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Tom [10]
3 years ago
8

Kevin is starting a saltwater aquarium with 36 fish. He wants to start with 11 times as many damselfish as clown fish. How many

of each fish will Kevin buy? How much will he pay for the fish?
Mathematics
1 answer:
Serga [27]3 years ago
4 0
33 damselfish and 3 clown fish and as for the price, you must say how much they cost
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One environmental group did a study of recycling habits in a California community. It found that 75% of the aluminum cans sold i
AysviL [449]

Answer:

a

  P( \^ p  >  0.775 ) =  0.12798

b

 P( 0.6718 < p  <  0.775 ) =0.87183

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.75

Considering question a  

     The sample size is  n  =  387

Generally the standard deviation of this sampling distribution is  

         \sigma  = \sqrt{ \frac{p(1 - p)}{ n } }    

=>      \sigma  = \sqrt{ \frac{0.75(1 - 0.75)}{ 387 } }    

=>      \sigma  = 0.022    

The sample proportion of cans that are recycled is

                 \^ p =  \frac{ 300}{387 }

=>              \^ p =  0.775

Generally the probability that 300 or more will be recycled is mathematically represented as

         P( \^ p  >  0.775 ) =  P( \frac{\^ p  -  p }{ \sigma }  >  \frac{0.775 - 0.75 }{ 0.022} )

\frac{\^ p  - p }{\sigma }  =  Z (The  \ standardized \  value\  of  \ \^ p  )

       P( \^ p  >  0.775 ) =  P( Z >  1.136  )

From the z table  the area under the normal curve to the left corresponding to  1.591   is

      P( Z >  1.136)  = 0.12798

=>    P( \^ p  >  0.775 ) =  0.12798

Considering question b

Generally the lower limit of  sample proportion of cans that are recycled is

                 \^ p_1 =  \frac{ 260 }{387 }

=>              \^ p_1  =  0.6718

Generally the upper limit of  sample proportion of cans that are recycled is

                 \^ p_2 =  \frac{ 300}{387 }

=>              \^ p_2  =  0.775

Generally probability that between 260 and 300 will be recycled is mathematically represented as

           P( 0.6718 < p  <  0.775 ) =  P( \frac{0.6718 - 0.75 }{ 0.022}<  \frac{\^ p  -  p }{ \sigma }

=>      P( 0.6718 < p  <  0.775 ) =  P( -3.55 <  Z < 1.136 )

=>        P( 0.6718 < p  <  0.775 ) = P(Z <  1.136 ) -  P( Z <  -3.55 )

From the z table  the area under the normal curve to the left corresponding to  1.136 and  -3.55  is

       P( Z <  -3.55 ) = 0.00019262

and

       P(Z <  1.136 )  = 0.87202

So

       P( 0.6718 < p  <  0.775 ) =  0.87202-  0.00019262

=>   P( 0.6718 < p  <  0.775 ) =0.87183

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