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Sagot :
To find the number of banks in the country according to the given function, we'll need to calculate [tex]\( f(x) \)[/tex] for the specific years in question. Here, [tex]\( x \)[/tex] represents the number of years after 1900.
### Part (a):
1. Calculate the number of banks in 1970:
- For 1970, [tex]\( x = 1970 - 1900 = 70 \)[/tex].
- Since [tex]\( 70 < 90 \)[/tex], we use the function [tex]\( f(x) = 84.7x + 12,363 \)[/tex].
- Substituting [tex]\( x = 70 \)[/tex]:
[tex]\[ f(70) = 84.7 \times 70 + 12,363 \][/tex]
[tex]\[ f(70) = 5,929 + 12,363 \][/tex]
[tex]\[ f(70) = 18,292 \][/tex]
- Therefore, the number of banks in 1970 is [tex]\( 18,292 \)[/tex].
2. Calculate the number of banks in 2000:
- For 2000, [tex]\( x = 2000 - 1900 = 100 \)[/tex].
- Since [tex]\( 100 \geq 90 \)[/tex], we use the function [tex]\( f(x) = -376.3x + 48,686 \)[/tex].
- Substituting [tex]\( x = 100 \)[/tex]:
[tex]\[ f(100) = -376.3 \times 100 + 48,686 \][/tex]
[tex]\[ f(100) = -37,630 + 48,686 \][/tex]
[tex]\[ f(100) = 11,056 \][/tex]
- Therefore, the number of banks in 2000 is [tex]\( 11,056 \)[/tex].
3. Calculate the number of banks in 2010:
- For 2010, [tex]\( x = 2010 - 1900 = 110 \)[/tex].
- Since [tex]\( 110 \geq 90 \)[/tex], we use the function [tex]\( f(x) = -376.3x + 48,686 \)[/tex].
- Substituting [tex]\( x = 110 \)[/tex]:
[tex]\[ f(110) = -376.3 \times 110 + 48,686 \][/tex]
[tex]\[ f(110) = -41,393 + 48,686 \][/tex]
[tex]\[ f(110) = 7,293 \][/tex]
- Therefore, the number of banks in 2010 is [tex]\( 7,293 \)[/tex].
Summary for part (a):
- Number of banks in 1970: [tex]\( 18,292 \)[/tex]
- Number of banks in 2000: [tex]\( 11,056 \)[/tex]
- Number of banks in 2010: [tex]\( 7,293 \)[/tex]
### Part (b):
Graph the function for [tex]\( 50 \leq x \leq 120 \)[/tex].
To graph the function:
- Split the range [tex]\( 50 \leq x \leq 120 \)[/tex] into two parts: [tex]\( 50 \leq x < 90 \)[/tex] and [tex]\( 90 \leq x \leq 120 \)[/tex].
- Use [tex]\( f(x) = 84.7x + 12,363 \)[/tex] for [tex]\( 50 \leq x < 90 \)[/tex].
- Use [tex]\( f(x) = -376.3x + 48,686 \)[/tex] for [tex]\( 90 \leq x \leq 120 \)[/tex].
1. Plot [tex]\( f(x) = 84.7x + 12,363 \)[/tex] for [tex]\( 50 \leq x < 90 \)[/tex]:
- Select a range of [tex]\( x \)[/tex] values, e.g., [tex]\( 50, 51, 52, \ldots, 89 \)[/tex].
- Calculate corresponding [tex]\( y = f(x) \)[/tex] values for each [tex]\( x \)[/tex].
2. Plot [tex]\( f(x) = -376.3x + 48,686 \)[/tex] for [tex]\( 90 \leq x \leq 120 \)[/tex]:
- Select a range of [tex]\( x \)[/tex] values, e.g., [tex]\( 90, 91, 92, \ldots, 120 \)[/tex].
- Calculate corresponding [tex]\( y = f(x) \)[/tex] values for each [tex]\( x \)[/tex].
3. Graph these points on a coordinate system:
- Use the [tex]\( x \)[/tex]-axis to represent the years after 1900.
- Use the [tex]\( y \)[/tex]-axis to represent the number of banks.
The graph will show a piecewise linear function with a positive slope for [tex]\( 50 \leq x < 90 \)[/tex] and a negative slope for [tex]\( 90 \leq x \leq 120 \)[/tex].
### Part (a):
1. Calculate the number of banks in 1970:
- For 1970, [tex]\( x = 1970 - 1900 = 70 \)[/tex].
- Since [tex]\( 70 < 90 \)[/tex], we use the function [tex]\( f(x) = 84.7x + 12,363 \)[/tex].
- Substituting [tex]\( x = 70 \)[/tex]:
[tex]\[ f(70) = 84.7 \times 70 + 12,363 \][/tex]
[tex]\[ f(70) = 5,929 + 12,363 \][/tex]
[tex]\[ f(70) = 18,292 \][/tex]
- Therefore, the number of banks in 1970 is [tex]\( 18,292 \)[/tex].
2. Calculate the number of banks in 2000:
- For 2000, [tex]\( x = 2000 - 1900 = 100 \)[/tex].
- Since [tex]\( 100 \geq 90 \)[/tex], we use the function [tex]\( f(x) = -376.3x + 48,686 \)[/tex].
- Substituting [tex]\( x = 100 \)[/tex]:
[tex]\[ f(100) = -376.3 \times 100 + 48,686 \][/tex]
[tex]\[ f(100) = -37,630 + 48,686 \][/tex]
[tex]\[ f(100) = 11,056 \][/tex]
- Therefore, the number of banks in 2000 is [tex]\( 11,056 \)[/tex].
3. Calculate the number of banks in 2010:
- For 2010, [tex]\( x = 2010 - 1900 = 110 \)[/tex].
- Since [tex]\( 110 \geq 90 \)[/tex], we use the function [tex]\( f(x) = -376.3x + 48,686 \)[/tex].
- Substituting [tex]\( x = 110 \)[/tex]:
[tex]\[ f(110) = -376.3 \times 110 + 48,686 \][/tex]
[tex]\[ f(110) = -41,393 + 48,686 \][/tex]
[tex]\[ f(110) = 7,293 \][/tex]
- Therefore, the number of banks in 2010 is [tex]\( 7,293 \)[/tex].
Summary for part (a):
- Number of banks in 1970: [tex]\( 18,292 \)[/tex]
- Number of banks in 2000: [tex]\( 11,056 \)[/tex]
- Number of banks in 2010: [tex]\( 7,293 \)[/tex]
### Part (b):
Graph the function for [tex]\( 50 \leq x \leq 120 \)[/tex].
To graph the function:
- Split the range [tex]\( 50 \leq x \leq 120 \)[/tex] into two parts: [tex]\( 50 \leq x < 90 \)[/tex] and [tex]\( 90 \leq x \leq 120 \)[/tex].
- Use [tex]\( f(x) = 84.7x + 12,363 \)[/tex] for [tex]\( 50 \leq x < 90 \)[/tex].
- Use [tex]\( f(x) = -376.3x + 48,686 \)[/tex] for [tex]\( 90 \leq x \leq 120 \)[/tex].
1. Plot [tex]\( f(x) = 84.7x + 12,363 \)[/tex] for [tex]\( 50 \leq x < 90 \)[/tex]:
- Select a range of [tex]\( x \)[/tex] values, e.g., [tex]\( 50, 51, 52, \ldots, 89 \)[/tex].
- Calculate corresponding [tex]\( y = f(x) \)[/tex] values for each [tex]\( x \)[/tex].
2. Plot [tex]\( f(x) = -376.3x + 48,686 \)[/tex] for [tex]\( 90 \leq x \leq 120 \)[/tex]:
- Select a range of [tex]\( x \)[/tex] values, e.g., [tex]\( 90, 91, 92, \ldots, 120 \)[/tex].
- Calculate corresponding [tex]\( y = f(x) \)[/tex] values for each [tex]\( x \)[/tex].
3. Graph these points on a coordinate system:
- Use the [tex]\( x \)[/tex]-axis to represent the years after 1900.
- Use the [tex]\( y \)[/tex]-axis to represent the number of banks.
The graph will show a piecewise linear function with a positive slope for [tex]\( 50 \leq x < 90 \)[/tex] and a negative slope for [tex]\( 90 \leq x \leq 120 \)[/tex].
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