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mylen [45]
3 years ago
8

add 7. double the result. subtract 8. divide by 2. su tract the orginial selected number 1st nuber is 3 second number is 4 the t

hird number is 9 and fourth number is 12
Mathematics
1 answer:
sertanlavr [38]3 years ago
8 0
Hi,


The equation looks like this...

n + 7 \times 2 - 8 \div 2 \\ 3 + 7 \times 2 - 8 \div 2 = 3 + 14 - 4 = 13 \\ 4 + 7 \times 2 - 8 \div 2 = 4 + 14 - 4 = 14 \\ 9 + 7 \times 2 - 8 \div 2 = 9 + 14 - 4 = 19 \\ 12 + 7 \times 2 - 8 \div 2 = 12 + 14 - 4 = 22

Hope this helps.
r3t40
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Answer:

The  value is  P(X \ge  15) =  0.5

Step-by-step explanation:

From the question we are told that

    The probability of the device failing during the warranty period is p =  0.005

    The  sample size is  n = 3000

     The random variable  considered is  x  =  15

Generally this is distribution is binomial given the fact that there is only two out comes hence

      X  which is a variable representing a randomly selected selected electronic follows a binomial distribution i.e

     X  \~ \  B(n , p)

Now the mean is mathematically evaluated as

      \mu  =  n *  p

=>   \mu  = 3000 *  0.005

=>    \mu  =15

The standard deviation is mathematically represented as

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=>  \sigma  =  \sqrt{3000 *  0.005 * (1 - 0.005 )}

=>   \sigma  =  3.86

Now given that n is very large, then it mean that we can successfully apply normal approximation on this  binomial distribution

So

    P(X \ge  15) =  P( \frac{X - \mu}{\sigma }  \ge \frac{x - \mu}{\sigma } )

Now  applying  Continuity Correction we have

   P(X \ge  (15-0.5)) =  P( \frac{X - \mu}{\sigma }  > \frac{(15 -0.5) - 15}{3.86 } )

Generally  \frac{X - \mu}{\sigma }   =  Z (The \ standardized \ value  \ of  \ X)

    P(X \ge  (15-0.5)) =  P(Z >-0.130 )

From the z-table  

       P(Z >-0.130 )  =0.5

Thus  

    P(X \ge  15) =  0.5

 

   

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