Answer
We have $3$ possibilities:
We could have bought
1) $4$ flash cameras, $16$ daylight cameras
2) $5$ flash cameras, $15$ daylight cameras
3)$6$ flash cameras, $14$ daylight cameras.
Work Step by Step
Let $x$ be the number of flash cameras bought. Then the number of daylight cameras bought is $20-x$. Then our inequality is: $65\leq4.25x+(20-x)2.75\leq75\\65\leq4.25x+55-2.75x\leq75\\10\leq1.5x\leq20\\3\lt\frac{10}{3}\leq x\leq\frac{20}{3}\lt 7$
Thus the integer possibilities for $x$ are $4,5,6$. Thus we have $3$ possibilities:
We could have bought
1) $4$ flash cameras, $16$ daylight cameras
2) $5$ flash cameras, $15$ daylight cameras
3)$6$ flash cameras, $14$ daylight cameras.
