Given that the concentration of bacteria in the refrigerated food is
[tex]10T^2-20T-6----\mleft\lbrace1\mright\rbrace[/tex]
and the temperature of the food is given by
[tex]T(t)=3t+2-----\mleft\lbrace2\mright\rbrace[/tex]
Therefore, N(T(t) is given by
[tex]N\mleft(T(t)\mright)=10(3t+2)^2-20(3t+2)^{}-6[/tex]
Then,
[tex]\begin{gathered} N(T(t))=10(3t+2)^2-20(3t+2)^{}-6 \\ =10(9t^2+6t+6t+4)-60t-40-6 \\ =10(9t^2+12t+4)-60t-46 \\ =90t^2+120t+40-60t-46 \\ =90t^2+60t-6 \end{gathered}[/tex]
Answer: The composition is
[tex]N(T(t))=90t^2+60t-6[/tex]
It can be interpreted as the concentration of bacteria in the food when outside of the refrigerator with time.