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Sholpan [36]
3 years ago
8

Which coordinates identify a location north of a city that has a latitude of 38. 0°n and a longitude of 25. 0°w?.

Mathematics
1 answer:
marissa [1.9K]3 years ago
8 0

Using translation concepts, y-coordinates greater than 38 identify a location north of the city.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.

Considering latitude and longitude, the coordinates of a city are given as follows:

(LON, LAT).

Hence, the given city has coordinates given by:

(25, 38)

As we move north, the latitude increases, hence, y-coordinates greater than 38 identify a location north of the city.

More can be learned about translation concepts at brainly.com/question/4521517

#SPJ1

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A randomly selected sample of college basketball players has the following heights in inches. See Attached Excel for Data. Compu
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The  confidence interval is  64.86

Step-by-step explanation:

From the question we are given the following data

   The following heights are

66, 65, 67, 62, 62, 65, 61, 70, 66, 66, 71, 63, 69, 65, 71, 66, 66, 69, 68, 62, 65, 67, 65, 71, 65, 70, 62, 62, 63, 64, 67, 67      

 The  sample size is n  =32

  The confidence level is k  = 95% = 0.95

The mean is evaluated as

          \= x = 66+ 65+ 67+ 62+ 62+ 65+ 61+ 70+ 66+ 66+ 71+63+ 69+ 65+ 71+ 66+ 66+ 69+ 68+ 62+ 65+ 67,+\\65+ 71+ 65+ 70+ 62+ 62+ 63+ 64+ 67+ 67 / 32

=>   \= x = \frac{2108}{32}

=>     \= x = 65.875

The standard deviation is evaluated as

           \sigma =  \sqrt{ v}

Now  

   v = ( 66-65.875 )^2+(65-65.875)^2+( 67-65.875)^2+ (62-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+( 61-65.875)^2+ (70-65.875)^2+ (66-65.875)^2+ (66-65.875)^2+ (71+63-65.875)^2+ (69-65.875)^2+ (65-65.875)^2+ (71-65.875)^2+( 66-65.875)^2+ (66-65.875)^2+ (69-65.875)^2+ (68-65.875)^2+ (62-65.875)^2+ (65-65.875)^2+ (67-65.875)^2,+\\(65-65.875)^2+ (71-65.875)^2+ (65-65.875)^2+ (70-65.875)^2+( 62-65.875)^2+( 62-65.875)^2+ (63-65.875)^2+ (64-65.875)^2+ (67-65.875)^2+ (67-65.875)^2 / 32

=>v=  8.567329

=>   \sigma  =  \sqrt{8.567329}

=>   \sigma  =  2.927

The level of significance is evaluated as

        \alpha  =  1 - 0.95

        \alpha  =  0.05

The degree of freedom is  evaluated as

    Df =  n- 1 \equiv  Df  =  32 -1 = 31

The critical values for the level of significance is obtained from the z -table as

      t_c = t_{\alpha/2 } , Df =  t _{0.05/2}, 31 =\pm 1.96

The confidence interval is evaluated as

       \mu  = \= x \pm t_c *  \frac{\sigma }{\sqrt{n} }

substituting values

        \mu =65.875 \pm 1.96* \frac{2.927}{\sqrt{32} }

       \mu =65.875 \pm 1.01415

=>    64.86

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Answer:

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Answer:

two and one half hours

Step-by-step explanation:

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