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Lady bird [3.3K]
3 years ago
5

The ratio of pages read to minutes for Hang is 5 : 8. Who has a greater ratio of pages read to minutes than Hang? Select all tha

t apply. A. Kira:2 pages read every 4 minutes B. Joe:6 pages read every 9 minutes C. Quon:5 pages read every 7 minutes D. Lulu:10 pages read every 13 minutes E.
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:

B, C, and D

Step-by-step explanation:

By doing 5÷8 = .625 you know that Hang can read .625 of a page per minute.

You can do this for the rest of the people finding that Kira reads .5 page per minute which is a slower/ smaller ratio, Joe rate is .66, Quon's ratio is .71 and Lulus rate is .76, the only rate that is smaller than Hangs is Kira,s, leaving you with Joe, Quon, and Lulu to be your answer

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Drupady [299]

Answer:

56°

Step-by-step explanation:

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3 0
3 years ago
Write 81.42 as a mixed number
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Nastasia [14]

Answer:

  a. 2 7/9 miles

  b. 21.6 minutes

  c. 8 1/3 miles

Step-by-step explanation:

The rate at which Catlin runs can be determined from the graph by finding the coordinates of a grid point that lies on the line. The line goes through the origin, so the relationship between time and distance is proportional.

We see that the point (miles, minutes) = (1 2/3, 12) lies on the graph, so the contant of proportionality is ...

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__

<h3>a.</h3>

The proportion can be written ...

  distance/(20 minutes) = (1 mile)/(7.2 minutes)

  distance = (20/7.2) miles = 2 7/9 miles

Catlin will run 2 7/9 miles in 20 minutes

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<h3>b.</h3>

The proportion can be written ...

  time/(3 miles) = (7.2 minutes)/(1 mile)

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Catlin was running 21.6 minutes

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<h3>c.</h3>

In 1 hour, Catlin will run 3 times as far as in 20 minutes. Using the result from part A, we find ...

  3(2 7/9 miles) = 8 1/3 miles 8 1/3 miles

Catlin will run 8 1/3 miles in one hour.

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<em>Additional comment</em>

Once we have a relationship between minutes and miles, we can write proportions with either of those numbers in the numerator or denominator. We find it convenient to put the unknown value in the numerator of a proportion, which is why the ratios are written differently in the different parts of the problem.

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2 years ago
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