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mote1985 [20]
3 years ago
7

Robert was able to travel 292 miles in 4 hours and uses 38 l of gasoline. if gas costs $2.09 per gallon, how much did the trip c

ost robert? (1 gal≈3.785 l).
Mathematics
2 answers:
Butoxors [25]3 years ago
7 0

cost of 1 gallon gas = $2.09

cost of 3.785 litres gas = $2.09 ( 1 gallon = 3.785 litres)

cost of 1 litre gas = 2.09/3.785

cost of 18 litres gas = 2.09/3.785 * 18 = $20.982

babunello [35]3 years ago
5 0

Answer:

Robert´s trip cost $20.98.

Step-by-step explanation:

Robert use 38 liters of gas to make his 292 miles trip. The gallon of gas cost is $2.09. First we have to convert the 38 liters of gas to gallons.

38 liters*\frac{1 gallon}{3.785 liters} =10.04 gallons

Having convert the units, we can calculate how much cost the gas, this is:

10.04gallons*2.09=20.98

Robert´s trip cost $20.98.

The distance and time data is used to confuse.

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a) P(R

And we can find this probability using the normal standard table or excel and we got:

P(z

b) P(R>110)=P(\frac{R-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{3.606})=P(Z>2.774)

And we can find this probability using the complement rule and the normal standard table or excel and we got:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the time for the step 1 and Y the time for the step 2, we define the random variable R= X+Y for the total time and the distribution for R assuming independence between X and Y is:

R \sim N(40+60 = 100,\sqrt{2^2 +3^2}= 3.606 s)  

Where \mu=65.5 and \sigma=2.6

We are interested on this probability

P(R

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{R-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(R

And we can find this probability using the normal standard table or excel and we got:

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Part b

P(R>110)=P(\frac{R-\mu}{\sigma}>\frac{110-\mu}{\sigma})=P(Z>\frac{110-100}{3.606})=P(Z>2.774)

And we can find this probability using the complement rule and the normal standard table or excel and we got:

P(z>2.774)=1-P(Z

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