모두CBT
간극비(e)와 간극률(n, %)의 관계를 옳게 나타낸 것은?
e=1−n/100n/100e = \dfrac{1 - n/100}{n/100}e=n/1001−n/100
e=n/1001−n/100e = \dfrac{n/100}{1 - n/100}e=1−n/100n/100
e=1+n/100n/100e = \dfrac{1 + n/100}{n/100}e=n/1001+n/100
e=1+n/1001−n/100e = \dfrac{1 + n/100}{1 - n/100}e=1−n/1001+n/100