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melomori [17]
3 years ago
7

Lila hired a chainsaw for a week.

Mathematics
2 answers:
Blababa [14]3 years ago
8 0
First, if she used it Monday to Friday, that’s 5 days. 5 x 37 = $185
Secondly, if she used the chainsaw on Saturday and Sunday too, and if 2 x $42 = $84, then she would have paid $84 for the Saturday and Sunday in which she used the chainsaw.

Add up both of cost of the days she used the chainsaw and you get a total of $269
I hope this helped :)
Allisa [31]3 years ago
4 0

Answer:

She payed $269 for the chainsaw :)

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Veena ate lunch at a deli. She ordered a turkey snadwich for $9.25 and a salad for 4.35. The tax was $7.5. What is the amount of
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A bag contains 56 marbles: 7 red, 8 green, 1l yellow, 17 brown, and 13 blue.
Thepotemich [5.8K]

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Step-by-step explanation:

3 0
2 years ago
Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:
Vikentia [17]

Answer:

a

   \= x  = 18.5  ,  \sigma =  5.15

b

 15.505 < \mu <  21.495

c

 14.93 < \mu <  22.069

Step-by-step explanation:

From the question we are are told that

    The  sample data is  21, 14, 13, 24, 17, 22, 25, 12

     The sample size is  n  = 8

Generally the ample mean is evaluated as

        \= x  =  \frac{\sum x  }{n}

        \= x  =  \frac{  21 + 14 + 13 + 24 + 17 + 22+ 25 + 12  }{8}

         \= x  = 18.5

Generally the standard deviation is mathematically evaluated as

         \sigma =  \sqrt{\frac{\sum (x- \=x )^2}{n}}

\sigma =  \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}

\sigma =  5.15

considering part b

Given that the confidence level is  90% then the significance level is evaluated as

         \alpha  =  100-90

         \alpha  = 10\%

         \alpha  = 0.10

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

     Z_{\frac{ \alpha }{2} }  =  1.645

The margin of error is mathematically represented as

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

=>    E =1.645  *  \frac{5.15 }{\sqrt{8} }

=>     E =  2.995

The 90% confidence interval is evaluated as

       \= x  -  E < \mu <  \= x +  E

substituting values

       18.5 -  2.995 < \mu <  18.5 +  2.995

       15.505 < \mu <  21.495

considering part c

Given that the confidence level is  95% then the significance level is evaluated as

         \alpha  =  100-95

         \alpha  = 5\%

         \alpha  = 0.05

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

     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{5.15 }{\sqrt{8} }

=>     E = 3.569

The 95% confidence interval is evaluated as

       \= x  -  E < \mu <  \= x +  E

substituting values

       18.5 - 3.569 < \mu <  18.5 +  3.569

       14.93 < \mu <  22.069

8 0
3 years ago
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