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ki77a [65]
3 years ago
8

two fifths of the instruments in the marching band are brass. One-eighth of the brass instruments are tubas. What fraction of th

e band is tubas?"
Mathematics
1 answer:
Shtirlitz [24]3 years ago
6 0
2/5 are brass, 1/8 of 2/5 are tubas
When you have the word "of" you're going to use multiplication. 
40% is brass (you know this by dividing 2 by 5 for the 2/5ths ratio), so you'll want to know what 1/8 of that is. 
40*(1/8) = 5 (this is already a percent)
You can also take 1/8 and multiply it by 2/5. You end up with the same thing, but it would be written as .05. To make this into a percent you multiply it by 100 to get 5%
Your final answer is 5% of the band are tubas. 

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Answer:

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3 years ago
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5 0
4 years ago
Read 2 more answers
An oval track is made by erecting semicircles on each end of a 44 m by 88 m rectangle. Find the length of the track and the area
Dmitry_Shevchenko [17]

Answer:

The length of the track is 247 m.

The area enclosed by the track is  5391 . 76 sq m.

Step-by-step explanation:

The length of the rectangle  = 88 m

The width of the rectangle  =  44 m

Now, as we know the semicircles are at the end of rectangular track.

So, the diameter of semicircles = Width of the rectangle

⇒ The diameter of the semi circle  = 44 m

⇒Radius of the semi circle  = 44/2 = 22 m

TOTAL LENGTH OF TRACK  

= 2 ( Length of rectangle)  + 2 (Circumference of 1 semicircle)

= 2 x (88 m)  + 2 (\frac{2\pi r}{2})

= 176 m + 2 (3.14)(22 m)  = 176 + 71.08 =  247 m

Hence, the length of the track is 247 m

TOTAL AREA OF TRACK  

= (Area of rectangle)  + 2 (Area of 1 semicircle)

= (Length x Width)  +    2 (\frac{\pi r^2}{2})

=  (44  m x 88 m) +  (3.14)(22 m)(22 m)   = (3872  + 1519.76) sq m

=  5391 . 76 sq m

Hence, the area enclosed by the track is  5391 . 76 sq m.

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