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sattari [20]
3 years ago
6

Monte performs an experiment using 2 identical graduated cylinders, each with a radius of 2 centimeters. The volume of the liqui

d in the first graduated cylinder is 188.4 cubic centimeters. The volume of the liquid in the second graduated cylinder is 314 cubic centimeters. What is the difference in the height of the liquid in the two cylinders? Use 3.14 for pi.
Enter your answer in the box.
Mathematics
2 answers:
elena55 [62]3 years ago
5 0
Volume of a cylinder is h x r^2 x pi

You have the volume and radius of each one, so taking 188.4=(2)^2 x 3.14 x h1, you can solve this equation for the height h1. Likewise, for the other graduated cylinder you take 314=2^2 x 3.14 x h2, and then solve for h2.

To get the difference in heights, just take h2-h1 after solving the equations for h2 and h1
soldier1979 [14.2K]3 years ago
4 0

Answer: I did the test and the answer is 10


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The amount of money spent on textbooks per year for students is approximately normal.
Ostrovityanka [42]

Answer:a

a

   336.04    <  \mu < 443.96

b

  The  margin of error will increase

c

The  margin of error will decreases

d

The 99% confidence interval is  0.4107 <  p  < 0.4293

Step-by-step explanation:

From the question we are  told that

   The sample size  n =  19

    The sample mean is  \= x  = \$\  390

    The  standard deviation is  \sigma =  \$ \  120

 

Given that the confidence level is  95% then the level of significance is mathematically represented as

           \alpha = 100 -  95

          \alpha  =  5 \%

          \alpha  =  0.05

Next we obtain the critical value of \frac{\alpha }{2} from the normal distribution table

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         Z_{\frac{\alpha }{2} } =  1.96

The  margin of error is mathematically represented as

      E =  Z_{\frac{\alpha }{2} } *  \frac{\sigma}{\sqrt{n} }

=>    E = 1.96 *  \frac{120}{\sqrt{19} }

=>   E = 53.96

The 95% confidence interval is  

     \= x  -  E  <  \mu < \= x  +  E

=>   390  -   53.96   <  \mu < 390  -   53.96

=>  336.04    <  \mu < 443.96

When the confidence level increases the Z_{\frac{\alpha }{2} } also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

         n  \  \ \alpha  \ \  \frac{1}{E^2 }

So if the sample size increases the margin of error decrease

The  sample proportion is mathematically represented as

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Given that the confidence level is 0.99 the level of significance is  \alpha =  0.01

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

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  Generally the margin of error is mathematically represented as

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

=>   E =  0.42 *  \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }

=>     E =  0.0093

The 99% confidence interval  is

     \r p  -  E <  p  < \r p  +  E

     0.42  -  0.0093 <  p  < 0.42  +  0.0093

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Answer:

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Answer:

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