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brilliants [131]
4 years ago
14

The population of a town 4 years ago was 50,000

Mathematics
1 answer:
tino4ka555 [31]4 years ago
8 0

Answer:

population will be 79806.

Step-by-step explanation:

Population of a town 4 years ago  was 50,000

Population of town had grown exponentially and present population is 63,124.

So the formula of the population growth will be

A_{n}=A_{o}(r)^{n-1}

Where A_{n} = Final population n = number of years.

           A_{o} = Initial population

           r = rate of increase in population per year.

so  63124 = 50,000 (r)^{4-1}

r^{3}=\frac{63124}{50000}= 1.26428

r=(1.26428)^{\frac{1}{3} }

= 1.0813

Now we have to find the population after 4 years from now.

A_{n} =  63124(1.08)^{4-1}

       =  63124(1.08)^{3}

       = 63124 × 1.264

       = 79806

Therefore, after 4 years from now population will be 79806.

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Answer:

$7,000 at a rate of 7% and $21,000 at a rate of 14%.

Step-by-step explanation:

Let x be amount invested at 7% and y be amount invested at 14%.

We have been given that a women's professional organization made two small-business loans totaling $28,000. We can represent this information in an equation as:

x+y=28,000...(1)

The interest earned at 7% in one year would be 0.07x and interest earned at 14% in one year would be 0.14x.

We are also told that the organization received from these loans was $3,430. We can represent this information in an equation as:

0.07x+0.14y=3,430...(2)

Form equation (1), we will get:

x=28,000-y

Upon substituting this value in equation (2), we will get:

0.07(28,000-y)+0.14y=3,430

1960-0.07y+0.14y=3,430

1960+0.07y=3,430

1960-1960+0.07y=3,430-1960

0.07y=1470

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y=21,000

Therefore, an amount of $21,000 was invested at a rate of 14%.

x=28,000-y

x=28,000-21,000

x=7,000

Therefore, an amount of $7,000 was invested at a rate of 14%.

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Answer:

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Step-by-step explanation:

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