t\le1.

Here, the line integral is

\displaystyle\int_{\mathcal C}x\,\mathrm dy-y\,\mathrm dx=\int_{\mathcal C}\langle-y,x\rangle\cdot\langle\mathrm dx,\mathrm dy\rangle
=\displaystyle\int_{t=0}^{t=1}\langle-y_1-(y_2-y_1)t,x_1+(x_2-x_1)t\rangle\cdot\langle x_2-x_1,y_2-y_1\rangle\,\mathrm dt
=\displaystyle\int_{t=0}^{t=1}(x_1y_2-x_2y_1)\,\mathrm dt
=(x_1y_2-x_2y_1)\displaystyle\int_{t=0}^{t=1}\,\mathrm dt
=x_1y_2-x_2y_1

as required.
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