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Fudgin [204]
3 years ago
15

Whats the answer I am stuck

Mathematics
2 answers:
lubasha [3.4K]3 years ago
8 0

15/28 I believe is the answer

IrinaK [193]3 years ago
8 0

Answer:

3 \frac{11}{28}

Step-by-step explanation:

Note that

2 \frac{1}{7} = 2 + \frac{1}{7} and

1 \frac{1}{4} = 1 + \frac{1}{4}

We can add the whole numbers together and the 2 fractions separately, that is

2 + 1 + \frac{1}{7} + \frac{1}{4}

To add the fractions we require them to have a common denominator

The lowest common multiple of 7 and 4 is 28.

Multiply the numerator/denominator of the first fraction by 4

Multiply the numerator/denominator of the second fraction by 7

= 3 + \frac{1(4)}{7(4)} + \frac{1(7)}{4(7)}

= 3 + \frac{4}{28} + \frac{7}{28}

= 3 + \frac{4+7}{28}

= 3 + \frac{11}{28}

= 3 \frac{11}{28}

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At the end of 2006, the population of Riverside was 400 people. The population for this small town can be modeled by the equatio
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The correct format of the question is

At the end of 2006, the population of Riverside was 400 people. The population for this small town can be modeled by the equation below, where t represents the number of years since the end of 2006 and P represents the number of people. P = 400 ( 1.2 )^ t Based on this model, approximately what was the increase in the population of Riverside at the end of 2009 compared to the end of 2006?

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(C) 1040

(D) 1440

Answer:

The increase in the population at the end of 2009 is 291 people

Step-by-step explanation:

We are given the equation as P = 400 ( 1.2 )^ t

where

P = No of People

t= No of Years

it is given that in the year 2006 the population is 400

this will only happen when we take t= 0

so for

Year    value of t

2006-    t = 0

2007-    t  = 1

2008-    t = 2

2009     t = 3

No of people in 2009 will be

P = 400(1.2)^3

     = 400*1.728

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Since the equation represents no of people so it can't be in decimals, Therefore the population will be 691

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                = 291

The increase in the population at the end of 2009 is 291 people.

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