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:
13.5 this is the awnser hope it helps
The function is written as:
f(x) = log(-20x + 12√x)
To find the maximum value, differentiate the equation in terms of x, then equate it to zero. The solution is as follows.
The formula for differentiation would be:
d(log u)/dx = du/u ln(10)
Thus,
d/dx = (-20 + 6/√x)/(-20x + 12√x)(ln 10) = 0
-20 + 6/√x = 0
6/√x = 20
x = (6/20)² = 9/100
Thus,
f(x) = log(-20(9/100)+ 12√(9/100)) = 0.2553
<em>The maximum value of the function is 0.2553.</em>
The Correct answer is 3
Explanation:
Using the defined function, f(a) will produce the same result when substituted for x:
F(x)=x^2-x^3
Setting this equal to 4, you can solve for a:
a2 – 5 = 4
a2 = 9
a = –3 or 3
