Find the GCF of 40 and 120, write 0.8333333... as a rational number, and complete the equivalent fraction 19/7 = 76/.

To find the Greatest Common Factor (GCF), we can list the prime factors of each number.
Prime factorization of 40:

$$40 = 2 \times 20 = 2 \times 2 \times 10 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5$$

Prime factorization of 120:

$$120 = 2 \times 60 = 2 \times 2 \times 30 = 2 \times 2 \times 2 \times 15 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$$

The common prime factors are $2^3$ and 5.
The GCF is the product of the lowest powers of the common prime factors.

$$\text{GCF}(40, 120) = 2^3 \times 5 = 8 \times 5 = 40$$ $$\boxed{\text{40}}$$

Let $x = 0.8333333...$
This can be written as $x = 0.8\overline{3}$.
Multiply by 10 to move the non-repeating part to the left of the decimal:

$$10x = 8.333333... \quad (1)$$

Multiply by 100 to move one full repeating cycle to the left of the decimal:

$$100x = 83.333333... \quad (2)$$

Subtract equation (1) from equation (2):

$$100x - 10x = 83.333333... - 8.333333...$$ $$90x = 75$$

Divide by 90:

$$x = \frac{75}{90}$$

Simplify the fraction by dividing both numerator and denominator by their greatest common factor, which is 15:

$$x = \frac{75 \div 15}{90 \div 15} = \frac{5}{6}$$ $$\boxed{\frac{5}{6}}$$

$\frac{19}{7} = \frac{76}{\square}$
To find the missing denominator, first determine what factor the numerator 19 was multiplied by to get 76.

$$19 \times n = 76$$ $$n = \frac{76}{19}$$ $$n = 4$$

Now, multiply the original denominator 7 by the same factor:

$$7 \times 4 = 28$$

So, the equivalent fraction is $\frac{19}{7} = \frac{76}{28}$.

$$\boxed{\text{28}}$$

Divide 16 by 3:

$$\frac{16}{3} = 5.333333...$$

To express this correct to 4 decimal places, we look at the fifth decimal place. If it is 5 or greater, we round up the fourth decimal place. If it is less than 5, we keep the fourth decimal place as it is.
The fifth decimal place is 3, which is less than 5. So, we keep the fourth decimal place as 3.

$$5.3333$$ $$\boxed{5.3333}$$