Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

the accompanying bar graph shows the resident population and the land a…

Question

the accompanying bar graph shows the resident population and the land area of a particular country for selected years from 1800 through 2019. use the information to find the population density of a certain country, to the nearest tenth, expressed in people per square kilometer, in 1900.
1 in² = 6.5 cm²
1 ft² = 0.09 m²
1 yd² = 0.8 m²
1 mi² = 2.6 km²
1 acre = 0.4 hectare (ha)
click here to view the bar graph.

the population density in 1900 was approximately
(round to the nearest tenth as needed.)

Explanation:

Response

To solve this problem, we need the population and land area data from the bar graph for the year 1900. Since the bar graph is not provided here, I'll assume typical values for the United States (a common example for such problems) in 1900: population was approximately 76,212,168 people, and land area was approximately 3,540,842 square miles.

Step 1: Convert land area from square miles to square kilometers

We know that \( 1 \, \text{mi}^2 = 2.6 \, \text{km}^2 \).
Let \( A_{\text{km}^2} \) be the land area in square kilometers.
\[
A_{\text{km}^2} = 3,540,842 \, \text{mi}^2 \times 2.6 \, \frac{\text{km}^2}{\text{mi}^2}
\]
\[
A_{\text{km}^2} = 3,540,842 \times 2.6 = 9,206,189.2 \, \text{km}^2
\]

Step 2: Calculate population density

Population density \( D \) is defined as \( D = \frac{\text{Population}}{\text{Land Area}} \).
Let \( P = 76,212,168 \) people and \( A = 9,206,189.2 \, \text{km}^2 \).
\[
D = \frac{76,212,168}{9,206,189.2}
\]
\[
D \approx 8.3 \, \text{people per square kilometer}
\]

Final Answer

The population density in 1900 was approximately \(\boldsymbol{8.3}\) people per square kilometer (assuming the population and land area values for the US in 1900).

Note: If the bar graph provides different values for population and land area, substitute those into the calculations above.

Answer:

To solve this problem, we need the population and land area data from the bar graph for the year 1900. Since the bar graph is not provided here, I'll assume typical values for the United States (a common example for such problems) in 1900: population was approximately 76,212,168 people, and land area was approximately 3,540,842 square miles.

Step 1: Convert land area from square miles to square kilometers

We know that \( 1 \, \text{mi}^2 = 2.6 \, \text{km}^2 \).
Let \( A_{\text{km}^2} \) be the land area in square kilometers.
\[
A_{\text{km}^2} = 3,540,842 \, \text{mi}^2 \times 2.6 \, \frac{\text{km}^2}{\text{mi}^2}
\]
\[
A_{\text{km}^2} = 3,540,842 \times 2.6 = 9,206,189.2 \, \text{km}^2
\]

Step 2: Calculate population density

Population density \( D \) is defined as \( D = \frac{\text{Population}}{\text{Land Area}} \).
Let \( P = 76,212,168 \) people and \( A = 9,206,189.2 \, \text{km}^2 \).
\[
D = \frac{76,212,168}{9,206,189.2}
\]
\[
D \approx 8.3 \, \text{people per square kilometer}
\]

Final Answer

The population density in 1900 was approximately \(\boldsymbol{8.3}\) people per square kilometer (assuming the population and land area values for the US in 1900).

Note: If the bar graph provides different values for population and land area, substitute those into the calculations above.