Suppose $f(x)$ has the following properties:
- $f(x)$ is continuous and positive for att $x\in [6,19]$
- $f(x)$ is decreasing for all $x\in [6,19]$
- ${\int }_6^{19}f(x)dx=102$,
- ${\int }_6^{12}f(x)dx=60$,

Use this information to find the least and greatest possible value of $f(12)$.
$$\le f(12)\le$$

Step 1 ：Given that $f(x)$ is continuous and positive for all $x\in [6,19]$

Step 2 ：Given that $f(x)$ is decreasing for all $x\in [6,19]$

Step 3 ：Given that ${\int }_6^{19}f(x)dx=102$

Step 4 ：Given that ${\int }_6^{12}f(x)dx=60$

Step 5 ：The integral from 12 to 19 is ${\int }_{12}^{19}f(x)dx=102-60=42$

Step 6 ：The least possible value of $f(12)$ occurs if $f(x)$ decreases rapidly after 12, making the area from 12 to 19 as small as possible

Step 7 ：The greatest possible value of $f(12)$ occurs if $f(x)$ decreases slowly after 12, making the area from 12 to 19 as large as possible

Step 8 ：The least average value of $f(x)$ from 12 to 19 is $\frac{42}{7}=6.0$

Step 9 ：The greatest average value of $f(x)$ from 12 to 19 is $\frac{42}{7}=6.0$

Step 10 ：The least possible value of $f(12)$ is equal to the least average value, which is 6.0

Step 11 ：The greatest possible value of $f(12)$ is equal to the integral from 6 to 12 divided by the number of units, which is $\frac{60}{6}=10.0$

Step 12 ：Final Answer: $\boxed{{\displaystyle 6\le f(12)\le 10}}$