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grin007 [14]
4 years ago
12

Andrea has just created a scale drawing of her bedroom room which is 18 feet long by 12 feet wide. The scale drawing shows the l

ength as 6 inches. What is the width on the scale drawing?
Mathematics
1 answer:
vivado [14]4 years ago
7 0

Width of the scale drawing is 4 inches

Step-by-step explanation:

We can solve this problem by using the rule of three.

In fact, the actual size of the bedroom are:

L=18 ft (length)

W=12 ft (width)

In the drawing instead, the actual sizes on scale are

L' = 6 in (length in the scale drawing)

W' (width in the scale drawing)

Therefore, applying the rule of three,

\frac{L'}{L}=\frac{W'}{W}

And solving for W', we find the width on the scale drawing:

W'=\frac{L'W}{L}=\frac{(6 in)(12 ft)}{18 ft}=4 in

So, the width on the scale drawing is 4 inches.

Learn more about scales and rule of three:

brainly.com/question/570757

#LearnwithBrainly

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You roll a standard number cube once find p(not multiple of 2)
GalinKa [24]

Answer: \frac{1}{2}

Step-by-step explanation:

Let's look at the numbers that we count.

The question says not multiple of 2, so we have to count to following numbers:

1 , 3 , 5

And in total there are 6 numbers, so our probability is

\frac{3}{6} = \frac{1}{2}

7 0
3 years ago
We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent
Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(c) P(X = −2) = 1/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

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Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

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                  P ( X = -2 ):  P ( X_2 = 1 ) * P ( X_1 = 3 )

                                 :  ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 1 / 9 )

   

                  ( X = -1 ):  { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

                 P ( X = -1 ):  P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 2 / 9 )

         

       ( X = 0 ):  { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } +  { X_2 = 3 , X_1 = 3 }

       P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 3 / 9 ) = ( 1 / 3 )

       

                    ( X = 1 ):  { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

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                                 :  ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

                                 : ( 2 / 9 )

                    ( X = 2 ):  { X_2 = 1 , X_1 = 3 }

                  P ( X = 2 ):  P ( X_2 = 3 ) * P ( X_1 = 1 )

                                    :  ( 1 / 3 ) * ( 1 / 3 )

                                    : ( 1 / 9 )                  

- The distribution Y = X_2,

                          P(Y=0) = 0

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                          P(Y=2) = 1/ 3

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