Answer
(a) The apparent weight is $1102N$
(b) The apparent weight is $931N$
(c) The apparent weight is $807.5N$
Work Step by Step
(a) As the person stands in the elevator moving upward with $a=1.8m/s^2$, she is under the influence of her weight $mg$, which points downward, and the normal force $F_N$, which points upward.
$$F_N-mg=ma$$
Here, the number registered in the scale is the value of $F_N$, which is her apparent weight. $$F_N=m(g+a)=95(9.8+1.8)=1102N$$
(b) The elevator moves upward at $a=0$, meaning $$F_N-mg=0$$ $$F_N=95\times9.8=931N$$
(c) The elevator moving downward only changes the sign of $a$ $(a=-1.3m/s^2)$; the person is still under these same two forces: $$F_N-mg=-ma$$ $$F_N=m(g-a)=95(9.8-1.3)=807.5N$$
