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ValentinkaMS [17]
3 years ago
13

In automobile mileage and gasoline-consumption testing, 13 automobiles were road tested for 300 miles in both city and highway d

riving conditions. The following data were recorded for miles-per-gallon performance.City: 16.2 16.7 15.9 14.4 13.2 15.3 16.8 16.0 16.1 15.3 15.2 15.3 16.2 Highway: 19.4 20.6 18.3 18.6 19.2 17.4 17.2 18.6 19.0 21.1 19.4 18.5 18.7 Use the mean, median, and mode to make a statement about the difference in performance for city and highway driving.
Mathematics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

Looking at the mean, the median and the mode, cars are more efficient on a highway than in a city

Step-by-step explanation:

First, we calculate the average (mean) performance by adding all values and dividing the sum by the number of values added.

Mean_{city} =\frac{(16.2+16.7+15.9+14.4+13.2+15.3+16.8+16.0+16.1+15.3+15.2+15.3+16.2)mpg }{13} =15.6 mpg

Mean_{highway} =\frac{(19.4+20.6+18.3+18.6+19.2+17.4+17.2+18.6+19.0+21.1+19.4+18.5+18.7 )mpg }{13} =18.9 mpg

Then, to know what the median is, we have to order from least to greatest and look the middle value, i.e. half of the values will be higher than the median and half will be lower.

For the mode, we have to look up what is the most repeated value in our list.

For city performances:

  1. 13.2
  2. 14.4
  3. 15.2
  4. 15.3
  5. 15.3
  6. 15.3
  7. 15.9
  8. 16
  9. 16.1
  10. 16.2
  11. 16.2
  12. 16.7
  13. 16.8  

The median value is 15.9 miles per gallon, and the mode is 15.3 miles per gallon.

For highway performances:

  1. 17.2
  2. 17.4
  3. 18.3
  4. 18.5
  5. 18.6
  6. 18.6
  7. 18.7
  8. 19
  9. 19.2
  10. 19.4
  11. 19.4
  12. 20.6
  13. 21.1

The median value is 18.7 miles per gallon, and the mode is 18.6 and 19.4 miles per gallon.

We can say then, that looking at the mean, the median and the mode, cars are more efficient on a highway than in a city and that the least-consuming car in a city still is worst  in terms of efficiency than the worst-performing in a highway.

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Please help. I got the 12 part can’t can’t seem to simplify the fraction bit.
Sati [7]
<h3>Answer: D)  12 & 3/8</h3>

Whole part = 12

Fractional part = 3/8

==============================================

Explanation:

For now, let's just focus on the fractional parts of each mixed number.

We have these fractions:

  • 1/4
  • 4/16
  • 3/8
  • 1/2

The 4/16 reduces to 1/4, so 4/16 = 1/4

So in reality, we have these fractional parts.

  • 1/4
  • 1/4
  • 3/8
  • 1/2

It's not a typo that I'm listing "1/4" twice.

The next step is to get everything to have the same denominator. That denominator is 8 (aka the LCD)

  • 1/4 = 2/8
  • 1/4 = 2/8
  • 3/8 = 3/8
  • 1/2 = 4/8

Adding those new fractions gets us:

(2/8)+(2/8)+(3/8)+(4/8) = (2+2+3+4)/8 = 11/8

Rewrite that as a mixed number

11/8 = (8+3)/8

11/8 = (8/8) + (3/8)

11/8 = 1 + 3/8

11/8 = 1 & 3/8

We get the result 1 & 3/8, where 1 is the whole part and 3/8 is the fractional part.

Keep in mind that all we've done so far is add up the fractional parts. We were ignoring the whole parts from each original mixed number.

If we add the whole parts (3,2,1,5) we get 3+2+1+5 = 11, which adds onto the previous whole part of 1 to get 12. So that's where the 12 comes from.

The fractional part 3/8 comes along for the ride to get 12 & 3/8 as the final answer, which is choice D.

-----------------

An alternative route you can take is to convert each original mixed number into an improper fraction, then add those improper fractions (don't forget to get the LCD), and lastly convert the result to a mixed number.

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Kisachek [45]

Answer:

The answer is 0.

Step-by-step explanation:

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460313
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Answer: she made 11 sparkelers and gavin made 21

Step-by-step explanation:

4 0
3 years ago
Calculus 3 chapter 16​
o-na [289]

Evaluate \vec F at \vec r :

\vec F(x,y,z) = x\,\vec\imath + y\,\vec\jmath + xy\,\vec k \\\\ \implies \vec F(\vec r(t)) = \vec F(\cos(t), \sin(t), t) = \cos(t)\,\vec\imath + \sin(t)\,\vec\jmath + \sin(t)\cos(t)\,\vec k

Compute the line element d\vec r :

d\vec r = \dfrac{d\vec r}{dt} dt = \left(-\sin(t)\,\vec\imath+\cos(t)\,\vec\jmath+\vec k\bigg) \, dt

Simplifying the integrand, we have

\vec F\cdot d\vec r = \bigg(-\cos(t)\sin(t) + \sin(t)\cos(t) + \sin(t)\cos(t)\bigg) \, dt \\ ~~~~~~~~= \sin(t)\cos(t) \, dt \\\\ ~~~~~~~~= \dfrac12 \sin(2t) \, dt

Then the line integral evaluates to

\displaystyle \int_C \vec F\cdot d\vec r = \int_0^\pi \frac12\sin(2t)\,dt \\\\ ~~~~~~~~ = -\frac14\cos(2t) \bigg|_{t=0}^{t=\pi} \\\\ ~~~~~~~~ = -\frac14(\cos(2\pi)-\cos(0)) = \boxed{0}

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1 year ago
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