If the arithmetic mean of the following data is 7, then a + b =
`{:(x_(i), 4,6,7,9),(f_(i)"",a,4,b,5):}`

Options

A. 4
B. 2
C. 3
D. cannot be determined

Explanation

To solve the problem step by step, we will find the values of \( a \) and \( b \) given that the arithmetic mean of the data is 7.

### Step 1: Set up the data
We have the following data:
- \( x_i = \{4, 6, 7, 9\} \)
- \( f_i = \{a, 4, b, 5\} \)

### Step 2: Calculate \( \Sigma (x_i f_i) \)
We need to calculate \( \Sigma (x_i f_i) \):
\[ \Sigma (x_i f_i) = 4a + 6 \cdot 4 + 7b + 9 \cdot 5 \]
Calculating the constants:
\[ = 4a + 24 + 7b + 45 \]
Combining like terms:
\[ = 4a + 7b + 69 \]

### Step 3: Calculate \( \Sigma f_i \)
Next, we calculate \( \Sigma f_i \):
\[ \Sigma f_i = a + 4 + b + 5 \]
Combining like terms:
\[ = a + b + 9 \]

### Step 4: Set up the equation for the mean
We know that the arithmetic mean is given by:
\[ \text{Mean} = \frac{\Sigma (x_i f_i)}{\Sigma f_i} \]
Given that the mean is 7, we can set up the equation:
\[ 7 = \frac{4a + 7b + 69}{a + b + 9} \]

### Step 5: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ 7(a + b + 9) = 4a + 7b + 69 \]
Expanding the left side:
\[ 7a + 7b + 63 = 4a + 7b + 69 \]

### Step 6: Simplify the equation
Now, we can simplify the equation:
\[ 7a + 7b + 63 - 7b = 4a + 69 \]
This simplifies to:
\[ 7a + 63 = 4a + 69 \]

### Step 7: Isolate \( a \)
Now, we isolate \( a \):
\[ 7a - 4a = 69 - 63 \]
This simplifies to:
\[ 3a = 6 \]
Dividing both sides by 3:
\[ a = 2 \]

### Step 8: Substitute \( a \) back into the equation
Now that we have \( a = 2 \), we can substitute it back into the equation for \( \Sigma f_i \):
\[ \Sigma f_i = 2 + b + 9 = b + 11 \]

### Step 9: Find \( a + b \)
To find \( a + b \):
\[ a + b = 2 + b \]
Since we have one equation with two unknowns, we can express \( b \) in terms of \( a + b \):
\[ a + b = 2 + b \]
This means that \( a + b \) can take any value depending on the value of \( b \).

### Conclusion
Thus, we conclude that \( a + b \) can take infinite values depending on \( b \), but we know \( a = 2 \).