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Rina8888 [55]
3 years ago
11

#2 , #4 , #6 , #8 , #10 please

Mathematics
1 answer:
erastovalidia [21]3 years ago
5 0
2)
(4.18  x 10^-4)(9 x 10^-4)
= (4.18 x 9)(10^-4 x 10^-4)
= 37.62 x 10^-8
= 3.762 x 10^-7

4)
(9.45 x 10^10) / (1.5 x 10^6)
=(9.45 / 1.5) (10^10 / 10^6)
= 6.3 x 10^4

6)
(9 x 10^-11) / (2.4 x 10^8)
= (9 / 2.4) (10^-11 / 10^8)
= 3.75 x 10^-19

8)
(9.5 x 10^11) + (6.3 x 10^9)
= 10^9 (950 + 6.3)
= 956.3 x 10^9
= 9.563 x 10^11

10)
(1.357 x 10^9) + 590,000
= (1.357 x 10^9) + (5.9 x 10^5)
= 10^5 (13,570 + 5.9)
=13,575.9 x 10^5
= 1.35759 x 10^9
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Answer:

The estimate is  P__{hat}} \pm E  = 0.37 \pm 0.0348

Step-by-step explanation:

From the question we are told that  

    The sample size is  n =  522

    The sample proportion of students  would like to talk about school is  \r p__{hat}} =  0.37

  Given that the confidence level is  90 % then the level of significance can be mathematically evaluated as

                  \alpha  =  100 - 90

                  \alpha  =  10\%

                  \alpha  =  0.10

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

               Z_{\frac{\alpha }{2} } =Z_{\frac{0.10}{2} } =  1.645

Generally the margin of error can be mathematically represented as

               E =  Z_{\frac{\alpha }{2} } *  \sqrt{\frac{\r P_{hat}(1- \r P_{hat} )}{n } }

=>            E = 1.645 *  \sqrt{\frac{0.37 (1- 0.37  )}{522 } }

=>             E = 0.0348

Generally the estimate the proportion of all teenagers who want more family discussions about school at 90% confidence level is  

                       P__{hat}} \pm  E

substituting values

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