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mrs_skeptik [129]
2 years ago
11

A bag contains five yellow tickets numbered one to five. The bag also contains five green tickets numbered one to five. You rand

omly pick a ticket. It is green or has a number greater than four. Find the probability of this occuring.
Mathematics
1 answer:
tamaranim1 [39]2 years ago
5 0

The probability of picking a ticket that is green or has a number greater than four is 3/5

<h3>How to determine the probability?</h3>

The given parameters are:

Yellow = 1 - 5

Green = 1 - 5

Total = 10

There are 2 cards whose numbers are greater than 4 i.e. Yellow 5 and Green 5

So, we have:

P(Number greater than 4) = 2/10

There are 5 green cards.

So, we have:

P(Green) = 5/10

Also, 1 green card is numbered greater than 4

So, we have:

P(Green greater than 4) = 1/10

The required probability is:

P = P(Green) + P(Number greater than 4) - P(Green greater than 4)

This gives

P = 5/10 + 2/10 - 1/10

Evaluate

P = 6/10

Simplify

P =3/5

Hence, the probability of picking a ticket that is green or has a number greater than four is 3/5

Read more about probability at:

brainly.com/question/11234923

#SPJ1

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Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant
Alex787 [66]

Find the intercepts for both planes.

Plane 1, <em>x</em> + <em>y</em> + 2<em>z</em> = 2:

y=z=0\implies x=2\implies (2,0,0)

x=z=0\implies y=2\implies(0,2,0)

x=y=0\implies 2z=2\implies z=1\implies(0,0,1)

Plane 2, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 8:

y=z=0\implies4x=8\implies x=2\implies(2,0,0)

x=z=0\implies4y=8\impliesy=2\implies(0,2,0)

x=y=0\implies z=8\implies(0,0,8)

Both planes share the same <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (<em>x</em>, <em>y</em>)-plane where <em>z</em> = 0, we see the bounded region projects down to the triangle in the first quadrant with legs <em>x</em> = 0, <em>y</em> = 0, and <em>x</em> + <em>y</em> = 2, or <em>y</em> = 2 - <em>x</em>.

So the volume of the region is

\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx

=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}

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3 years ago
Grant attended 75% of his Student Council meetings. If there were 40 meetings, how many did Grant NOT attend?
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Answer:

10

Step-by-step explanation:

take your total and multiply it by the percentage.

40 × .75 = 30. he attended 30 meetings. take your total and subtract your new value.

40 - 30 = 10 meetings not attended

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A linear equation is RISE over RUN (y/x) so the starting point would be -6 and from there its a straight line across
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Find the value of x.
svetoff [14.1K]

Answer:

68

Step-by-step explanation:

1) Find the interior angle using relations of angles in straight line I.e ( sum of angles in a straight line is 180 ) and we know the sum of all the interior angle of quadrilateral is 360 degree .

2) Solve further for x.

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Is it no real solutions?
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