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sukhopar [10]
3 years ago
9

It has been estimated that the population of Indianapolis is growing at a rate of 0.9% per year. Assuming that the population gr

owth continues at this rate and the population of Indianapolis was 1,988,817 in January of 2016, estimate the expected population of Indianapolis in January of 2019. (Please help me with this question. I’m genuinely stuck.)
Mathematics
1 answer:
sweet [91]3 years ago
8 0

Hi there!

Ok, so, here are the important facts:

Yearly population growing rate : 0.9%

Population in January 2016 : 1 988 817

Population in January 2019 (3 years later) : x


First off, you need to calculate 0.9% of the population in January 2016 to figure out the number of people it represents :

Saying 0.9% of 1 988 817 is like saying 0.9% times 1 988 817!


\frac{0.9}{100} = \frac{?}{1 988 817}


To solve this, you can use the cross product method :

(1 988 817 × 0.9) ÷ 100 = ?

1 789 935.3 ÷ 100 = ?

17 899.353 = ?


So 0.9% represents a growing rate of 17 899.353 people (we'll keep the decimals just to be more precise in the end).


Now, since you want to know the expected population in January 2019, 3 years later, you need to multiply 17 899.353 by 3 and then add the result to the population number in 2016 :


(17 899.353 × 3) + 1 988 817 = x

53 698.059 + 1 988 817 = x

2 042 515.059 = x

2 042 515 ≅ x (Since we are talking about a population, which represents a certain amout of people, you need to round the number)


Your answer is : The expected population of Indianapolis in January 2019 would be of 2,042,515 people.


There you go! I really hope this helped, if there's anything just let me know! :)

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Answer:

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Step-by-step explanation:

Given - Amanda went to the store to purchase ink pens. She found three    

            kinds of pens. The first cost $4 each; the price of the second kind

            was 4 for $1; and the cost for the third kind was 2 for $1. She bought

            20 pens and she bought at least one of each kind. (It is possible to

            buy only 1 of the pens that are "4 for $1" or "2 for $1".) The cost was

            $20.

To find - When she got back to her office, Amanda decided to turn this into

              a math problem for me. She asked: how many of each kind did I

              buy?

Proof -

Let Amanda buy first kind of pen = x

                      second kind of pen = y

                           third kind of pen = z

As given,

She bought total pen = 20

⇒x + y + z = 20          ...............(1)

Now,

As given,

cost for first kind pen = $4 for 1 pen

As she bought x pens of first kind , so

Cost of x pens of first kind = $4x

Now,

The price of the second kind was 4 for $1

⇒Cost of second kind = $\frac{1}{4} for 1 pen

As she bought y pens of send kind , so

Cost of y pens of second kind = $\frac{1}{4}y

Now,

The price of the third kind was 2 for $1

⇒Cost of third kind = $\frac{1}{2} for 1 pen

As she bought z pens of send kind , so

Cost of z pens of third kind = $\frac{1}{2}z

Now,

As given, The cost was $20

⇒4x + \frac{1}{4}y + \frac{1}{2}z = 20

⇒16x + y + 2z = 80             .....................(2)

∴ we get 2 equations

x + y + z = 20                   .....................(1)

16x + y + 2z = 80             .....................(2)

Now,

Subtract equation (1) from equation (2) , we get

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z = 60 - 15x

y = 14x - 40

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If x = 1, then y = 14 - 40 = -26

Not possible

If x = 2 , then y = 14(2) - 40 = -12

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If x = 3, then y = 14(3) - 40 = 2 and z = 60 - 15(3) = 15

Possible.

If x = 4,  then y = 14(4) - 40 = 16 and z = 60 - 15(4) = 0

Not Possible.

If x = 5, then y = 14(5) - 40 = 30 and z = 60 - 15(5) = -15

Not Possible.

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x = 3, y = 2, z = 15

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Amanda buy second kind of pen = y = 2

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