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stepan [7]
3 years ago
7

Abdul and his wife are each starting a saving Plan. Abdul Will initially set aside $50 and then add $50.85 every week to the sav

ings. the amount A (in dollars) saved this way is given by the function A= 50.85n+50 where N is the number of weeks he has been saving . His wife will not set an initial amount side but will add $75.65 to the savings every week the amount B (in dollars) saved using this plan is given by the function B=75.65N. Let T be the total amount (in dollars) saved using both plants combined right in equation relating T to N. simplify your answer as much as possible.
Mathematics
1 answer:
slamgirl [31]3 years ago
6 0

Answer:T(n)=126.5n+50

Explanation: We are adding two linear functions: A the original function describing Abdul's deposits, and B the deposits made by his wife.

A(n) = 50.85n+50\\B(n)=75.65n\\T(n)=A(n)+B(n) = 50.85n+50 + 75.65n= (50.85+ 75.65)n+50=126.5n+50\\T(n)=126.5n+50

The resulting function is a larger slope due to the added deposit rate.


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Answer:

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 The standard error is  SE   =0.020

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  200

     The number of defective is  k =  18

The null hypothesis is  H_o  :  p  =  0.08

The  alternative hypothesis is  H_a  :  p > 0.08

Generally the sample proportion is mathematically evaluated as

            \r p =  \frac{18}{200}

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Given that the confidence level is  95% then the level  of significance is mathematically evaluated as

        \alpha  =  100 -  95

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Next we obtain the critical value of  \frac{ \alpha }{2} from the normal distribution table, the value is  

        Z_{\frac{\alpha }{2} } =  1.96

Generally the standard of error is mathematically represented as

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

substituting values

         SE   =  \sqrt{\frac{0.09  (1 -  0.09)}{200} }

        SE   =0.020

The  margin of error is  

       E =  Z_{\frac{ \alpha }{2} }  * SE

=>    E =  1.96  *  0.020

=>   E =  0.0397

The  95% confidence interval is mathematically represented as

     \r p  -  E  <  \mu <  p <  \r p  + E

=>   0.09 - 0.0397  <  \mu <  p < 0.09 + 0.0397

=>  0.0503  <   p < 0.1297

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