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inysia [295]
3 years ago
8

Artemis has six boards cut in various lengths as shown she wants to use three of them to frame a triangular window in her new cl

ubhouse how many unique triangles can Artemis make with the lengths she has
Mathematics
1 answer:
SOVA2 [1]3 years ago
3 0

Question:

The sizes of the length of the boards are as follows;

Number of boards with length, 2 ft   = 2

Number of boards with length, 6 ft   = 2

Number of boards with length, 9 ft   = 2.

Answer:

The answer to the question is;

The number of unique triangles Artemis can make with the lengths she has is four.

Step-by-step explanation:

The sizes of the six boards are as follows

2 by 9 ft length boards

2 by 6 ft length boards

2 by 2 ft length boards

Therefore since the sum of the lengths of the other two sides of a triangle must be greater than the third side, we have the following possible unique triangles

9 ft by 9 ft by 6 ft ........(1)

9 ft by 9 ft by 2 ft ........(2)

6 ft by 6 ft by 9 ft ........(3)

6 ft by 6 ft by 2 ft ........(4)

Therefore we have four unique triangles.

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Troyanec [42]

Answer:

E(X)=3.125

Step-by-step explanation:

We are given that two four sided dice.

Then , the sample space

{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)}

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Let the random variable X represent the maximum value of the two dice

Outcomes                            X                                 P(X)

(1,1)                                          1                                 1/16

(1,2),(2,1),(2,2)                                   2                        3/16

(1,3),(2,3),(3,1),(3,2),(3,3)                    3                           5/16  

(1,4),(3,4) ,(2,4),(4,1),(4,2),(4,3),(4,4)    4                            7/16

Using the probability formula

P(E)=\frac{Favorable\;outcomes}{Total\;number\;of\;outcomes}

Now,

E(X)=\sum_{i=1}^{n}x_iP(x_i)

E(x)=1(1/16)+2(3/16)+3(5/16)+4(7/16)

E(x)=\frac{1+6+15+28}{16}

E(x)=\frac{50}{16}=3.125

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