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:
cool :))))))))) good luck
93+36+m=180 combine like terms
93+36=129 rewrite
129+m=180 subtract 129 from both sides
180-129=51 rewrite
M=51 and to check this just plug 51 in for m
93+36+51=180
93+36=129
129+51=180
180=180
Hope this helps have a nice nite!
Answer:
here's what we know:
a + b = 72
a - b = 26
We can use the elimination method to solve this problem:
a + b = 72
a - b = 26
---------------
2a = 98
Divide both sides by 2:
a = 49
So we now have:
49 + b = 72
and:
49 - b = 26
Subtract 49 from both sides:
b = 23
and
-b = -23 (multiply both sides by -1) → b = 23
Step-by-step explanation:
I believe your answer is--- > The slope = 1; x= -2; y=2
