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prohojiy [21]
2 years ago
13

An oil tanker is leaking at a constant rate per minute. After 11 minutes, 931 gallons of oil remain in the tanker. After 18 minu

tes, 633 gallons of oil remain in the tanker. Write the function f(t) that describes this scenario.
Mathematics
1 answer:
bekas [8.4K]2 years ago
5 0

The  function f(t) that describes this scenario will be A = -42.57t + 1399.28

  • Let the amount of oil remaining after time t be A
  • Let the time taken be "t"

This required linear expression will be given as A= mt + b

Writing the gallons of oil and the time taken as a coordinate point (A, t), these are given as (11, 931) and (18, 633)

m is the rate of change = 633-931/18-11

m = -298/7

m = -42.57

Get the intercept;

633 = -42.57(18) + b

633 = -766.29 + b

b = 1,399.28

Heence the  function f(t) that describes this scenario will be A = -42.57t + 1399.28

Learn more on functions here; brainly.com/question/17431959

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Identify the class​ width, class​ midpoints, and class boundaries for the given frequency distribution.
Crank

Answer:

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

c.

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

M=\frac{lower class limit+upper class limit}{2}

Class

Interval Midpoints

45-48 \frac{45+48}{2}=46.5

49-52 \frac{49+52}{2}=50.5

53-56 \frac{53+56}{2}=54.5

57-60 \frac{57+60}{2}=58.5

61-64 \frac{61+64}{2}=62.5

65-68 \frac{65+68}{2}=66.5

69-72 \frac{69+72}{2}=70.5

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

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Because when u plug in x-value(0) in the graph the y=-3
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