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jeka57 [31]
3 years ago
8

Can you help me plz and thank you

Mathematics
1 answer:
UNO [17]3 years ago
3 0
Honestly I don’t know this one but I think it’s like you start at 35 on the number line then you subtract 4 then add 2 then subtract 5. So I think it’s B but I’m not sure.
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The solution to 5 - 3x> 35 is either x>-10 or -10> x. which solution is correct
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It should be x < -10 because when you divide by a negative number the inequality sign flips.

3 0
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An electronics store usually sells computers priced at $1500 each. If the customer orders the computer over the Internet, he has
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a box of crackers has a volume of 48 cubic inches. what is the volume of a similar box that is smaller by a scale factor of 2/3
earnstyle [38]
For this case what you need to know is that the original volume of the cookie box is:
 V = (w) * (l) * (h)
 Where,
 w: width
 l: long
 h: height.
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 V = (w) * (l) * (h) = 48 in ^ 3
 The volume of a similar box is:
 V = (w * (2/3)) * (l * (2/3)) * (h * (2/3))
 We rewrite:
 V = ((w) * (l) * (h)) * ((2/3) * (2/3) * (2/3))
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 Answer:
 the volume of a similar box that is smaller by a scale factor of 2/3 is:
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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
2 years ago
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