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Anarel [89]
2 years ago
7

An urn contains 5 are red balls and 6 blue balls. Suppose 5 are randomly selected without replacement. What is the probability t

hat exactly 3 are red? Answer correct to four decimal places.
Mathematics
1 answer:
Viefleur [7K]2 years ago
5 0

Answer:

27.94%

Step-by-step explanation:

The statement tells us that we have 5 red balls and 6 blue balls, that is, there are 11 in total (5 + 6)

So the probability of red balls = 5/11 and blue balls probability = 6/11

Let X be the number of red balls of those 5 selected balls.

 Then X follows a binomial distribution with the following parameters:

n = 5

p = 5/11

q = 6/11

P (X) = nCx * p ^ (x) * q ^ (n -x)

required probability is P (X = 3), replacing:

P (X = 3) = 5C3 * (5/11) ^ (3) * (6/11) ^ (5 -3)

P (X = 3) = 5! / (3! (5-3)!) * 0.02794

P (X = 3) = 10 * 0.02794

P (X = 3) = 0.2794

Which means that the probability is 27.94%

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56+8x
Luba_88 [7]
Get the a99 ph0tomath
4 0
2 years ago
Evaluate the triple integral ∭ExydV where E is the solid tetrahedon with vertices (0,0,0),(5,0,0),(0,9,0),(0,0,4).
Elan Coil [88]

Answer: \int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV = 1087.5

Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.

An equation of a plane is found with a point and a normal vector. <u>Normal</u> <u>vector</u> is a perpendicular vector on the plane.

Given the points, determine the vectors:

P = (5,0,0); Q = (0,9,0); R = (0,0,4)

vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)

vector QR = (0,9,0) - (0,0,4) = (0,9,-4)

Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:

n = PQ × QR = \left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right]\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right]

n = 36i + 0j + 45k - (0k + 0i - 20j)

n = 36i + 20j + 45k

Equation of a plane is generally given by:

a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0

Then, replacing with point P and normal vector n:

36(x-5) + 20(y-0) + 45(z-0) = 0

The equation is: 36x + 20y + 45z - 180 = 0

Second, in evaluating the triple integral, set limits:

In terms of z:

z = \frac{180-36x-20y}{45}

When z = 0:

y = 9 + \frac{-9x}{5}

When z=0 and y=0:

x = 5

Then, triple integral is:

\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx

Calculating:

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx

\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}}  \, dy } \, dx

\frac{1}{45} \int\limits^5_0  {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx

\frac{1}{45} \int\limits^5_0  {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4}  } \, dx

\frac{1}{45} [30375-60750+118462.5-39150]

\int\limits^5_0 {\int\limits {\int\ {xyz}  \, dy } \, dx = 1087.5

<u>The volume of the tetrahedon is 1087.5 cubic units.</u>

3 0
3 years ago
An isosceles triangle has an angle that measures 98°. Which other angles could be in that isosceles triangle? Choose all that ap
Tamiku [17]

an isosceles triangle is a triangle with two twin sides, and therefore two twin angles.

if one angle is 98°, well, we can't have another 98° angle because that makes 98+98 = 196°, and the sum of all interior angles in a triangle cannot be more than 180°.

so the other remaining angles must be the twin smaller angles, namely 180 - 98 = 82, and each twin takes half of that, namely 41°.


Check the picture below.

4 0
2 years ago
Help me with this plz
inessss [21]

Answer:

6/7

Step-by-step explanation:

x +(-x) =0

5+1= 6

2+5= 7

hopefully that's all correct

6 0
2 years ago
Determine the maximum value of the objective function, P.
Ad libitum [116K]

Answer:

Maximum (250,125) Answer

Step-by-step explanation:

This is more of a graphing problem than it is anything else.

Begin by graphing all 4 given equations.

When you do that, mark the intersection points of at least 2 lines. In this case it is exactly 2 lines for each intersecting point.

6x + 4y <= 2000 and 2x + 4y <= 1000 intersect at (250,125)

x=>0 and y=>0 intersect at (0,0)

6x + 4y <=2000 and x => 0 intersect at 333.333

2x + 4y <=1000 and y>=0 intersect at  0,250.

Any other intersection points fall outside the range of the givens. The shaded part we are interested in is sort of a very dark green/blue. It is the interior of the quadrilateral determined by the 4 vertices that are marked.

Now all you have to do is determine the maximum point using P=15x + 12y

For 0,0                 P = 15*0 + 12,0 = 0

For 0,250            P = 15*0 + 12*250 = 3000

For 333.3333,0   P = 15*333.3333 + 12*0 = 5000 rounded.

For 250,125        P = 15*250 + 12*125 = 5250  Which is the maximum

Answer (250,125) produces the maximum value Answer

5 0
3 years ago
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