Questions: HW Score: 97.6%, 9.75 of 10 points
Points: 0.75 of 1
The graph shows the projections in total enrollment at degree granting institutions from Fall 2003 to Fall 2012.
The linear model, y=2145x+15.59, provides the approximate enrollment, in millions, between the years 2003 and 2012, where x=0 corresponds to 2003, x=1 to 2004, and so on, and y is in millions of students.
(a) Use the model to determine projected enrollment for Fall 2008.
The projected enrollment for Fall 2008 is millions. (Type an integer or decimal rounded to the nearest tenth as needed.)
Transcript text: HW
Question 10, 1.2.45
Part i of 4
HW Score: $97.6 \%, 9.75$ of 10 points
Points: 0.75 of 1
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The graph shows the projections in total enrollment at degree granting institutions from Fall 2003 to Fall 2012
The linear model, $y=2145 x+15.59$, provides the approximate enrollment, in millions, between the years 2003 and 2012, where $x=0$ corresponds to 2003, $x=1$ to 2004, and so on, and $y$ is in millions of students
(a) Use the model to determine projected enrollment for Fall 2008
The projected enrollment for Fall 2008 is $\square$ millions
(Type an integer or decimal rounded to the nearest tenth as needed.)
Solution
Solution Steps
Step 1: Identify the given linear model and the year for projection
The given linear model is \( y = 0.2145x + 15.59 \), where \( x = 0 \) corresponds to 2003. We need to determine the projected enrollment for Fall 2008.
Step 2: Determine the value of \( x \) for the year 2008
Since \( x = 0 \) corresponds to 2003, we calculate \( x \) for 2008 as follows:
\[ x = 2008 - 2003 = 5 \]
Step 3: Substitute \( x = 5 \) into the linear model
Substitute \( x = 5 \) into the equation \( y = 0.2145x + 15.59 \):
\[ y = 0.2145(5) + 15.59 \]
Step 4: Calculate the projected enrollment
Perform the calculation:
\[ y = 0.2145 \times 5 + 15.59 \]
\[ y = 1.0725 + 15.59 \]
\[ y = 16.6625 \]
Final Answer
The projected enrollment for Fall 2008 is approximately 16.7 million students.