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The problem can be solved by setting up a system of linear equations.

Let:
- $ x $ be the number of children
- $ y $ be the number of adults

We have two pieces of information:
1. The total number of people is 303:  
   \[
   x + y = 303
   \]
2. The total amount of money collected is $1276. Since children pay $2.75 and adults pay $6.60, we can write:  
   \[
   2.75x + 6.60y = 1276
   \]

Now, we have the system of equations:
\[
\begin{aligned}
x + y &= 303 \\
2.75x + 6.60y &= 1276
\end{aligned}
\]

### Step 1: Solve the first equation for one variable.
From the first equation, solve for $ y $:
\[
y = 303 - x
\]

### Step 2: Substitute $ y $ into the second equation.
Substitute $ y = 303 - x $ into the second equation:
\[
2.75x + 6.60(303 - x) = 1276
\]
Now simplify and solve for $ x $:
\[
2.75x + 6.60(303) - 6.60x = 1276
\]
\[
2.75x + 1999.8 - 6.60x = 1276
\]
\[
-3.85x + 1999.8 = 1276
\]
\[
-3.85x = 1276 - 1999.8
\]
\[
-3.85x = -723.8
\]
\[
x = \frac{-723.8}{-3.85} = 188
\]

So, $ x = 188 $, meaning there are 188 children.

### Step 3: Find $ y $.
Substitute $ x = 188 $ back into the first equation:
\[
188 + y = 303
\]
\[
y = 303 - 188 = 115
\]

So, there are 115 adults.

### Final answer:
- Number of children = 188
- Number of adults = 115

Would you like further clarification or additional details?

Here are some related questions to explore:
1. How would the solution change if the total admission fees collected were higher?
2. What if the adult fee was increased to $7.50? How would that affect the number of children and adults?
3. Can you solve the same problem using substitution or graphing methods?
4. What happens if the park introduces a family discount for a group of 4?
5. How would this system work if more types of tickets (like senior tickets) were added?

**Tip:** Always check if the solution makes sense by plugging the values back into both original equations to ensure correctness!

The admission fee at an amusement park is $2.75 for children and $6.60 for adults. On a certain day, 303 people entered the park, and the admission fees collected totaled $1276. How many children and how many adults were admitted?

Learn how to solve a real-world algebra problem involving a system of equations. The problem focuses on determining the number of children and adults admitted to an amusement park based on admission fees and the total number of people. Suitable for students in Grades 8-10.

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