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madreJ [45]
3 years ago
7

The perimeter of a basket ball court is 108 m and the length is 6 m long than twice the width. What are the length and the width

?
Mathematics
2 answers:
Yanka [14]3 years ago
8 0

38 is the correct awnser

Gnoma [55]3 years ago
3 0
Let's say

P = perimeter, L : Length and w : width

Then

P=2(L+w)=2L+2w=108

But 

L=6+2w

So

2L+2w=108\\2(6+2w)+2w=108\\12+4w+2w=108\\12+6w=108\\6w=108-12\\6w=96\\w=16

L=6+2w\\L=6+2(16)\\L=6+32\\L=38

We verify :

2L+2w=108\\2(38)+2(16)=108\\76+32=108\\108=108

Length = L = 38
Width = w = 16

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Answer:

a

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b

P(B) = \frac{15}{36}

c

P(U) = \frac{11}{36}

Step-by-step explanation:

From the question we are told that

   The  number of dice is  n  =  3

Generally  for all three dice to show the same number the second and the third dice must have the same outcome  as the outcome of the first dice.

This means that the number of outcome the first die can  have is  6  (6 sides), the number of outcome which the second and the third dice can have is  1 (they must match the first )

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Generally for two of the three dice show the same number after the first roll then the number of outcome for  one of the  dice would be is  6 , the number of outcome for another one must be   1(i.e it must match the first ) , the number of outcome for the remaining one must be   5 (i.e it can show any of the remaining  5  sides which the first and second dice are not showing )

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=>  P(B) = \frac{15}{36}

Generally from the question we are told that if two dice match, the player re-rolls the die that does not match.

Now the probability that  the die that did not match the first  time will match the second time is

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Generally if that one die does not show the same number in the second round , the probability that it will match  in the third round is

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Generally the probability that he wins (i.e when all three are showing the same number ) and  exactly two is showing the same number is mathematically represented as

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=>    P(K) =  \frac{1}{6} * \frac{15}{36}  +  \frac{5}{6} *  \frac{1}{6} * \frac{15}{36}

=>  P(K)= \frac{165}{1296}

So the probability that the player wins, conditioned on exactly two of the three dice showing the same number after the first roll is mathematically represented as  

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=>  P(U) = \frac{11}{36}

     

 

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