In Ms Barclay's classroom 2/5 students play chess. 5/6 of the students play Sudoku. If there are 30 students, in her class how m
1 answer:
Answer:
12 students play chess
and 25 students play Sudoku
Step-by-step explanation:
There are 30 students in Ms Barclay's classroom.
2/5 students play chess.
i.e. play chess
i.e. 12 students play chess
5/6 of the students play Sudoku.
i.e. play Sudoku
i.e. 25 students play Sudoku
Hence, 12 students play chess
and 25 students play Sudoku
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Answer:
-1.14
Step-by-step explanation:
The given information in statement is
mean=μ=69
standard deviation=σ=3.5
Let X be the Ishaan's exam score
X=65
The Z score can be computed as
z=-1.1429
z=-1.14 (rounded to two decimal places).
Thus, the computed z-score for Ishaan's exam grade is -1.14.
Answer:
a) n(none) = 25
b) n(PE but not Bio) = 25
c) n(ENG but not both BIO and PE) = 55
d) n(students that did not take Eng or Bio) = 40
e) P( Students did not take exactly two subjects) = 0.65
Step-by-step explanation:
From the Venn diagram drawn:
a) Number of students that took none
n(Freshmen) = 135
n(all three) = 5
n (PE and Bio) = 10
n(PE and Eng) = 15
n(Bio and Eng) = 7
n (PE and Bio only) = 10 - 5 = 5
n(PE and Eng only) = 15 - 5 = 10
n(Bio and Eng only) = 7 - 5 = 2
n(PE only) = 35 - 5 - 5 - 10 = 15
n(Bio only) = 42 - 5 - 5 - 2 = 30
n(Eng only) = 60 - 10 - 5 -2 = 43
n(Freshmen) = n(PE only) + n(Bio only) + n(Eng only) + n(PE and Bio only) + n(PE and Eng only) + n(Bio and Eng only) + n(all three) + n(none)
135 = 15 + 30 + 43 + 5 + 10 + 2 + 5 + n(none)
135 = 110 + n(none)
n(none) = 135 - 110
n(none) = 25
b)Number of students that too PE but not Bio
n(PE but not bio)= n(PE only) + n(PE and Eng only)
n(PE but not Bio) = 15 + 10
n(PE but not Bio) = 25
c) Number of students that took ENG but not both BIO and PE
n(ENG but not both BIO and PE) = n(Eng only) + n(Eng and Bio only) + n(Eng and PE only) = 43 + 2 + 10
n(ENG but not both BIO and PE) = 55
d) Number of students that did not take ENG or BIO
n( students that did not take Eng or Bio) = n(PE only) + n(none)
n(students that did not take Eng or Bio) = 15 + 25
n(students that did not take Eng or Bio) = 40
e) Probability that a randomly-chosen student from this group did not take exactly two subjects
n( Students that did not take exactly two subjects) = n(PE only) + n(Bio only) + n(Eng only)
n( Students that did not take exactly two subjects) = 15 + 30 + 43
n( Students that did not take exactly two subjects) = 88
P( Students did not take exactly two subjects) = 88/135
P( Students did not take exactly two subjects) = 0.65
