Answer:
a) 0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.
b) 0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.
Step-by-step explanation:
Question a:
20% of employees engage in physical exercise.
This 20% is composed by:
8% of 60%(males)
x% of 100 - 60 = 40%(females).
Then, x is given by:




0.38 = 38%
Probability of being a female who engages in exercise:
40% are female, 38% of 40% engage in exercise. So
0.38*0.4 = 0.152
0.152 = 15.2% probability that this person is a female who engages in physical exercise activities during the lunch hour.
B. If we choose an employee at random from this corporation,what is the probability that this person is a female who does not engage in physical exercise activities during the lunch hour?
40% are female, 100% - 38% = 62% of 40% do not engage in exercise. So
0.62*0.4 = 0.248
0.248 = 24.8% probability that this person is a female who does not engage in physical exercise activities during the lunch hour.
Average (mean) = (sum of all the data) / (# of data)
sum of all the data = (average)(# of data)
Thus for 100 students with an average of 93,
sum of all data = (93)(100) = 9300
and for 300 students with an average of 75,
sum of all data = (75)(300) = 22500
Therefore you would expect the overall average to be
(9300 + 22500) / (100 + 300) = 79.5 %
Now if there are x # of advanced students and y # of regular students, then
x + y = 90 (total # of students) and 93x + 75y = 87(x + y) (overall average)
The second equation can be simplified to x - 2y = 0
Subtracting the two equations yields
x = 60 and y = 90
Therefore you would need 60 advanced and 30 regular students.
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➷ 3.5 + x = 7
We need to isolate x
Subtract 3.5 from both sides:
x = 3.5
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Answer:
8
Step-by-step explanation:
15x-31+9×+11+×=180
25×-20=180
25×=200
×=8
15(8)-31=89
9(8)+11=83
8