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krek1111 [17]
3 years ago
9

Tim mixed 2 parts vinegar and 5 parts water to make a cleaning liquid how many parts of vinegar did Tim mix with each part of wa

ter?
Mathematics
2 answers:
irga5000 [103]3 years ago
7 0
5 parts of water plus 2 parts of vinegar.

1 part of water =  (2/5) parts of vinegar

                          = 0.40 parts of vinegar.

So Tim mixed 0.40 parts of vinegar with each (1) part of water.  
Licemer1 [7]3 years ago
5 0

Answer:  Tim mixed 0.4 parts of vinegar with each part of water.  

Step-by-step explanation:  Given that Tim mixed 2 parts vinegar and 5 parts water to make a cleaning liquid.

We are to find the number of parts of vinegar that Tim mix with each part of water.

We will be using the UNITARY method to solve the problem.

Number of parts of vinegar mix with 5 parts of water = 2.

Therefore, the number of parts of vinegar mix with 1 part of water i given by

\dfrac{2}{5}=0.4.

Thus, Tim mixed 0.4 parts of vinegar with each part of water.  

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

      P(U |D ) = 0.198

b

   P(O\ n \ B) = 0.188

c

  P(O | B) =   0.498

Step-by-step explanation:

The total number of deaths is mathematically represented as

      T   =  16 +  23 + \cdots  +  16

        T   =623

The total number of deaths in 1996 - 2000 is mathematically represented as

     T_a =  16 +  23+ \cdots + 30

      T_a = 235

The total number of deaths in 2001 - 2005 is mathematically represented as

     T_b =  17 +  16 + \cdots + 23

      T_b = 206

The total number of deaths in 2006 - 2010 is mathematically represented as

     T_c =  15 +  17 + \cdots + 16

      T_d = 182

Generally the the probability that it would occur under the tree given that the death was  after  2000 is mathematically represented as

     P(U |D ) = \frac{P(A \ n\  U )}{P(A)}

Here  P(A \ n\  U ) represents the probability that it was after 2000 and it was under the tree and this is mathematically represented as

         P(A \ n\  U )   = \frac{Z}{ T}

Here Z is the total number of death under the tree after 2000 and it is mathematically represented as

         Z =  35 +  42

=>       Z =  77

=>       P(A \ n\  U )   = \frac{77}{ 623}

=>      

Also

     P(A) is the probability of the death occurring after 2000  and this is mathematically represented as

        P(A) =  \frac{T_b  +  T_c}{ T}

=>      P(A) =  \frac{ 206+  182}{623}

=>  

So

         P(U |D ) = \frac{\frac{77}{ 623} }{ \frac{ 206+  182}{623}}

=>      P(U |D ) = 0.198

Generally the probability that the death was from camping or being outside and was before 2001 is mathematically represented as

      P(O | B) = \frac{T_z}{ T}

Here T_z is the total number of death outside / camping before 2001  and the value is  117  

So

            P(O \ n \ B) = \frac{117}{623}

=>          P(O\ n \ B) = 0.188

Generally the probability that the death was from camping or being outside given that it was before 2001 is mathematically represented as

       P(O | B) =  \frac{ P( O \ n \ B)}{ P(B)}

Here P(B) is the probability that it was before 2001 , this is mathematically represented as  

          P(B ) =  \frac{T_a}{T}

=>       P(B ) =  \frac{235}{623}

So

          P(O | B) =  \frac{ \frac{117}{623}}{ \frac{235}{623}}

=>       P(O | B) =   0.498

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