#### Answer

The frequency observed by the bat is $83.6~kHz$

#### Work Step by Step

We can find the speed of sound in the air:
$v = 331+0.6~T = 331+(0.6)(10.0^{\circ}C) = 337~m/s$
To find the frequency received by the moth, we can use the equation for the Doppler effect when the source is approaching and the observer is moving away:
$f_o = \left(\frac{v-v_o}{v-v_s}\right)~f$
$f_o = \left(\frac{337~m/s-1.20~m/s}{337~m/s-4.40~m/sv_s}\right)~(82.0~kHz)$
$f_o = 82.79~kHz$
To find the reflected frequency, we can let $f_o$ be the frequency of the source and use the equation for the Doppler effect when the source is moving away and the observer is approaching:
$f_r = \left(\frac{v+v_o}{v+v_s}\right)~f_o$
$f_r = \left(\frac{337~m/s+4.40~m/s}{337~m/s+1.20~m/s}\right)~(82.79~kHz)$
$f_r = 83.6~kHz$
The frequency observed by the bat is $83.6~kHz$