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joja [24]
3 years ago
12

Question

Mathematics
1 answer:
meriva3 years ago
4 0

Question:

In a neighbourhood pet show, each of the animals entered is equally likely to win. if there are 7 dogs, 6 cats, 3 birds, and 2 gerbils entered, what is the probability that a bird will win the top prize?

Answer:

Probability that a bird will win the top prize is 0.167

Step-by-step explanation:

Given:

The number of  dogs  = 7

The number of  cats = 6

The number of  birds  = 3

The number of gerbils = 2

To Find:

Probability that a bird will win the top prize = ?

Solution:

Let us first find the total number of  pets .

The Total number of pets = 7 + 6 + 3  + 2 =  18

Now the probability of  a bird will win the top prize is

=> \frac{\text {Number of birds}}{\text{ Total number of pets}}

=>\frac{3}{18}

=> \frac{1}{6}

=>0.167

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{4x-2y+5z=6 <br> {3x+3y+8z=4 <br> {x-5y-3z=5
lesya692 [45]

There are three possible outcomes that you may encounter when working with these system of equations:


  •    one solution
  •    no solution
  •    infinite solutions

We are going to try and find values of x, y, and z that will satisfy all three equations at the same time. The following are the equations:

  1. 4x-2y+5z = 6
  2. 3x+3y+8z = 4
  3. x-5y-3z = 5

We are going to use elimination(or addition) method

Step 1: Choose to eliminate any one of the variables from any pair of equations.

In this case it looks like if we multiply the third equation by 4 and  subtracting it from equation 1, it will be fairly simple to eliminate the x term from the first and third equation.

So multiplying Left Hand Side(L.H.S) and Right Hand Side(R.H.S) of 3rd equation with 4 gives us a new equation 4.:

4. 4x-20y-12z = 20      

Subtracting eq. 4 from Eq. 1:

(L.HS) : 4x-2y+5z-(4x-20y-12z) = 18y+17z

(R.H.S) : 20 - 6 = 14

5. 18y+17z=14

Step 2:  Eliminate the SAME variable chosen in step 2 from any other pair of equations, creating a system of two equations and 2 unknowns.

Similarly if we multiply 3rd equation with 3 and then subtract it from eq. 2 we get:

(L.HS) : 3x+3y+8z-(3x-15y-9z) = 18y+17z

(R.H.S) : 4 - 15 = -11

6. 18y+17z = -11

Step 3:  Solve the remaining system of equations 6 and 5 found in step 2 and 1.

Now if we try to solve equations 5 and 6 for the variables y and z. Subtracting eq 6 from eq. 5 we get:

(L.HS) : 18y+17z-(18y+17z) = 0

(R.HS) : 14-(-11) = 25

0 = 25

which is false, hence no solution exists



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2 years ago
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lbvjy [14]

Answer:

RQ = 18

Step-by-step explanation:

KR is an angle bisector and divides the opposite side into segments that are proportional to the other two sides, that is

\frac{KP}{KQ} = \frac{PR}{RQ} , substitute values

\frac{50}{45} = \frac{38-x}{x} ( cross- multiply )

45(38 - x) = 50x ← distribute left side

1710 - 45x = 50x ( add 45x to both sides )

1710 = 95x ( divide both sides by 95 )

18 = x

That is RQ = 18

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