{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"mmlu-pro","formal_name":"MMLU-Pro","introduction":"MMLU-Pro rebuilds MMLU with ten answer options instead of four and removes items that no longer separate models. Questions span fourteen academic and professional subjects.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","indexing_mode":"index","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[],"history":"","construction":"","limitations":"","saturation":"","scores":[]}},"task_id":"31708c5a-e8ed-5177-b20e-082876353508","task_key":"default--test--12155","task_revision_id":"1","upstream_id":"12155","short_description":"Medium 1 has k_1 = 1.5 and extends to the left of theyz plane. Medium 2 hask_2=…","config":"default","split":"test","body":"{\"category\":\"engineering\",\"options\":\"['E_1 = x̂_1 - \\\\\\\\^{y}_2 + ẑ, E_2 = x̂0.6 - \\\\\\\\^{y}2 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂1.5 - \\\\\\\\^{y}3 + ẑ1), D_2 = \\\\\\\\epsilon_0(x̂1.5 - \\\\\\\\^{y}5 + ẑ1.25)', 'E_1 = x̂_3 - \\\\\\\\^{y}_3 + ẑ, E_2 = x̂1.3 - \\\\\\\\^{y}3 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂2 - \\\\\\\\^{y}4.5 + ẑ_1.5), D_2 = \\\\\\\\epsilon_0(x̂2 - \\\\\\\\^{y}7.5 + ẑ2.5)', 'E_1 = x̂_3 - \\\\\\\\^{y}_4 + ẑ, E_2 = x̂1.8 - \\\\\\\\^{y}4 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂4.5 - \\\\\\\\^{y}6 + ẑ2.25), D_2 = \\\\\\\\epsilon_0(x̂4.5 - \\\\\\\\^{y}10 + ẑ3)', 'E_1 = x̂_2 - \\\\\\\\^{y}_3.5 + ẑ, E_2 = x̂1.4 - \\\\\\\\^{y}3.5 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂3 - \\\\\\\\^{y}5.25 + ẑ1.75), D_2 = \\\\\\\\epsilon_0(x̂4.2 - \\\\\\\\^{y}8.75 + ẑ3.5)', 'E_1 = x̂_1 - \\\\\\\\^{y}_3 + ẑ, E_2 = x̂1.1 - \\\\\\\\^{y}3 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂4 - \\\\\\\\^{y}4.5 + ẑ_1.5), D_2 = \\\\\\\\epsilon_0(x̂4 - \\\\\\\\^{y}7.5 + ẑ2.5)', 'E_1 = x̂_2 - \\\\\\\\^{y}_2 + ẑ, E_2 = x̂1.2 - \\\\\\\\^{y}2 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂3 - \\\\\\\\^{y}3.5 + ẑ_1.5), D_2 = \\\\\\\\epsilon_0(x̂3 - \\\\\\\\^{y}6.5 + ẑ2.5)', 'E_1 = x̂_1.5 - \\\\\\\\^{y}_3 + ẑ, E_2 = x̂1 - \\\\\\\\^{y}3 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂2.25 - \\\\\\\\^{y}4.5 + ẑ1.5), D_2 = \\\\\\\\epsilon_0(x̂2.5 - \\\\\\\\^{y}7.5 + ẑ2)', 'E_1 = x̂_2.5 - \\\\\\\\^{y}_2.5 + ẑ, E_2 = x̂1.5 - \\\\\\\\^{y}2.5 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂3.75 - \\\\\\\\^{y}3.75 + ẑ1.875), D_2 = \\\\\\\\epsilon_0(x̂3.75 - \\\\\\\\^{y}6.25 + ẑ2.5)', 'E_1 = x̂_2 - \\\\\\\\^{y}_3 + ẑ, E_2 = x̂1.2 - \\\\\\\\^{y}3 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂3 - \\\\\\\\^{y}4.5 + ẑ_1.5), D_2 = \\\\\\\\epsilon_0(x̂3 - \\\\\\\\^{y}7.5 + ẑ2.5)', 'E_1 = x̂_2 - \\\\\\\\^{y}_4 + ẑ, E_2 = x̂1.5 - \\\\\\\\^{y}4 + ẑ, D_1 =\\\\\\\\epsilon_0(x̂3 - \\\\\\\\^{y}6 + ẑ1.5), D_2 = \\\\\\\\epsilon_0(x̂4.5 - \\\\\\\\^{y}10 + ẑ2.5)']\",\"question\":\"Medium 1 has k_1 = 1.5 and extends to the left of theyz plane. Medium 2 hask_2= 2.5 and extends to the right of the yzplane. E_1 = (x̂2 - \\\\^{y}3 + ẑ) Vm^-1 FindD_1,E_2, andD_2.\",\"src\":\"stemez-Electromagnetics\"}","display_format":"text","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}