#### Answer

$\frac{m_1}{m_2} = 0.48$

#### Work Step by Step

Let $F$ be the force provided by one rubber band.
From graph 1, we can see that the acceleration of $m_1$ is $5a$ when there are three rubber bands. We can set up an equation.
$3F = (m_1)(5a)$
$m_1 = \frac{3F}{5a}$
From graph 2, we can see that the acceleration of $m_2$ is $4a$ when there are five rubber bands. We can set up an equation.
$5F = (m_2)(4a)$
$m_2 = \frac{5F}{4a}$
We can find the mass ratio $\frac{m_1}{m_2}$ as:
$\frac{m_1}{m_2} = \frac{3F/5a}{5F/4a}$
$\frac{m_1}{m_2} = \frac{12Fa}{25Fa}$
$\frac{m_1}{m_2} = \frac{12}{25}$
$\frac{m_1}{m_2} = 0.48$