Answer :
To determine the relationship between [tex]\(x\)[/tex] (years since 2005) and [tex]\(y\)[/tex] (average annual salary in thousand dollars), let's analyze the information given:
1. The average annual salary in 2005 ([tex]\(x = 0\)[/tex]) was [tex]$70,000, which is equivalent to 70 thousand dollars. Therefore, the initial salary \(y_0 = 70\). 2. In 2006 (\(x = 1\)), the average annual salary increased to $[/tex]82,000, or 82 thousand dollars. Thus, [tex]\(y_1 = 82\)[/tex].
We are given that the salary increases by the same factor each year. This indicates an exponential growth relationship. The general form of an exponential growth equation is:
[tex]\[ y = y_0 \times r^x \][/tex]
where
- [tex]\(y_0\)[/tex] is the initial value (70 in this case),
- [tex]\(r\)[/tex] is the common ratio of increase,
- [tex]\(x\)[/tex] is the number of years since 2005,
- [tex]\(y\)[/tex] is the salary after [tex]\(x\)[/tex] years.
To find the common ratio [tex]\(r\)[/tex], we can use the salaries given for 2005 and 2006:
[tex]\[ y_1 = y_0 \times r \][/tex]
Plugging in the values we have:
[tex]\[ 82 = 70 \times r \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r = \frac{82}{70} = 1.17 \][/tex]
So the relationship between [tex]\(x\)[/tex] and [tex]\(y\)[/tex] should be:
[tex]\[ y = 70 \times (1.17)^x \][/tex]
Let's examine the options given and see which one matches this relationship:
1. [tex]\( y = 70(1.17)^x \)[/tex]
2. [tex]\( y = 82(1.17) \)[/tex]
3. [tex]\( y + 70 - 27 \)[/tex]
4. [tex]\( y = 82(2.2)^r \)[/tex]
The correct relationship is:
[tex]\[ y = 70(1.17)^x \][/tex]
Therefore, the correct model that represents the relationship between [tex]\(x\)[/tex] (years since 2005) and [tex]\(y\)[/tex] (average annual salary in thousand dollars) is:
[tex]\[ y = 70(1.17)^x \][/tex]
1. The average annual salary in 2005 ([tex]\(x = 0\)[/tex]) was [tex]$70,000, which is equivalent to 70 thousand dollars. Therefore, the initial salary \(y_0 = 70\). 2. In 2006 (\(x = 1\)), the average annual salary increased to $[/tex]82,000, or 82 thousand dollars. Thus, [tex]\(y_1 = 82\)[/tex].
We are given that the salary increases by the same factor each year. This indicates an exponential growth relationship. The general form of an exponential growth equation is:
[tex]\[ y = y_0 \times r^x \][/tex]
where
- [tex]\(y_0\)[/tex] is the initial value (70 in this case),
- [tex]\(r\)[/tex] is the common ratio of increase,
- [tex]\(x\)[/tex] is the number of years since 2005,
- [tex]\(y\)[/tex] is the salary after [tex]\(x\)[/tex] years.
To find the common ratio [tex]\(r\)[/tex], we can use the salaries given for 2005 and 2006:
[tex]\[ y_1 = y_0 \times r \][/tex]
Plugging in the values we have:
[tex]\[ 82 = 70 \times r \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r = \frac{82}{70} = 1.17 \][/tex]
So the relationship between [tex]\(x\)[/tex] and [tex]\(y\)[/tex] should be:
[tex]\[ y = 70 \times (1.17)^x \][/tex]
Let's examine the options given and see which one matches this relationship:
1. [tex]\( y = 70(1.17)^x \)[/tex]
2. [tex]\( y = 82(1.17) \)[/tex]
3. [tex]\( y + 70 - 27 \)[/tex]
4. [tex]\( y = 82(2.2)^r \)[/tex]
The correct relationship is:
[tex]\[ y = 70(1.17)^x \][/tex]
Therefore, the correct model that represents the relationship between [tex]\(x\)[/tex] (years since 2005) and [tex]\(y\)[/tex] (average annual salary in thousand dollars) is:
[tex]\[ y = 70(1.17)^x \][/tex]