Respuesta :
In order to solve this question, let's develop each of the given expressions to find their solutions:
I. Notice that the numbers are presented in the form of an integer and a fraction part. We can rewrite them as:
[tex](-2\frac{2}{5}\cdot3)\colon3\frac{3}{5}=\lbrack-(2+\frac{2}{5})\cdot3\rbrack\colon(3+\frac{3}{5})[/tex]Now, to solve the sums, we need to write both numbers in the sum with the same denominator, that is, 5:
[tex]\begin{gathered} \lbrack-(2+\frac{2}{5})\cdot3\rbrack\colon(3+\frac{3}{5})=\lbrack-(\frac{10}{5}+\frac{2}{5})\cdot3\rbrack\colon(\frac{15}{5}+\frac{3}{5}) \\ =(-\frac{12}{5}\cdot3)\colon(\frac{18}{5}) \end{gathered}[/tex]Then, we need to use the property:
(a/b) : (c/d) = (a/b) . (d/c)
And we obtain:
[tex](-\frac{12}{5}\cdot3)\colon(\frac{18}{5})=(-\frac{12\cdot3}{5})\cdot(\frac{5}{18})=\frac{-36}{18}=-2[/tex]II. 28 + (-2.4 : 0.08) = 28 - 240/8 = 28 - 30 = -2
III. We solve this in the same way we did for the first one:
[tex]\begin{gathered} 2\frac{1}{4}\lbrack\frac{7}{3}+(-1\frac{4}{9})\rbrack=(2+\frac{1}{4})\cdot\lbrack\frac{7}{3}-(1+\frac{4}{9})\rbrack=(\frac{8}{4}+\frac{1}{4})\cdot\lbrack\frac{7}{3}-(\frac{9}{9}+\frac{4}{9})\rbrack \\ \\ =\frac{9}{4}\cdot(\frac{7}{3}-\frac{13}{9})=\frac{9}{4}\cdot(\frac{21}{9}-\frac{13}{9})=\frac{9}{4}\cdot\frac{8}{9}=\frac{8}{4}=2 \end{gathered}[/tex]Therefore, I and II only have a solution of -2.
Answer: option G
