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Gwar [14]
2 years ago
12

8x = 96 How would I solve for x

Mathematics
2 answers:
Vaselesa [24]2 years ago
6 0

How to solve your problem

fomenos2 years ago
5 0

x = 12

Step-by-step explanation:

8x = 96 \\  \\ divide \:  \: both \:  \: sides

8x \div 8 = 96 \div 8 \\  \\ divide \\  \\  = x = 12 \\  \\ hope \:  \: it \:  \: helps \:

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7-ounce package costs $2.59. A 12 ounce package costs $4.56. An 18 ounce package costs $6.30. Which package is the best buy?
Yanka [14]

Answer:

18 ounce package for $6.30

Step-by-step explanation:

For each one, you divide the money by the package to see how much it is per ounce.

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3 years ago
Evaluate the expression when C=6 .<br> C^2-7
pashok25 [27]

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Step-by-step explanation:

6 0
3 years ago
A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex
Olenka [21]

Answer:

a

 The  90% confidence interval that  estimate the true proportion of students who receive financial aid is

     0.533  <  p <  0.64

b

   n = 1789

Step-by-step explanation:

Considering question a

From the question we are told that

      The sample size is  n = 200

      The number of student that receives financial aid is k = 118

Generally the sample proportion is  

      \^ p = \frac{k}{n}

=>   \^ p = \frac{118}{200}

=>   \^ p = 0.59

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{0.59 (1- 0.59)}{200} }

=>  E = 0.057

Generally 90% confidence interval is mathematically represented as  

      \^ p -E <  p <  \^ p +E

  =>  0.533  <  p <  0.64  

Considering question b

From the question we are told that

    The margin of error  is  E = 0.03

From the question we are told the confidence level is  99% , hence the level of significance is    

      \alpha = (100 - 99 ) \%

=>   \alpha = 0.01

Generally from the normal distribution table the critical value  of   is  

   Z_{\frac{\alpha }{2} } = 2.58

Generally the sample size is mathematically represented as      

        [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )

=>      n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )

=>      n = 1789

8 0
3 years ago
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