Answer:

Step-by-step explanation:
We are given a joint probability table.
There are four different graders in a school
1. Grade Ninth
2. Grade Tenth
3. Grade Eleventh
4. Grade Twelfth
Field trip refers to the students who will attending the amusement park field trip.
No field trip refers to the students who will not be attending the amusement park field trip.
We want to find out the probability that the selected student is an eleventh grader given that the student is going on a field trip.

Where P(eleventh and FT) is the probability of students who are in eleventh grade and will be going to field trip

Where P(FT) is the probability of students who will be going to field trip

So the required probability is

Answer: About 191 students scored between a 60 and an 80.
Step-by-step explanation:
Given : A set of 200 test scores are normally distributed with a mean of 70 and a standard deviation of 5.
i.e.
and 
let x be the random variable that denotes the test scores.
Then, the probability that the students scored between a 60 and an 80 :

The number of students scored between a 60 and an 80 = 0.9544 x 200
= 190.88 ≈ 191
Hence , about 191 students scored between a 60 and an 80.
Answer:
Mean: 17
Median: 16
Mode: 10
Step-by-step explanation:
8, 10, 10, 11, 16, 17, 19, 21, 41
Median: (the middle number in the number list/data set) 16
Mode: 10
Mean: 8 + 10 + 10 + 11 + 16 + 17 + 19 + 21 + 41 = 153/9 = 17
Answer:
x=-13 The solution is in the picture above. if you need more solutions let me know. :)
Answer:
g = 7 yards
Step-by-step explanation:
you have 7
yards for the angle adjacent 30 deg.
tan 60 =
tan 60 = 7
/ g
7 * root(3) / g = root(3)
7/g = 1
g = 7 yards