First, we need to work out the total number of students who were being surveyed.
We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population
That means total number of students being surveyed is 13+13=26 students
Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13
The probability distribution is shown in the table below. Let

be the number of pets and

is the probability of owning the number of pets
Answer:
ST = 4
Step-by-step explanation:
A segment joining the midpoints of 2 sides of a triangle is half the length of the third side.
ST =
PQ substitute values
- 32 + 9x =
(- 5x + 28) ← multiply both sides by 2 to clear the fraction
- 64 + 18x = - 5x + 28 (add 5x to both sides )
- 64 + 23x = 28 ( add 64 to both sides )
23x = 92 ( divide both sides by 23 )
x = 4
Then
ST = - 32 + 9x = - 32 + 9(4) = - 32 + 36 = 4
If b is in the first position then c can be in any 1 of the remaining 6 positions.
If we start with ab then the letter c can be in any one of 5 positions and if we have aab there are 4 possible positions for c and so on.
So the total number of possible sequences where b comes first = 6+5+4+3+2+1 = 21.
The same argument applies when c comes before b so that gives us 21 ways also.
So the answer is 2 *21 = 42 different sequences.
A more direct way of doing this is to use factorials:-
answer = 7! / 5! = 7 * 6 = 42.
( We divide by 5! because of the 5 a's.)
Answer:
x = 12
Step-by-step explanation: