Answer:
0.3137 ; 0.2228
Step-by-step explanation:
Given a normal distribution :
Morning class :
Mean(Mm) = 71%
Standard deviation (Sm) = 12%
Afternoon class:
Mean(Ma) = 78%
Standard deviation (Sa) = 8%
M = Mm - Ma = (71 - 78) = - m7
S = √Sm + Sa = √12² + 8² = √208
A. What is the probability that a randomly selected student in the morning class has a higher final exam mark than a randomly selected student from an afternoon class?
P(morning > afternoon) = p(morning - afternoon > 0)
Using:
Z = (0 - (-7)) / S
Z = 7 / √208
Z = 0.4853628
P(Z > 0.49) = 0.3137
B)
What is the probability that the mean mark of four randomly selected students from a morning class is greater than the average mark of four randomly selected students from an afternoon class?
Using:
Z = (4 - (-7)) / S
Z = 11 / √208
Z = 0.7627127
P(Z > 0.49) = 0.2228
Answer:
x+-(1)/(2),-3
Step-by-step explanation:
Thank you:)
The population Its 860 u give only 8 points
1) Determine the GCF of the numbers 96 and 88
=> Decompose each number in their prime numbers:
=> 96 = (2^5)(3)
=> 88 = (2^3) (11)
=> GCF of 96 and 88 = 2^3 = 8
2) Determine the GCF of the letters, x^2 and x
=> x
3) Conclude the GCF of the terms is 8x
4) Now you can factor the expression by dividing each term by the GCF, 8x:
96 x^2 / (8x) = 12x
88x / (8x) = 11
So, the factored form is (8x) (12x + 11)
