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OverLord2011 [107]
3 years ago
11

Log4(3^-5)=x solve for x

Mathematics
1 answer:
andrew-mc [135]3 years ago
3 0

Answer:

.0024776131

Step-by-step explanation:

type it into a TI-84 and remember than any negative exponent turns the number into a fraction

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 3) is 3v + 4

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A number is chosen at random from the first 100 positive integers. Find the probability that the number is divisible by 5. 1/10
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The number of multiples of 5 from 1 to 100 is:

1≤5n≤100

1/5≤n≤20, since n must be an integer

1≤n≤20

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. Lightning Strikes It has been said that the probability of being struck by lightning is about 1 in 750,000, but under what cir
mrs_skeptik [129]

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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