Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Professional athletes are some of the highest paid people in the world. The average major league baseball player's salary has climbed from $1,998,000 in 2000 to $2,866,500 in 2006

Professional Athletes Are Some Of The Highest Paid People In The World The Average Major League Baseball Players Salary Has Climbed From 1998000 In 2000 To 2866 class=

Sagot :

Answer:

[tex]L(y)=144750y+1998000[/tex]

The slope means that each year the average professional baseball player's salary increased by $144,750 for every year after 2000.

The y-intercept means that in year 0 (2000) the average professional baseball player's salary was $1,998,000

The predicted average salary in 2007 is $3,011,250

(b)

[tex]E(y)=1998000\cdot1.062^y[/tex]

The initial value represents the average professional baseball player's salary in year 0 (2000), which was $1,998,000.

The growth factor means that the rate of change increases each year by 1.062 times the previous year's increase.

The predicted average salary in 2007 is $3,044,233.94

Explanation:

The problem gives us two pieces of information:

In year 2000, we call t = 0, the average salary was $1,998,000

In year 2006, we call t = 6, the average salary was $2,866,500

If we want to make a function of the average salary variation over the years, we have two points that must lie in the equation of that function:

(0, 1998000) and (6, 2866500)

For (a) we need to assume that is linear growth. The equation of a line is:

[tex]y=mx+b[/tex]

Where:

m is the slope

b is the y-intercept. In this case, since we established the year 2000 as t = 0, b = 1998000

Given two points P and Q, we can find the slope by the formula:

[tex]\begin{gathered} \begin{cases}P=(x_P,y_P){} \\ Q=(x_Q,y_Q)\end{cases} \\ . \\ m=\frac{y_Q-y_P}{x_Q-x_P} \end{gathered}[/tex]

Then, if we call:

P = (0, 1998000)

Q = (6, 2866500)

[tex]m=\frac{2866500-1998000}{6-0}=\frac{868500}{6}=144750[/tex]

Thus, the equation of the linear growth model is:

[tex]L(t)=144750t+1998000[/tex]

Now, we can use this to find a prediction for 2007. 2007 is 7 years since 2000; thus t = 7

[tex]L(7)=144750\cdot7+1998000=1013250+1998000=3011250[/tex]

In (b) we assume an exponential growth. The formula for the exponential growth is:

[tex]y=a(1+r)^t[/tex]

Where:

a is the initial value. In this case, the average salary in 2000, $1,998,000

r is the ratio of growth. We need to find this value

t is the time in years

Then, we can use the point (6, 2866500), and the fact that a = 1998000:

[tex]2866500=1998000(1+r)^6[/tex]

And solve:

[tex]\begin{gathered} \frac{2866500}{199800}=(1+r)^6 \\ . \\ \sqrt[6]{\frac{637}{444}}=1+r \\ . \\ 1.062=1+r \end{gathered}[/tex]

We call the term "1 + r" growth factor.

Now, we can write the formula:

[tex]E(t)=1998000\cdot1.062^t[/tex]

To find a prediction of the average salary in 2007, we use the function and t = 7:

[tex]E(7)=1998000\cdot1.062^7=1998000\cdot1.5236=3044233.937[/tex]