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pentagon [3]
3 years ago
15

Joshua owns a trucking company. For every truck that goes out, Joshua must pay the driver $25 per hour of driving and also has a

n expense of $2 per mile driven for gas and maintenance. On one particular day, the driver drove an average of 50 miles per hour and Joshua's total expenses for the driver, gas and truck maintenance were $625. Write a system of equations that could be used to determine the number of hours the driver worked and the number of miles the truck drove. Define the variables that you use to write the system.
Mathematics
1 answer:
lara [203]3 years ago
6 0

Answer:

number of miles=a

number of hours=y

Step-by-step explanation:

<h3>$25 per how y of driving</h3><h3>$2 per how many a per mile</h3><h3>a  x y = the total price </h3><h3 /><h3 /><h3 /><h3>plz mark me as the brainliest  </h3>
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Answer:

96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is  [335.89 , 356.10].

Step-by-step explanation:

We are given that a group of researchers at the BYU Off-Campus Housing department want to estimate the mean monthly rent that unmarried BYU students paid during winter 2018.

During March 2018, they randomly sampled 314 BYU students and found that on average, students paid $346 for rent with a standard deviation of $86.

So, the pivotal quantity for 95% confidence interval for the average age is given by;

           P.Q. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

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So, 96% confidence interval for the population mean monthly rent, \mu is ;

P(-2.082 < t_3_1_3 < 2.082) = 0.96

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P( -2.082 \times {\frac{s}{\sqrt{n} } } < {\bar X - \mu} < 2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

P( \bar X -2.082 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ) = 0.96

96% confidence interval for \mu = [ \bar X -2.082 \times {\frac{s}{\sqrt{n} } } , \bar X +2.082 \times {\frac{s}{\sqrt{n} } } ]

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Therefore, 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is [335.89 , 356.10].

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