1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ExtremeBDS [4]
3 years ago
7

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 19.1 20.3 18.0 18.3 18.9 17.1 16.9 18.3 18.7 20.8 19.1 18.2 18.4 Use the mean, median, and mode to make a statement about the difference in performance for city and highway driving. Compute the mean for the miles-per-gallon performance for city and highway driving. (Round your answers to two decimal places.) city 15.9 miles per gallon highway 15.3 miles per gallon Is the mean mileage better on the highway than in the city? Yes No Compute the median for the miles-per-gallon performance for city and highway driving. city miles per gallon highway miles per gallon Is the median mileage better on the highway than in the city? Yes No Compute the mode(s) for the miles-per-gallon performance for city and highway driving. (Enter your answers as a comma-separated list.) city miles per gallon highway miles per gallon Is the modal mileage better on the highway than in the city? Yes No Make a statement about the difference in performance for city and highway driving conditions. Miles-per-gallon performance is in highway driving condition than in city driving conditions.

Mathematics
1 answer:
Aloiza [94]3 years ago
3 0

Answer:

A. No

City driving mean requires less miles than Highway driving mean.

B. No

City driving median requires less miles than Highway driving median.

C. No

City driving mode requires less miles than Highway driving mode

D.

The driving conditions in the City is better than the driving conditions in the Highway because all measures of centers are lesser than those of Highway.

Step-by-step explanation:

You might be interested in
A. Find the amplitude.
Feliz [49]

Answers:

  • a) Amplitude = 2
  • b) Period = pi
  • c) Vertical shift = -2, which means it has been shifted down 2 units.
  • d) Horizontal shift = 3pi/8, this shifting is to the right.
  • e) There is <u>  one  </u> cycle between 0 and 2pi.
  • f) The equation of the graph is y = 2*sin(2(x-3pi/8))-2

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

Explanations:

Part (a)

The highest point is when y = 0 and the lowest point is when y = -4. The vertical distance between the peak and valley is 4 units, which cuts in half to 2. This is the amplitude. It's the vertical distance from the center to either the peak or valley.

Note: Amplitude is always positive as it measures a distance.

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

Part (b)

For x > 0, the first valley or lowest point occurs between 0 and pi/4. It appears to be the midpoint of the two values. So that would be (0+pi/4)/2 = pi/8.

The next valley occurs between pi and 5pi/4. Compute the midpoint to get (pi+5pi/4)/2 = (4pi/4+5pi/4)/2 = (9pi/4)/2 = 9pi/8

So we have the graph go from one valley x = pi/8 to the next valley over x = 9pi/8. This is a distance of pi units because 9pi/8-pi/8 = 8pi/8 = pi

The graph repeats itself every pi units, so the period is pi.

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

Part (c)

The midline is normally through y = 0, aka the x axis. However, the graph shows the midline is through y = -2. This means the graph has been shifted down 2 units.

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

Part (d)

This will depend on whether you use sine or cosine. This is entirely because cosine is a phase-shifted version of sine, and vice versa. I'll go with sine.

The parent sine function y = sin(x) goes through the origin (0,0)

However, as part (c) mentioned, we shifted the graph 2 units down. So we have y = sin(x)-2. But plugging x = 0 into this leads to the point (0,-2)

This doesn't match what the graph says. The graph shows the point (3pi/8, -2) on the red curve. The x coordinate 3pi/8 is the midpoint of pi/4 and pi/2

This must mean we need to shift the sine graph 3pi/8 units to the right.

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

Part (e)

Start at the lowest point when x = pi/8. If you start the cycle here, then it ends when x = 9pi/8. See part (b).

So far we've completed one cycle. If we start at x = 9pi/8, then the next valley or lowest point is slightly beyond or to the right of x = 2pi. This means we run out of room and we haven't completed a full cycle.

Overall, one full cycle is between 0 and 2pi.

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

Part (f)

Again I'm going to use sine instead of cosine. Refer back to part (d).

The general sine curve equation is

y = A*sin(B(x-C))+D

where

  • |A| = amplitude
  • B handles the period, specifically T = 2pi/B where T is the period. We can solve for B to get B = 2pi/T
  • C = horizontal phase shift
  • D = vertical shift, and ties together with the midline equation

In this case, we found that

  • |A| = 2
  • T = pi leads to B = 2pi/T = 2pi/pi = 2
  • C = 3pi/8
  • D = -2

So,

y = A*sin(B(x-C))+D

will update to

y = 2*sin(2(x-3pi/8))-2

which is one way to express the equation of the red curve. Optionally you can distribute the 2 through to (x-3pi/8).

6 0
2 years ago
The expression radical 50+radical 32 is equivalent to
leva [86]
<span>5<span>2–√ i would believe</span></span>
7 0
3 years ago
Plzz hel am timed which number is composite
mezya [45]
15 is compostit bevuasw (3x5=15) (1x15)
8 0
3 years ago
Read 2 more answers
Find the slope of the line passing through the points(3, -4) and q(-7,1). then find the equation of the line passing through tho
IRISSAK [1]
Slope is -4-1/-7-3= -5/-10=1/2 it is y2-y1/x2-x1
the equation is y-y1= m(x-x1)= y-(-4) =1/2(x-3) =y+4 =1/2(x-3)
5 0
2 years ago
1/3 of Murka’s age is twice as much as Ivan’s age. What is the ratio of Murka’s age to Ivan’s age?
Korvikt [17]

Answer:

6:1

Step-by-step explanation:

4 0
3 years ago
Other questions:
  • Suppose 45 students in an organization are male and 55 are female. Ten females and 15 males in the organization are graduating s
    7·1 answer
  • What would $25 deposited 56 yrs ago be worth today???
    13·1 answer
  • The LCM of 6, 12, 18
    12·1 answer
  • A circle is placed in a square with a side length of 18 mm, as shown below. Find the area of the shaded region.
    9·1 answer
  • Factor completely.
    15·1 answer
  • Please help me with this problem soon
    5·1 answer
  • In the triangle below, what is the length of the side opposite the 30 degree angle?
    5·1 answer
  • Would be great if someone could solve this for me I’m not too sure how to do it .
    14·1 answer
  • You go out to eat at the Outback Steakhouse. Your meal, drink and desert cost $38.56. You are
    8·1 answer
  • The table shows the purchases made by two customers at a meat counter. You want to buy 2 pounds of sliced ham and 3 pounds of sl
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!