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Ray Of Light [21]
3 years ago
8

Problem 8. Over the next 100 years, real GDP per capita in Groland is expected to grow at an average annual rate of 2.0%. In Pol

and, however, growth is expected to be somewhat slower, at an average annual growth rate of 1.5%. If both countries have a real GDP per capita today of $20,000, how will their real GDP per capita differ in 100 years?
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Answer:

Their GDP will differ by $56252.01

Step-by-step explanation:

Consider the provided information.

It is given that Groland is expected to grow at an average annual rate of 2.0%.

The GDP per capita today of Groland is $20,000,

In 100 years  Groland's real per capita GDP will be:

20000 \times (1 + 0.02) = 20000 \times (1.02)^{100}= 144892.92

Hence, in 100 year Groland's real per capita GDP will be $144892.92

It is given that Poland's is expected to grow at an average annual rate of 1.5%.

The GDP per capita today of Poland is $20,000,

In 100 years  Poland's real per capita GDP will be:

20000 \times (1 + 0.015) = 20000 \times (1.015)^{100}= 88640.91

Hence, in 100 year Poland's real per capita GDP will be $88640.91

Now compare how their real GDP per capital differ in 100 years.

$144892.92-$88640.91=$56252.01

Therefore, their GDP will differ by $56252.01

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After 6 months the simple interest earned annually on an investment of $3000 was $811. Find the interest rate to the nearest ten
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Need help solving the equation, trying to figure whether I multiply or divide.​
USPshnik [31]

Answer:

m∠B = 43°

b = 26.1

c = 38.3

Step-by-step explanation:

By applying triangle sum in the given triangle,

47° + 90° + m∠B = 180°

137° + m∠B = 180°

m∠B = 43°

By applying sine ratio to the angle measuring 47°.

sin(47°) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

            = \frac{28}{c}

c = \frac{28}{\text{sin}(47^{\circ})}

c = 38.28

c ≈ 38.3

By applying cosine rule,

cos(47°) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

             = \frac{b}{c}

             = \frac{b}{38.3}

b = 38.3[cos(47°)]

b = 26.1

5 0
3 years ago
What is [[3^3-7)·5)/10]+((1+3+6+9)-)]
yanalaym [24]

Answer:

\left(\left(3^3-7\right)\frac{5}{10}\right)+\left(\left(1+3+6+9\right)-1\right)=28

Step-by-step explanation:

As far as I am able to observe from the statement of your question, the expression is:

\left[[\left(3^3-7\right)\cdot \frac{5}{10}\right]+\left(\left(1+3+6+9\right)-1\right)

So, lets solve this expression, which anyways would clear your concept

Considering the expression

\left[[\left(3^3-7\right)\cdot \frac{5}{10}\right]+\left(\left(1+3+6+9\right)-1\right)

\mathrm{Remove\:parentheses}:\quad \left(a\right)=a

=\left(3^3-7\right)\frac{5}{10}+1+3+6+9-1

Lets first solve \left(3^3-7\right)\frac{5}{10}

As 3^3=27

=\frac{5}{10}\left(27-7\right)

=\frac{1}{2}\left(27-7\right)

=20\cdot \frac{1}{2}

=10

So,

\left(3^3-7\right)\frac{5}{10}+1+3+6+9-1 = 10+1+3+6+9-1  

                                                 =  28

Therefore,

\left(\left(3^3-7\right)\frac{5}{10}\right)+\left(\left(1+3+6+9\right)-1\right)=28

Keywords: Expression solving

Learn more about solving expression from brainly.com/question/4687406

#learnwithBrainly

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Answer:

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Step-by-step explanation:

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