Question

Write a graph exponential function is to model the number of students at school a.m. school be as a function of number of years in about how many years will the number of students at both schools be approximately the same explain how you can use a graph to determine the answer school A 240 students decreasing an annual rate of 2% Scooby, 180 students increasing at an annual rate of 3% Write the function for the number of students at school A after x years f(x)= simplify your answer used integers are decimals for any numbers in the expression

Answer 1

To model the number of students at each school over time, we can use exponential functions based on the given information.

**School A:**
- Initial number of students: 240
- Decreasing at an annual rate of 2%

The function for the number of students at School A after \( x \) years can be modeled as:
\[
f(x) = 240 \times (1 - 0.02)^x = 240 \times (0.98)^x
\]

**School B:**
- Initial number of students: 180
- Increasing at an annual rate of 3%

The function for the number of students at School B after \( x \) years can be modeled as:
\[
g(x) = 180 \times (1 + 0.03)^x = 180 \times (1.03)^x
\]

Now, we have the two functions:
1. \( f(x) = 240 \times (0.98)^x \) for School A
2. \( g(x) = 180 \times (1.03)^x \) for School B

### Finding when the number of students will be approximately the same
To find when these two functions are approximately equal, we need to solve the equation:
\[
240 \times (0.98)^x = 180 \times (1.03)^x
\]

### Solving the equation
We can rearrange the equation:
\[
\frac{240}{180} = \frac{(1.03)^x}{(0.98)^x}
\]

This simplifies to:
\[
\frac{4}{3} = \left(\frac{1.03}{0.98}\right)^x
\]

Taking the natural logarithm of both sides:
\[
\ln\left(\frac{4}{3}\right) = x \cdot \ln\left(\frac{1.03}{0.98}\right)
\]

Now, we can solve for \( x \):
\[
x = \frac{\ln\left(\frac{4}{3}\right)}{\ln\left(\frac{1.03}{0.98}\right)}
\]

### Calculating the values
Using a calculator:
1. Calculate \( \ln(4/3) \):
\[
\ln\left(\frac{4}{3}\right) \approx 0.2877
\]

2. Calculate \( \ln(1.03/0.98) \):
\[
\ln\left(\frac{1.03}{0.98}\right) \approx \ln(1.0306122449) \approx 0.0304
\]

3. Now substitute these values into the equation for \( x \):
\[
x \approx \frac{0.2877}{0.0304} \approx 9.46
\]

Thus, after about **9.46 years**, the number of students at both schools will be approximately the same.

### Using a Graph to Determine the Answer
To graphically determine when the two numbers of students will be the same, you would:

1. Plot the function \( f(x) \) and \( g(x) \) on the same set of axes.
2. Look for the intersection point of the two curves. The \( x \)-coordinate of the intersection point will give you the approximate number of years until the number of students at the two schools is the same.

In conclusion, the functions for the number of students at School A and School B are as follows:
- School A: \( f(x) = 240 \times (0.98)^x \)
- School B: \( g(x) = 180 \times (1.03)^x \)

They will be approximately the same after about 9.46 years.