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Anni [7]
3 years ago
8

A certain hybrid car has a mileage rating of 57 miles per gallon. if the car makes a trip of 293 miles, how many gallons of gaso

line will be used?
Mathematics
1 answer:
dlinn [17]3 years ago
4 0
To determine the number of gallons of gasoline that is used, we need to know the rate of usage of gasoline. This rate would describe the number of gasoline in units of volume that is being used per distance in units of length. In this case, we need the rate in units of miles per gallon. From what is asked and the given values, we simply divide the rate to the the total distance that was traveled by the car. We calculate as follows:

Gallons of gasoline = 293 miles / 57 miles / gallon
Gallons of gasoline = 5.14 gallons

Therefore, about 5 gallons of gasoline was consumed by the hybrid car for a distance of 57 miles.
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V125BC [204]

Answer:

Hi there, I like your pfp!

The correct answer should be 11 1/2!

Step-by-step explanation:

-48/-8 = 6 4/8 = 6 1/2

6 1/2 + 5 =

11 1/2

Anyways, hope this helped!

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2 years ago
Please answer fast!
rosijanka [135]

Multiply the three numbers.

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3 years ago
Evaluate the expression: m-3n when m=8, n=2<br><br> A. 10<br> B. 2<br> C. 14<br> D. 13
kvv77 [185]
I think it’s B. 2

8-3(2)
6 0
3 years ago
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Alice purchased paint in a bucket with a radius of 9 in. and a height of 9.5 in. The paint cost $0.09 per in3.
ratelena [41]
Volume = (pi)(radius^{2}(height)
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3 0
3 years ago
A chemist examines 12 sedimentary samples for bromide concentraction. The mean bromide concentration for the sample date is 0.43
Umnica [9.8K]

Answer:

a) z = 1.645

b) Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

Step-by-step explanation:

Population is approximately normal, so we can find the normal confidence interval.

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645. This is the critical value, the answer for a).

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.645*\frac{0.0325}{\sqrt{12}} = 0.015

The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.

b)

Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

4 0
3 years ago
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