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AnnyKZ [126]
2 years ago
7

Keith rented a truck for one day. There was a base fee of $17.99, and there was an additional charge of 97 cents for each mile d

riven. Keith had to pay $137.30 when he returned the truck. For how many miles did he drive the truck? miles
Mathematics
2 answers:
MrMuchimi2 years ago
8 0

Answer:

216 miles

Step-by-step

ex15.95+.73x=173.63

.73x=173.63-15.95

.73x=157.68

x=157.68/.73

x=216 miles driven.

PROOF;

15.95+.73*216=173.63

15.95+157.68=173.63

173.63=173.63

denis23 [38]2 years ago
6 0

Answer: Keith drove the truck for 123 miles

Step-by-step explanation:

17.99+0.97m=137.30

Subtract 17.99 on both sides:

17.99-17.99+0.97m=137.30-17.99

0.97m=119.31

Divide 0.97 on both sides:

0.97m/0.97=119.31/0.97

m=123 miles

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Answer:

Step-by-step explanation:

x+5=0

x=-5

-5) 16   80   1    5

  |       -80   0  -5

 ____________

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quotient=16x²+1

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8 0
3 years ago
A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation A pizza delivery company buys
zubka84 [21]

Answer:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

Step-by-step explanation:

Assuming this statement to complete the problem "with a standard deviation 5 mpg"

We have the following info given:

\mu = 34 represent the mean

\sigma= 5 represent the deviation

We have a sample size of n = 54 and we want to find this probability:

P(33.3 < \bar X< 34.3)

And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

And we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

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