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PSYCHO15rus [73]
3 years ago
6

Solve the system of equations 5x+9y+9z=5 4x+9y+6z=10 2x+2y+5z=9

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
4 0

Answer:

{x,y,z} = {-116,28,37}

Step-by-step explanation:

// Solve equation [3] for the variable  z  

 

 [3]    5z = -2x - 2y + 9

 [3]    z = -2x/5 - 2y/5 + 9/5

__________________________________________________________

// Plug this in for variable  z  in equation [1]

  [1]    5x + 9y + 9•(-2x/5-2y/5+9/5) = 5

  [1]    7x/5 + 27y/5 = -56/5

  [1]    7x + 27y = -56

__________________________________________________________

// Plug this in for variable  z  in equation [2]

  [2]    4x + 9y + 6•(-2x/5-2y/5+9/5) = 10

  [2]    8x/5 + 33y/5 = -4/5

  [2]    8x + 33y = -4

__________________________________________________________

// Solve equation [2] for the variable  y  

 

 [2]    33y = -8x - 4

 [2]    y = -8x/33 - 4/33

__________________________________________________________

// Plug this in for variable  y  in equation [1]

  [1]    7x + 27•(-8x/33-4/33) = -56

  [1]    5x/11 = -580/11

  [1]    5x = -580

__________________________________________________________

// Solve equation [1] for the variable  x  

  [1]    5x = - 580  

  [1]    x = - 116

__________________________________________________________

// By now we know this much :

   x = -116

   y = -8x/33-4/33

   z = -2x/5-2y/5+9/5

__________________________________________________________

// Use the  x  value to solve for  y  

   y = -(8/33)(-116)-4/33 = 28

__________________________________________________________  

// Use the  x  and  y  values to solve for  z  

 z = -(2/5)(-116)-(2/5)(28)+9/5 = 37

╦────────────────────────────╦

│Hope this helped  _____________________│    

│~Derelis ____________________________ │

╨___________________________________╨                  

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Answer:

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Step-by-step explanation:

For each children, there are only two possible outcomes. Either they carry the disease, or they do not. The probability of a children carrying the disease is independent of any other children, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

The probability that their offspring will develop the disease is approximately .25.

This means that p = 0.25

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This means that n = 3

Question a:

This is P(X = 3). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 3) = C_{3,3}.(0.25)^{3}.(0.75)^{0} = 0.0156

0.0156 = 1.56% probability that all children will develop the disease.

Question b:

This is P(X = 1). So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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0.1406 = 14.06% probability that the third children will develop the disease, given that the first two did not.

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Answer:

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Step-by-step explanation:

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