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xxTIMURxx [149]
3 years ago
13

You are at a trade show where you are selling baseball cards for $4 a card and buying baseball cards for $6 a card. If you start

ed the day with $200, how much money do you have at the end of the day if you sold c baseball cards and bought b baseball cards? A) 4c - 6b B) 200 + 4c - 6b C) 200 - 4c + 6b D) 200 - 2(c + b) E)
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
7 0

Answer:

B) 200 + 4c - 6b

Step-by-step explanation:

Be,

c = number of cards sold at $4

b = number of cards purchased at $6

x = money at the beginning of the day = $200

y = money at the end of the day

The variables x and c owe positive values because they represent possession or income of money, while b must be negative because purchases are expenses.

Then, the following equation is formed:

y = x + 4c - 6b

Substituting x for its value, you get

y = 200 + 4c - 6b

Hope this helps!

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Read 2 more answers
ducational Television In a random sample of people, said that they watched educational television. Find the confidence interval
SIZIF [17.4K]

Complete Question

Educational Television In a random sample of 200 people, 159 said that they watched educational television. Find the 90% confidence interval of the true proportion of people who watched educational television. Round the answers to at least three decimal places.

Answer:

The 90% confidence interval is   0.748  <  p <  0.842      

Step-by-step explanation:

From the question we are told that

     The sample size is  n = 200

     The number of people that watched the educational television is k =  159

Generally the sample proportion is mathematically represented as

               \^ p =  \frac{159}{200}

=>               \^ p =  0.795

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90) \%

=>   \alpha = 0.10

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

   Z_{\frac{\alpha }{2} } =  1.645

Generally the margin of error is mathematically represented as  

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

=>     E =  1.645  * \sqrt{\frac{0.795 (1- 0.795)}{200} }

=>     E = 0.04696

Generally 95% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

=>    0.795  -0.04696  <  p <  0.795 + 0.04696

=>    0.748  <  p <  0.842      

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