Simplify $3\cdot\frac{11}{4}\cdot \frac{16}{-55}$.
Note that 55 and 11 have a common factor of 11. Also, 4 and 16 have a common factor of 4. Since there is one negative sign among all the factors, our result will be negative. We get \[
3\cdot\frac{11}{4}\cdot \frac{16}{-55}=-3\cdot\frac{\cancel{11}}{\cancel{4}}\cdot \frac{\cancelto{4}{16}}{\cancelto{5}{55}} \quad =-\frac{3\cdot 4}{5}=\boxed{-\frac{12}{5}}.
\]
-\frac{12}{5}
Level 3
Prealgebra
Problem: Simplify $3\cdot\frac{11}{4}\cdot \frac{16}{-55}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first multiply the numerators and the denominators separately. The numerator is \(3 \cdot 11 \cdot 16\) and the denominator is \(4 \cdot -55\). 

Calculate the numerator: 
1. \(3 \cdot 11 = 33\)
2. \(33 \cdot 16 = 528\)

Now calculate the denominator:
1. \(4 \cdot -55 = -220\)

Now we have the fraction \(\frac{528}{-220}\). 

Next, simplify this fraction by finding the greatest common divisor (GCD) of 528 and 220. 

The prime factorization of 528 is:
1. \(528 = 2^4 \cdot 3 \cdot 11\)

The prime factorization of 220 is:
1. \(220 = 2^2 \cdot 5 \cdot 11\)

The common factors are \(2^2\) and \(11\). Thus, the GCD is:
1. \(2^2 \cdot 11 = 44\)

Now divide both the numerator and the denominator by the GCD:
1. \(528 \div 44 = 12\)
2. \(-220 \div 44 = -5\)

This gives us the simplified fraction \(\frac{12}{-5}\), which can be rewritten as \(-\frac{12}{5}\). 

Thus, the final answer is \(-\frac{12}{5}\).<|END_OF_LATENT|> The final answer is $-\frac{12}{5}$.