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Nataly_w [17]
3 years ago
11

Pls explain how to solve it! (Will mark brainylist)

Mathematics
1 answer:
Tasya [4]3 years ago
8 0

Answer:

1,1

Step-by-step explanation:

x=1

y=1

if this is substitution then x IS 7y-6 so then you can replace the x in the other problem with 7y-6 but since it is 5x you will need to do 5(7y-6) which is 35y-30-4y=1

add 30 to both sides to get 35y-4y=31

subtract 4y from 35y to get 31y=31 get the y alone by dividing 31 from both sides to get y=1. then replace the y in the problem where the x is already alone to get x=7×1-6

7x1 is 7 so then 7-6=1 and that gives us x=1.

Hope this helped

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Complete Question

Fewer young people are driving. In 1995, 63.9% of people under years 20 old who were eligible had a driver's license. Bloomberg reported that percentage had dropped to 41.7% in 2016. Suppose these results are based on a random sample 1,200 of people under 20 years old who were eligible to have a driver's license in 1995 and again in 2016.

a. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 1995?

Margin of error(to four decimal places)

Interval estimate (to four decimal places)

b. At 95% confidence, what is the margin of error and the interval estimate of the number of eligible people under 20 years old who had a driver's license in 2016?

Margin of error(to four decimal places)

Interval estimate    to   (to four decimal places)

Answer:

a

  0.6120 <  p <  0.639 + 0.6670

b

  0.3900 <  p < 0.4440

Step-by-step explanation:

Considering question a

   The sample proportion is 1995 is  \^ p_1 = 0.639

    The sample size is  n = 1200

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the margin of error is mathematically represented as  

     E =  Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }

=>   E =  1.96  * \sqrt{\frac{0.639 (1- 0.639)}{1200} }

=>   E =  0.027

Generally 95% Interval estimate is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.639 -0.027 <  p <  0.639 + 0.027

=>    0.6120 <  p <  0.639 + 0.6670

Considering question b

   The sample proportion is 1995 is  \^ p_2 = 0.417

    The sample size is  n = 1200

From the question we are told the confidence level is  95% , hence the level of significance is    

      \alpha = (100 - 95 ) \%

=>   \alpha = 0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

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Generally the margin of error is mathematically represented as  

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=>   E =  1.96  * \sqrt{\frac{0.417 (1- 0.417)}{1200} }

=>   E =  0.027

Generally 95% Interval estimate is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.417 -0.027 <  p <  0.417 + 0.027

=>    0.3900 <  p <   0.4440

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