Answer:
[tex]T_A_v_g=9.918192559$^{\circ}C[/tex]
Explanation:
The problem tell us that the temperature as function of time in downtown mathville is given by:
[tex]T(t)=10-5*sin(\frac{\pi}{12t})[/tex]
The average temperature over a given interval can be calculated as:
[tex]T_a_v_g=\frac{T_o+T_f}{2}[/tex]
Where:
[tex]T_o=Initial\hspace{3}temperature\\T_f=Final\hspace{3}temperature[/tex]
So, the initial temperature in this case, would be the temperature at noon, and the final temperature would be the temperature at midnight:
Therefore:
[tex]T_o=T(12)=10-5*sin(\frac{\pi}{12*12}) =9.890925575$^{\circ}C[/tex]
[tex]T_f=T(24)=10-5*sin(\frac{\pi}{12*24}) =9.945459543$^{\circ}C[/tex]
Hence, the average temperature between noon and midnight is:
[tex]T_A_v_g=\frac{9.890925575+9.945459543}{2}=9.918192559$^{\circ}C[/tex]