I just want to know if i solved this correctly FlightFly Trait DataMaleFemalePerformanceMaleFemaleDataWing length(mm)Wing area(mm²)Length of theeyestalk (mm)Total body massTurn duration4.46 + 0.1384.24 + 0.138142 + 38.1122 + 34.0(ms)Turn amplitude(°)Turn pathradius (mm)4.21 ± 0.3233.88 ± 0.32392 + 33.586 ± 33.44.12 ± 0.2072.83 ± 0.1326 + 3.05+ 3.0Turn rate7.12 + 1.0336.83 ± 1.034685 ± 282.6733 + 235.3(deg s)(mg)Mass of the eyeHorizontal0.24 + 0.0530.25 ± 0.0497.3 + 3.36.3 ± 3.0speed (cm s')(mg)Eye-stalk mass(mg)*Table 1 shows the mean trait values for maleand female flies, + one standard error.*Table 2 shows the mean value for male and0.11 + 0.0310.07 ± 0.020female flights, + one standard error.What data will you use to answer the questions?Calculate the moment of inertia for male and female flies below.Remember the equation I = mR2Where m = mass of the eye, R = length of eyestalk, and I = moment of inertiaMale| = mR?| = (0.24)4.122| = 4.07Female|= mR2| = (0.25)2.832| = 2.00

Given: We have to find moment of inertia for male and female.

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