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Fudgin [204]
3 years ago
14

It is known that the population mean for the verbal section of the SAT is 500 with a standard deviation of 100. In 2006, a sampl

e of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511. The point estimate of the mean for this group is _____ and the 95 percent confidence interval for this group is _____.
Physics
1 answer:
kvasek [131]3 years ago
4 0

Answer

given,

SAT is 500 with a standard deviation of 100.

a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.

sample mean = \dfrac{standard\ deviation}{\sqrt{n}}

                      = \dfrac{100}{\sqrt{400}}

                      = 5

95% confidence level is achieved within +/- 1.960 standard deviations.

1.960 standard deviations  x  5 is equal to +/- 9.8

confidence interval = 511 - 9.8   ---  511 + 9.8

                                = 501.2-----520.8

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3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
3 years ago
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