Pertanyaan

Disajikan tabel berikut : Nilai Frekuensi 20 – 24 7 25 – 29 11 30 – 34 10 35 – 39 9 40 – 44 5 Modus dari data tersebut adalah .....

Disajikan tabel berikut :

Nilai	Frekuensi
20 – 24	7
25 – 29	11
30 – 34	10
35 – 39	9
40 – 44	5

Modus dari data tersebut adalah .....

28,0
28,5
29,0
29,5
29,8

A. Acfreelance

Master Teacher

Jawaban terverifikasi

Pembahasan

$M o d u s space a d a l a h space d a t a space y a n g space s e r i n g space m u n c u l comma space j i k a space d a t a space k e l o m p o k space m a k a space k e l a s space M o d u s space a d a l a h space k e l a s space d e n g a n space f r e k u e n s i space t e r b e s a r comma space l a l u space n i l a i space m o d u s d a p a t space d i c a r i space m e n g g u n a k a n space space M subscript o equals T subscript b plus open parentheses fraction numerator d subscript 1 over denominator d subscript 1 plus d subscript 2 end fraction close parentheses i D i m a n a space colon T subscript b space a d a l a h space t e p i space b a w a h space k e l a s space M o d u s d subscript 1 space a d a l a h space s e l i s i h space f r e k u e n s i space m o d u s space d e n g a n space f r e k u e n s i space k e l o m p o k space s e b e l u m n y a d subscript 2 space a d a l a h space s e l i s i h space f r e k u e n s i space m o d u s space d e n g a n space k e l o m p o k space s e t e l a h n y a p space a d a l a h space p a n j a n g space i n t e r v a l space d a t a$

$K e l a s space m o d u s space a d a l a h space 28 minus 29 comma space m a k a space colon T subscript b space end subscript equals space 24 comma 5 d subscript 1 space end subscript equals space 11 minus 7 equals 4 d subscript 2 space end subscript equals space 11 minus 10 equals 1 p space equals space 5 M subscript o equals 24 comma 5 plus open parentheses fraction numerator 4 over denominator 4 plus 1 end fraction close parentheses 5 M subscript o equals 24 comma 5 plus open parentheses 4 over 5 close parentheses 5 M subscript o equals 24 comma 5 plus 4 M subscript o equals 28 comma 5$