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Digiron [165]
4 years ago
6

jacob painted 30% of a banner. luis painted an equal amount of the same banner. the rest of the banner remains unpainted. write

a fraction to describe the amount of banner that remains unpainted?
Mathematics
1 answer:
Arisa [49]4 years ago
3 0
Answer= 2/5

Jacob painted 30% of the banner

Luis painted an equal amount, so we can infer that Luis painted 30% of the banner.

Jacob+Luis=
30%+30%=60%

Jacob and Luis painted 60% of the banner. To find the amount of the banner that remains unpainted, simply subtract 60% from 100%...

100%-60%=40%

Remember that percent means 'out of one hundred', so 40% is the same as 40/100. we can simplify 40/100 to 2/5

2/5 of the banner remains unpainted
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Answer:

Step-by-step explanation:

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A(2, 2) → A'(2+4, 2+1)

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3 years ago
use elimination to solve. For a production of sweeney Todd in Las Vegas, orchestra seats cost $42 and mezzanine seats cost $25.
lidiya [134]

They bought 8 orchestra seats and 6 mezzanine seats.

Step-by-step explanation:

Cost of one orchestra seat = $42

Cost of one mezzanine seat = $25

No. of people = 14

Amount spent = $486

Let,

x be the number of orchestra seats

y be the number of mezzanine seats

According to given statement;

x+y=14     Eqn 1

42x+25y=486    Eqn 2

Multiplying Eqn 1 by 25;

25(x+y=14)\\25x+25y=350\ \ \ \ Eqn\ 3

Subtracting Eqn 3 from Eqn 2 to eliminate y;

(42x+25y)-(25x+25y)=486-350\\42x+25y-25x-25y=136\\17x=136

Dividing both sides by 17

\frac{17x}{17}=\frac{136}{17}\\x=8

Putting x=8 in Eqn 1

8+y=14\\y=14-8\\y=6

They bought 8 orchestra seats and 6 mezzanine seats.

Keywords: linear equation, elimination method

Learn more about elimination method at:

  • brainly.com/question/11258952
  • brainly.com/question/11343416

#LearnwithBrainly

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4 years ago
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Answer:

See figure attached and explanation below.

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And for this case we can use the followinf R code to create the frequency histogram.

> x<-c(0.74, 0.32, 1.66, 3.59, 4.55, 6.47, 9.99,0.7, 0.37, 0.76, 1.9, 1.77, 2.42, 1.09, 2.03, 2.69,2.41, 0.54, 8.32, 5.70, 0.75, 1.96, 3.36, 4.06,12.48)

> length(x)

[1] 25

> hist(x, prob=TRUE)

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