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3 years ago

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The population of a country has been decreasing for several decades. In 1990, the population was about 167 million people. In 20

10, the population was about 163 million people. Determine the percent decrease in the country's population during this time.
The country's population decreased by about ___% during this time period (Round to one decimal place as needed)

1 answer:

kogti [31]3 years ago

7 0

Answer: The country's population decreased by about 2.4% during this time period.

Step-by-step explanation:

Since we have given that

In 1990, the number of people = 167 million

In 2010, the number of people = 163 million

We need to find the percent decrease in the country's population during this time.

Percentage would be

$\dfrac{New-Original}{Original}\times 100\\\\=\dfrac{163-167}{167}\times 100\\\\=\dfrac{-4}{167}\times 100\\\\=-2.39\%\\\\=-2.4\%$

Hence, the country's population decreased by about 2.4% during this time period.

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Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

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$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

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$k * \frac{x^2}{2*2} [0,2]= 1$

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$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

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Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

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$f(x,y) = 1$

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$\int\limits^a_b \int\limits^a_b {f(x,y)}$

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$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

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