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lozanna [386]
2 years ago
6

Corey used the regression equation y = 1.505x − 88.21, where x is the temperature and y is the number of swimmers, to determine

a possible outside temperature when 80 swimmers are at City Pool.
1. y = 1.505(80) − 88.21

2. y = 120.4 − 88.21

3. y = 32.19

4. If there are 80 swimmers at the pool, the temperature is likely to be 32.2°F.

Analyze Corey’s work to find his error. What is Corey’s mistake?
Mathematics
2 answers:
solniwko [45]2 years ago
9 0

Answer:

A. He substituted 80 for x instead of y.

Step-by-step explanation:

BaLLatris [955]2 years ago
7 0

Answer: Corey substituted the number of swimmers into the variable (x) representing the temperature. Then he didn't wonder if the polar bear club really has 80 members.


Step-by-step explanation:


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  The  probability is  P(\= X < 33) = 0.8531

2

  The  probability is  P(\= X > 30) = 0.3520

Step-by-step explanation:

From the  question we are told that

     The population mean is  \mu =  28.29

      The standard deviation is \sigma  =  33.493

       The sample size is  n  = 56

Generally the standard error for the  sample  mean (\= x ) is mathematically evaluated as

        \sigma _{\=x} =  \frac{\sigma}{\sqrt{n} }

substituting values  

       \sigma _{\=x} =  \frac{33.493}{\sqrt{56} }

      \sigma _{\=x} = 4.48

Apply central limit theorem[CLT] we have  that

        P(\= X < 33) =  [z <  \frac{33 -  \mu }{\sigma_{\= x}} ]

substituting values

       P(\= X < 33) =  [z <  \frac{33 -  28.29 }{4.48} ]

       P(\= X < 33) =  [z <  1.05 ]

From the z-table  we have that  

       P(\= X < 33) = 0.8531

For the second question

    Apply central limit theorem[CLT] we have  that

    P(\= X > 30 ) =  [z >  \frac{30 -  \mu }{\sigma_{\= x}} ]

substituting values

   P(\= X < 33) =  [z >  \frac{30 -  28.29 }{4.48} ]

From the z-table  we have that  

     P(\= X < 30) = 0.6480

Thus  

     P(\= X > 30) = 1- P(\= X < 30) = 1- 0.6480

     P(\= X > 30) = 0.3520

3 0
2 years ago
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