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yKpoI14uk [10]
3 years ago
5

17,280 meal worms

Mathematics
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

a) 30 000 000

b) -100 000 000

c) -160 000 000

d) 27 500 000

Step-by-step explanation:

You want to know how many bacteria are killed every 6 hours which is 1/10 so 50 000 000:10= 5 000 000 every 6 hours.

a) how many times does 6 hours fit in 1 day) 24÷6=4 so 4×5 000 000= 20 000 000 so then you subtract that from 50 000 000 because you want to know what remains which means 30 000 000 bacteria remain.

b) we already calculated 1 day so 30 000 000×5=150 000 000. Then you subtract that from the 50 000 000. 50 000 000- 150 000 000= -100 000 000

c) 1 day× 7= 30 000 000×7= 210 000 000- 50 000 000= -160 000 000

d) 5 000 000÷2= 2 500 000= 30 000 000- 2 500 000= 27 500 000 6 hours divided by 2 equals 3 hours so you also divide the bacteria by 2.

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Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

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

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

  Here  \^ p is the sample proportion  of persons with a college degree.

So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

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

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 }  <  Z  < \frac{-0.47 + 0.52}{0.02106 }]

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  P(Z <  2.37 ) = 0.9911

and

  P(Z <  - 2.37 ) = 0.0089

So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

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