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Ede4ka [16]
4 years ago
7

The 20-inch-diameter wheels of one car travel at a rate of 24 revolutions per minute,

Mathematics
1 answer:
nydimaria [60]4 years ago
8 0
One revolution of a wheel is equal to it's circumference:

C = πd

C: circumference
d: diameter

<u>20" wheel</u>
d = 20"
C = (20π) in/rev
v = (24 rev/min)(20π in/rev) = 480π in/min

<u>30" wheel</u>
d = 24"
C = (30π) in/rev
v = (18 rev/min)(30π in/rev) = 540π in/min

Ratio =  \frac{velocity of 30" wheel}{velocity of 20" wheel} =  \frac{v_{2}}{v_{1}}

\frac{v_{2}}{v_{1}} =  \frac{540\pi}{480\pi}  = \frac{9}{8}












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A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and
Naya [18.7K]
<h2>The number of small buses used = 5</h2><h2>The number of big buses used  = 4</h2>

Step-by-step explanation:

Let us assume the total number of small buses needed = x

The capacity of 1 small bus  = 40

So, the capacity of x buses  = 40(x)  = 40 x

Let us assume the total number of big buses needed = y

The capacity of 1 big bus  = 50

So, the capacity of y buses  = 50(y)  = 50 y

Also, the total students travelling = 400

So, the number of students traveling by (Small bus + Big bus)  = 400

⇒ 40 x + 50 y = 400 ..... (1)

Also, the total number of drivers available  = 9

⇒ x +  y = 9  ..... (2)

Also, x  ≤ 8,   y ≤ 10

Now, solving both equations, we get:

40 x + 50 y = 400 ..... (1)

x +  y = 9  ⇒ y = (9-x) put in (1)

40 x + 50 y = 400  ⇒  40 x  + 50 (9-x)  = 400

or, 40 x  + 450 - 50 x  = 400

or, - 10 x  =- 50

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3 years ago
Miranda found 4 pieces of an ancient ceramic bowl.
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Answer:

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3.- the size of the pieces

4.- if the pieces can fit together

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Listed below are the ACT scores of 40 randomly selected students at a major university. (Use technology if you wish, Excel) 18 1
tiny-mole [99]

Answer:

A) See the picture

B) 14

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Step-by-step explanation:

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B) Because the question B and C have to be responded using a frequency table with 8 classes the answer is 14; the method of using cumulative frequency tables should only be considered as a way of estimation, that is because you obtain values that depend on your choice of class intervals. The way to get a better answer would be to use all the scores in the distribution

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