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notka56 [123]
4 years ago
9

Tom and his friend are driving to Tameka is taking the same 432-mile round trip as Tom to Louisville, but renting a car at a spe

cial rate of $99.99 with unlimited miles. If her car gets 35 mpg and cost of gasoline is $2.79, what will the rental, including gas, cost her?
Mathematics
1 answer:
Akimi4 [234]4 years ago
6 0

Answer: The correct answer is $ 134.42 for the gas and the rental.

Step-by-step explanation: The trip is 432 miles and Tameka's car gets 35 miles to the gallon. So we have to figure out how many gallons we would use for the entire trip by dividing 432 by 35= 12.34 gallons of gas.

We know gas is $2.79 a gallon so we multiply 2.79 (cost per gallon) x 12.34 ( our total gallons used) =$34.43 in gasoline.

The rental costs $99.99 so we add that cost and the cost of the gas $34.43= $134.42 the total cost of the rental and the gas.

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Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:
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Answer:

a

   \= x  = 18.5  ,  \sigma =  5.15

b

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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

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