
(1)设甲种鱼苗x尾,乙种鱼苗(6000-x)尾.根据题意得 0.5x+0.8(6000-x)=3600,解得:x=4000,乙种鱼苗的数量为:6000-x=2000(尾).答:甲种鱼苗4000尾,乙种鱼苗2000尾;(2)由题意,得90%×4000+96%×20006000×100%=92%.答:理论成活率为92%.

(1)设甲种鱼苗x尾,乙种鱼苗(6000-x)尾.根据题意得 0.5x+0.8(6000-x)=3600,解得:x=4000,乙种鱼苗的数量为:6000-x=2000(尾).答:甲种鱼苗4000尾,乙种鱼苗2000尾;(2)由题意,得90%×4000+96%×20006000×100%=92%.答:理论成活率为92%.