I believe the answer is 11W-0 or just plain 11W
I'm sorry but I know the answer is deffinetly c or d
4136/8= 517
The others
7116/8=889.5
4309/8=538.625
9406/8=1175.75
Answer:
0.0032
The complete question as seen in other website:
There are 111 students in a nutrition class. The instructor must choose two students at random Students in a Nutrition Class Nutrition majors Academic Year Freshmen non-Nutrition majors 17 18 Sophomores Juniors 13 Seniors 18 Copy Data. What is the probability that a senior Nutrition major and then a junior Nutrition major are chosen at random? Express your answer as a fraction or a decimal number rounded to four decimal places.
Step-by-step explanation:
Total number of in a nutrition class = 111 students
To determine the probability that the two students chosen at random is a junior non-Nutrition major and then a sophomore Nutrition major, we would find the probability of each of them.
Let the probability of choosing a junior non-Nutrition major = Pr (j non-N)
Pr (j non-N) = (number of junior non-Nutrition major)/(total number students in nutrition class)
There are 13 number of junior non-Nutrition major
Pr (j non-N) = 13/111
Let the probability of choosing a sophomore Nutrition major = Pr (S N-major)
Pr (S N-major)= (number of sophomore Nutrition major)/(total number students in nutrition class)
There are 3 number of sophomore Nutrition major
Pr (S N-major) = 3/111
The probability that the two students chosen at random is a junior non-Nutrition major and then a sophomore Nutrition major = 13/111 × 3/111
= 39/12321
= 0.0032
Add the three angles and set them to 180
4x-13+15+x+18=180
5x+20=180
5x=160
x=32
Then plug in to get A and C
A=4*32-13
A=115
C=32+18
C=50