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finlep [7]
3 years ago
8

We can calculate the depth d of snow, in centimeters, that accumulates in Harper's yard during the first h hours of a snowstorm

using the equation d=5h. How many hours does it take for 11 centimeter of snow to accumulate in Harper's yard? How many centimeters of snow accumulate per hour? centimeters
Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
8 0
<h2>Answer:</h2>
  1. How many hours does it take for 11 centimeter of snow to accumulate in Harper's yard? 2.2 hours which is 2 hours 12 minutes
  2. How many centimeters of snow accumulate per hour? 5 centimeter
<h2>Step-by-step explanation:</h2>

We know, that d means depth. Also, we know that h means hour.

d=5h basically means this: depth is equal to 5cm times hours

We have to questions to answer here. We already answered the second question. To answer first question, we just divide 11/5 ...That is 2.2 hours. That is 2 hours and 12 minutes.

  1. How many hours does it take for 11 centimeter of snow to accumulate in Harper's yard? 2.2 hours which is 2 hours 12 minutes
  2. How many centimeters of snow accumulate per hour? 5 centimeter

I hope I helped you. I will be grateful, if you mark my answer as the brainliest.


Andrews [41]3 years ago
6 0

Answer:

2.2 hours takes to accumulate 11 centimeter of snow in Harper's yard and snow accumulate per hour is 5 cm.

Step-by-step explanation:

Given:

d = 5h

where d is depth of snow in centimeters and h is hours of a snowstrom.

To find: Time when the depth of snow is 11 cm.

             Depth of snow accumulate per hour.

Consider,

d = 5h

put d = 11

11 = 5h

h=\frac{11}{5}

h = 2.2 hours

h = 2 hours 12 minutes

Now, Put h = 1

d = 5 × 1

d = 5 cm

Therefore, 2.2 hours takes to accumulate 11 centimeter of snow in Harper's yard and snow accumulate per hour is 5 cm.

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a

   $10,151   <  \mu  $11448.12

b

  n =  158

Step-by-step explanation:

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    The  sample size is  n  =  19

    The  sample mean is  \= x  =$10,800

     The  standard deviation is  \sigma  =$1095  

    The  population mean is  \mu  =$225

     

Given that the confidence level is 99%  the level of significance is mathematically represented as

       \alpha  =  100 -99

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        Z_{\frac{0.01}{2} } =  2.58

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

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    MOE  =  2.58*  \frac{1095 }{\sqrt{19} }

     MOE  =  648.12

The  99% confidence interval for the population mean yearly premium is mathematically represented as

    \= x -MOE  <  \mu <  \= x +MOE

substituting values

   10800 -648.12  <  \mu <  10800 + 648.12

    10800 -648.12  <  \mu <  10800 + 648.12

   $10,151   <  \mu  $11448.12

The  largest sample needed is mathematically evaluated as

      n =  [\frac{Z_{\frac{\alpha }{2}    } *  \sigma }{\mu} ]

substituting values

       n =  [    \frac{ 2.58  *  1095}{225} ]^2

      n =  158

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